The Trappist-1 Solar System: Seven Earth-sized Planets Potentially Harboring Life

Trappist-1 system vs. Jupiter
Figure 1:  The red dwarf Trappist-1 and its seven planets, compared to the Sun and its solar system (above), and also to Jupiter and its moons (below).  Note that the horizontal dimension displays the period of each orbiting body.  Image credit: Nature

A. Introduction

The recent discovery (Ref. 1) of seven exoplanets orbiting the red dwarf star Trappist-1 is an exciting and potentially key development in the search for extraterrestrial life in the universe.  All of the seven planets are Earth-sized, and at least three lie in the temperate “Goldilocks” zone, neither too near nor too far from their host star for surface oceans to exist.  Given the mass and temperature of these planets, their atmospheres are likely to contain nitrogen, oxygen, water vapor, and carbon dioxide, just like on Earth.  As an added bonus, the small size of the host star will help astronomers determine the chemical composition of each planet’s atmosphere.

The goal of this post is to show students how scientists use elementary physics and trigonometry to derive many of the physical properties of this and other extrasolar systems.  We present straightforward explanations appropriate to an introductory physics course, and offer exercises suitable for a homework assignment or recitation class to help students appreciate the significance of the Trappist-1 discovery.  We hope these materials will be useful to you, and that your students will enjoy and benefit from them.

B.  Background

In February 2017, an international team of astronomers, led by Michaël Gillon of Belgium, announced the discovery (Ref. 1) of seven Earth-sized exoplanets orbiting the red dwarf Trappist-1, so named for the instrument in Chile (TRAPPIST: Transiting Planets and Planetesimals Small Telescope) first employed in their work.  Observations continued for one year using an array of ground-based telescopes located in Chile, Morocco, Hawaii, Spain, and South Africa, plus the orbiting Spitzer Space Telescope.  The exoplanet discoveries were made by the transit photometry technique:  when a planet eclipses its host star, i.e., as it passes between the star and Earth, it blocks a tiny fraction of starlight from reaching Earth.  The resulting light curve (flux vs. time, see Fig. 2) and its period P (time between successive transits), plus estimates of the star’s mass \boldsymbol{M_{\ast }} and radius \boldsymbol{R_{\ast }}, suffice to determine the planet’s radius \boldsymbol{R_{P}} and the radius of its orbit  \boldsymbol{a}.  This information can then be used to estimate the surface temperature \boldsymbol{T} and the potential for life on the planet.

light curves
Figure 2:  Light curves from Trappist-1 planets, taken using NASA’s Spitzer Space Telescope.  Each curve shows a dip in brightness as a planet eclipses the star, and each is labeled with that planet’s orbital period (days).  Image credit: M. Gillon et al/ESO

Astronomers determine stellar masses by studying binary stars.  For each binary system, they measure the distance from Earth by parallax, and measure the apparent brightness of each star to calculate its luminosity \boldsymbol{L_{\ast }}.  Mass is found by studying the orbital motion of the binary, as described in PPE, Section 8.11.  By compiling data on many binaries, an empirical mass-luminosity relation \boldsymbol{L_{\ast }=f(M_{\ast })} can be constructed.  A mass-radius relation can be constructed in a similar way, using data from eclipsing binary stars and long baseline interferometry.  For now, let’s accept the mass and radius of Trappist-1 given in Ref. 1: \boldsymbol{M_{\ast }/M_{\bigodot }=0.0802\pm .0073}\boldsymbol{R_{\ast }/R_{\bigodot }=0.117\pm 0.0036}, where the subscript \boldsymbol{\bigodot } indicates the Sun.  Table 1 is adopted from Ref. 1, and lists many of the properties of the Trappist-1 exosystem.

Data Table
Table 1:  Parameters of the Trappist-1 system, extracted from Table 1 of Ref. 1

1.   A commonly quoted mass-luminosity relation, suitable for some small-mass main sequence stars, is \boldsymbol{L_{\ast }/L_{\bigodot }=(M_{\ast }/M_{\bigodot })^{2.62}}, where the subscript \boldsymbol{\bigodot } indicates the Sun.  The measured luminosity of Trappist-1 is \boldsymbol{L_{\ast }/L_{\bigodot }=5.24\times 10^{-4}}.  Show that this leads to a value of \boldsymbol{M_{\ast }} that agrees with the value quoted in Ref. 1.

C.  Kepler’s Laws and Orbital Radii

Assume that the mass of each planet is much less than that of the host star, \boldsymbol{M_{p}\ll M_{\ast }}, and that each planet is in a near-circular orbit of radius \boldsymbol{a\gg R_{\ast }}, just like the planets of our own solar system.  Kepler’s 3rd law is

\LARGE \boldsymbol{\frac{P^{2}}{a^{3}}=\frac{4\pi ^{2}}{GM_{\ast }}}


where P is the orbital period.  After calculating \boldsymbol{M_{\ast }} from its luminosity, and measuring P , we can use Eqn. 1 to derive a.

2.  Astronomers have recently (February 2017) discovered seven exoplanets orbiting the star Trappist-1, an ultracool red dwarf about 12 pc from Earth.  This is an exciting discovery because at least three  planets are in the “temperate zone” of the host star, with surface temperatures favorable for life.  The table below lists properties of six bodies that orbit either Trappist-1 (\boldsymbol{M_{\ast }\simeq 0.08M_{\bigodot }}) or one other central body with mass \boldsymbol{M\simeq 0.001M_{\bigodot }}.  Fill in the missing numbers.  (Hint: Use ratios rather than converting to meters, seconds and kilograms.)  Can you identify M and the bodies in orbit about it?

Question 2

D.  Analyzing the Light Curve

Fig. 2 contains a great deal of information that can be used to further characterize the Trappist-1 system.  Even if \boldsymbol{M_{\ast }} and \boldsymbol{R_{\ast }} are not known a priori, the period P and the shape of the transiting planet’s light curve provide enough information to determine \boldsymbol{R_{p}}, the orbital radius a and inclination i, and the density of the host star \boldsymbol{\rho _{\ast }}.  Combining the star’s density with the appropriate mass-radius relation for red dwarfs yields \boldsymbol{M_{\ast }} and \boldsymbol{R_{\ast }}.  This strategy uses Kepler’s 3rd law plus simple kinematics and trigonometry, making it amenable to first-year physics students.  It is described in detail by Seager and Mallén-Ornelas in Ref. 2, and we will follow their treatment closely.

Figure 3
Figure 3:  (a) A sideways view of a planetary transit.  If the inclination i = 90 deg, the orbit is viewed edge-wise by obervers on Earth.  The impact parameter b < 1 for a transit to occur.  (b) A view of the transit from Earth.  The time between the first and last “contact” between the stellar and planetary discs is tt.  The planet is fully within the stellar disc for time tf.  Delta-F is the change in flux reaching Earth due to the eclipsing planet.  Adapted from Fig. 1 of Ref. 2

As usual, assume that the exoplanet’s mass is much less than the star’s, and that it is in a circular orbit of radius \boldsymbol{a\gg R_{\ast }}.  Fig. 3a is a “side view” of the orbit, perpendicular to the line of sight from Earth to the star.  The inclination angle is defined so that if \boldsymbol{i=90\: \textup{deg}}, observers on Earth view the orbit edge-on.  The impact parameter b is defined by \boldsymbol{bR_{\ast }=acosi}.  Clearly, \boldsymbol{b<1} for a transit to occur, and since \boldsymbol{a\gg R_{\ast }}, this implies that \boldsymbol{i\simeq 90\; \textup{deg}}.

From Fig. 3b, the amount of light blocked by the planet is proportional to its area, so the fractional change in flux reaching Earth during a transit is

\LARGE \boldsymbol{\frac{\Delta F}{F}=\frac{R_{p}^{2}}{R_{\ast }^{2}}}.


In practice, the straight line segments of the light curve shown in Fig. 3b are rounded by limb-darkening (as seen in Fig. 2): less light reaches us from areas near the edge of the stellar disc than from equal areas near the center of the disc.  For simplicity, we will ignore this effect.

3.  Fig. 2 displays the light curves of the seven planets orbiting Trappist-1.  Which planet has the largest radius?  (Ans: Trappist-1b)

Since \boldsymbol{a\gg R_{\ast }}, we may approximate the trajectory of the planet as a straight line as it moves across the face of the star.  For a circular orbit, the planet’s speed is \boldsymbol{v=2\pi a/P}.  By the Pythagorean theorem, the distance traveled by the planet in time \boldsymbol{t_{t}} is \boldsymbol{d_{t}=vt_{t}=2\sqrt{(R_{\ast }+R_{p})^{2}-b^{2}R_{\ast }^{2}}}, so

\LARGE \boldsymbol{t_{t}=\frac{P}{\pi a}\sqrt{(R_{\ast }+R_{p})^{2}-b^{2}R_{\ast }^{2}}}.


4.  Derive the corresponding equation for \large \boldsymbol{t_{f}}
(Ans: \large \boldsymbol{t_{t}=\frac{P}{\pi a}\sqrt{(R_{\ast }-R_{p})^{2}-b^{2}R_{\ast }^{2}}})

If \boldsymbol{R_{\ast }} and \boldsymbol{M_{\ast }} are not known a priori, then the only information available is the light curve and its period.  Nevertheless, it is still possible to proceed.  From Eqn. 3 and the corresponding expression for  \boldsymbol{t_{f}} (Question 4), it is easy to show that

\LARGE \boldsymbol{(t_{t}^{2}-t_{f}^{2})^{\frac{1}{2}}=\frac{2P}{\pi a}R_{\ast }\left ( \frac{\Delta F}{F} \right )^{\frac{1}{4}}},

where we have used Eqn. 2 to express \boldsymbol{R_{p}/R_{\ast }=\sqrt{\Delta F/F}}.  Rearranging the above expression,

\LARGE\boldsymbol{\frac{a}{R_{\ast }}=\frac{2P}{\pi }\frac{(\Delta F/F)^{\frac{1}{4}}}{(t_{t}^{2}-t_{f}^{2})^{\frac{1}{2}}}}


\LARGE \boldsymbol{\frac{a^{3}}{R_{\ast }^{3}}=\frac{8P^{3}}{\pi^{3} }\frac{(\Delta F/F)^{\frac{3}{4}}}{(t_{t}^{2}-t_{f}^{2})^{\frac{3}{2}}}}.


Combining Eqn. 4 with \boldsymbol{M_{\ast }=4\pi ^{2}a^{3}/GP^{2}} (Kepler’s 3rd law) yields an expression for the stellar “density”:

\LARGE \boldsymbol{\rho _{\ast }=\frac{M_{\ast }}{R_{\ast }^{3}}=\frac{32P}{\pi G}\frac{(\Delta F/F)^{\frac{3}{4}}}{(t_{t}^{2}-t_{f}^{2})^{\frac{3}{2}}}}.

It is convenient to re-express this in terms of the solar “density”

\LARGE \boldsymbol{\frac{\rho _{\ast }}{\rho _{\bigodot }}=\frac{32R_{\bigodot }^{3}}{\pi GM_{\bigodot }}\: \frac{(\Delta F/F)^{\frac{3}{4}}}{(t_{t}^{2}-t_{f}^{2})^{\frac{3}{2}}}\: P}


where the pre-factor \boldsymbol{32R_{\bigodot }^{3}/\pi GM_{\bigodot }=3.46\times 10^{-3}\: {\textrm{day}}^{2}}.  This is the key result of this section.

For classroom purposes, we can crudely express the mass-radius relation as \boldsymbol{R_{\ast }=M_{\ast }^{x}}. Choosing \boldsymbol{x=0.85}, we can now calculate the mass and radius of Trappist-1 in solar mass units:

\LARGE\boldsymbol{\frac{M_{\ast }}{M_{\bigodot }}=\frac{\rho _{\ast }}{\rho _{\bigodot }}\frac{R_{\ast }^{3}}{R_{\bigodot }^{3}}=\frac{\rho _{\ast }}{\rho _{\bigodot }}\left ( \frac{M_{\ast }}{M_{\bigodot }} \right )^{2.55}}


\LARGE \boldsymbol{\frac{M_{\ast }}{M_{\bigodot }}=\left ( \frac{\rho _{\ast }}{\rho _{\bigodot }} \right )^{-0.65}\; and\; \; \; \frac{R_{\ast }}{R_{\bigodot }}=\left ( \frac{\rho_{\ast } }{\rho_{\bigodot } } \right )^{-0.55}},


where \boldsymbol{\rho _{\bigodot }\equiv M_{\bigodot }/R_{\bigodot }^{3}=5.90\times 10^{3}\: \mathrm{kg/m}^{3}}.  The validity of these results depends sensitively on the mass-radius relation chosen.  Admittedly, the exponent \boldsymbol{x=0.85} was chosen to yield results consistent with those of Ref. 1.  In the literature, \boldsymbol{x=0.80} is more commonly quoted.

5.  According to Ref. 1, the transit duration of Trappist-1d is \boldsymbol{t_{t}=49.1\: \textrm{min}}, the period is \boldsymbol{P=4.05\: \textrm{days}}, and the transit depth is \boldsymbol{\Delta F/F=0.0037}.

a.  Use the values of \boldsymbol{M_{\ast }} and \boldsymbol{R_{\ast }} given in section B to compute \boldsymbol{\rho _{\ast }/\rho _{\bigodot }} and then find \boldsymbol{t_{f}}.  Check your answer against the light curve in Fig. 2.  (Ans: \boldsymbol{t_{f}\approx 0.88t_{t}=43.3\: \textrm{min}})

b.  Find the radius of the planet’s orbit.  (Ans: \boldsymbol{a=2.14\times 10^{-2}\: \textup{\textrm{AU}}})

c.  Find the radius of the planet.  Express \boldsymbol{R_{p}} as a fraction of \boldsymbol{R_{Earth}}.  (Ans: \boldsymbol{R_{p}=0.78\, R_{Earth}})

6.  Researchers at Trump University recently reported the discovery of an eighth planet orbiting Trappist-1.  In a series of tweets, team leaders provided a summary of their observations taken with the privately funded Andrew Jackson Telescope in Palm Beach, FL:  \boldsymbol{P=15.0\pm 0.2\: \textrm{min}}\boldsymbol{\Delta F/F=0.005\pm 0.0001}\boldsymbol{t_{t}=86.4\pm 0.2\: \textrm{min}}, and \boldsymbol{t_{f}=55.4\pm 0.2\: \textrm{min}}.  Their measurements have not yet been confirmed.  If you were a program director at the National Science Foundation, would you approve a grant to the Trump exoplanet program?

E.  Too Hot, Too Cold, or Just Right?

A blackbody is an object that absorbs all of the radiation incident on its surface.  (Since it reflects no light, it appears black.)  But a blackbody is not just an ideal absorber of radiant energy, it is an ideal emitter of energy as well.  The power S emitted by a blackbody per unit surface area is given by the Stefan-Boltzmann Law,

\LARGE \boldsymbol{S=\sigma _{B}T^{4}},


where T is the surface temperature in kelvin, and \boldsymbol{\sigma _{B}=5.67\times 10^{-8}\: \textrm{W/m}^{2}\textrm{-K}^{4}} is called the Stefan-Boltzmann constant.  The Sun and a cloudless Earth behave approximately as blackbodies.  The power emitted by the Sun, its luminosity, is \boldsymbol{L_{\bigodot }=4\pi \sigma _{B}R_{\bigodot }^{2}T_{\bigodot }^{4}=3.83\times 10^{26}\: \textrm{W}}.  At a distance r from the Sun, this power is spread evenly over a sphere of area \boldsymbol{4\pi r^{2}}.  A spherical planet of radius \boldsymbol{R_{P}} presents a cross-sectional area \boldsymbol{\pi R_{P}^{2}} to the Sun, so if it acts as a blackbody, it intercepts and absorbs energy at a rate \boldsymbol{W_{abs}=L_{\bigodot }R_{P}^{2}/4r^{2}}.  This energy warms the planet’s surface to a temperature , causing it to emit energy at a rate \boldsymbol{W_{emit}=4\pi R_{P}^{2}\cdot \sigma _{B}T^{4}}. When \boldsymbol{W_{emit}=W_{abs}}, the temperature reaches a steady-state value given by

\LARGE \boldsymbol{T=\left ( \frac{L_{\bigodot }}{16\pi \sigma _{B}r^{2}} \right )^{\frac{1}{4}}}.


7.  The Earth is 1 AU (\boldsymbol{1.5\times 10^{11}\: \textrm{m}}) from the Sun.  What is the steady-state temperature on Earth, according to Eqn. 8?  (Ans: 278 K.  Keep in mind, though, that this ignores the greenhouse effect, which raises the temperature significantly.)

To apply Eqn. 8 to an extrasolar system, simply replace \boldsymbol{L_{\bigodot }} by the luminosity of the host star \boldsymbol{L_{\ast }}.

8.  The period of Trappist-1d is 4.05 days.  Use the star’s mass \boldsymbol{M_{\ast }/M_{\bigodot }=0.08} and luminosity \boldsymbol{L_{\ast }/L_{\bigodot }=5.24\times 10^{-4}} to calculate the steady-state surface temperature on the planet.  Based only on your answer, what is the likelihood of finding liquid water on this planet?  (Ans: T ≈ 300 K.  However, if the planet has an Earth-like atmosphere, the greenhouse effect would raise its temperature too high to harbor liquid water on its surface.  This is mentioned briefly in Ref. 1.)

9.  What is the likelihood of finding intelligent life on Trappist-1i, the planet claimed by researchers at Trump University?


1.   Michaël Gillon et al, Nature 542, 456 (2017)

2.  S. Seager and G. Mallén-Ornelas, Ap. J. 585, 1038 (2003)

Work, Energy and the “Satellite Drag Paradox”

Figure 1: A view of the International Space Station taken from Space Shuttle Endeavor.  Every few months, the ISS’s orbit must be boosted to maintain an altitude of about 400 km.  Image credit: NASA

A.  Introduction

Here is a lovely application of energy conservation and the “work-energy theorem” that will surely intrigue your first-year physics students.  A satellite in low Earth orbit experiences a drag force \boldsymbol{\vec{F}_{d}} as it passes through the upper  reaches of the planet’s atmosphere.  (See PPE, Section 6.10, p. 188 ff.)  The drag force is antiparallel to the satellite’s velocity \boldsymbol{\vec{v}}, so it robs the spacecraft of energy and causes it to spiral inward toward the Earth’s surface.  However, assuming a gradual decay and a near-circular trajectory, it is easy to show (see below) that the speed of the spacecraft must increase as the orbital radius decays.  It seems unphysical that by applying a drag (friction) force to a body, its speed will increase!

This “paradox” has been around since the 1950’s, but to my knowledge it is not treated in introductory physics textbooks.  One of the clearest and most complete solutions was given by B. D. Mills in 1959, and we will use his approach in the following paragraphs.  As usual, the presentation will be simplified as much as possible to make it tractable for introductory students.

B.  The Satellite Drag Paradox

Let the satellite be moving on a low Earth near-circular trajectory with a slowly decreasing orbital radius \boldsymbol{r(t)} due to the drag force \boldsymbol{\vec{F}_{d}}.  Mills (Ref. 1) shows that for a weak drag force, the tangential speed of the spacecraft is at all times nearly equal to that of a body moving in a perfectly circular orbit of the same instantaneous radius.  (See also Ref. 5.)  Let \boldsymbol{M} and \boldsymbol{m} denote the mass of the Earth and satellite, respectively.  From Newton’s Second Law,  \boldsymbol{mv^{2}/r=GMm/r}, the tangential speed \boldsymbol{v_{c}} is

\large \boldsymbol{v_{c}=\sqrt{GM/r}}


and it increases as the orbital radius decreases.  The kinetic energy of a body \large \boldsymbol{m} in circular orbit is found using Eqn. 1:



where \boldsymbol{U=-GMm/r} is the potential energy, so the total energy of the orbiting body is

\large \boldsymbol{E=K+U=\frac{1}{2}U=-\frac{GMm}{2r}}.


These results should be familiar to students.

1.  Using Eqns. 2 and 3, verify that as \boldsymbol{r} decreases, \boldsymbol{K} increases even though  \boldsymbol{U} and \boldsymbol{E} decrease.

If the spacecraft’s orbit is initially elliptical (as opposed to circular), it will soon circularize due to the atmospheric drag force.  The drag force increases dramatically at lower altitudes, so it will be greatest when the spacecraft passes through perigee. Imagine that the drag force produces an instantaneous negative impulse at perigee, with change in velocity \boldsymbol{\Delta v} antiparallel to the spacecraft’s velocity.  This reduces the total energy of the spacecraft, shrinking its semi-major axis without changing its perigee position \boldsymbol{r_{p}}.  As a result, the apogee distance \boldsymbol{r_{a}} shrinks, and as \boldsymbol{r_{a}\rightarrow r_{p}}, the orbit becomes more circular.  So even if the initial orbit is significantly elliptical, atmospheric drag will soon circularize it, and the analysis presented below will be applicable.

For a spacecraft experiencing a drag force \boldsymbol{\vec{F}_{d}}, the rate of change of \boldsymbol{E} is equal to \boldsymbol{\vec{F}_{d}\cdot \vec{v}}, where \boldsymbol{\vec{v}} is the total velocity of the spacecraft, including the inspiral motion.  This follows because \boldsymbol{\vec{F}_{d}} is the only external force acting on the Earth-satellite system.  Taking the time derivative of Eqn. 3, and noting that \boldsymbol{\vec{F}_{d}} is antiparallel to the velocity \boldsymbol{\vec{v}}, we find

\large \boldsymbol{\frac{dE}{dt}=\frac{GMm}{2r^{2}}\frac{dr}{dt}=-F_{d}v}


\large \boldsymbol{\frac{dr}{dt}=-\frac{2r^{2}}{GMm}F_{d}v}.

Figure 2 illustrates the forces acting on the satellite, and the components of its velocity: \boldsymbol{\vec{v}=v_{c}\hat{i}+\frac{dr}{dt}\hat{j}}.  For satellites passing through the thin upper atmosphere of a planet, \boldsymbol{\left | \frac{dr}{dt} \right |\ll v_{c}}, so we can approximate \boldsymbol{\vec{F}_{d}} to be antiparallel to  \boldsymbol{\vec{v}_{c}}, as shown.  Thus, the above result may be written as

\large \boldsymbol{\frac{dr}{dt}=-\frac{2r^{2}}{GMm}F_{d}v_{c}}.


Figure 2:  The orbiting spacecraft experiences a drag force Fd as well as a gravitational force Fg.  Its velocity has a small radial component dr/dt in addition to the tangential speed vc.  

Using the work-energy theorem \boldsymbol{\Delta K=\vec{F}\cdot \Delta \vec{r}}, where \boldsymbol{\vec{F}} is the total force acting on the body, we obtain

\large \boldsymbol{\frac{dK}{dt}=\vec{F}\cdot \vec{v}=F_{g}\frac{dr}{dt}-F_{d}v_{c}}


where \boldsymbol{F_{g}=-GMm/r^{2}} is the gravitational force exerted on the spacecraft by the Earth.  Eqn. 5 reveals the source of the apparent paradox: while the drag force decreases \boldsymbol{K}, the satellite gains speed as it falls toward Earth.  To evaluate \boldsymbol{dK/dt}, substitute Eqn. 4 into Eqn. 5:

\large \boldsymbol{\frac{dK}{dt}=-\frac{GMm}{r^{2}}\left ( -\frac{2r^{2}}{GMm}F_{d}v_{c} \right )-F_{d}v_{c}=F_{d}v_{c}}.


Surprisingly, the drag term appears in Eqn. 6 without a minus sign!  As described by Mills, it is “as if the air drag force, reversed, were pushing the satellite.”

2.   In your own words, resolve the paradox noted by Mills.

C.  Application: The International Space Station (ISS)

The International Space Station (Fig. 1) is maintained in a near-circular orbit of approximate altitude 400 km.  Figure 3 (Ref. 2) is a plot of the ISS’s mean altitude vs. time, and shows how the altitude is periodically boosted (the near-vertical line segments) and how it decays between boosts.  For an on-board demonstration of what ISS astronauts experience during a “reboost,” see Ref. 4.

Figure 3:  Mean altitude of the ISS vs. time.  The vertical lines indicate orbital boosts lasting only a few minutes.  Image credit:  Chris Peat,

In November 2016, the ISS was boosted to a mean altitude of 406.5 km, and then fell steadily to 404.5 km over the subsequent three month interval (92 d).

3.  What is the orbital speed of the ISS at an altitude of about 400 km?  (Ans: 7.66 km/s)

4.  From November 2016 to February 2017, the altitude decreased by 2 km.  Verify that \boldsymbol{\left | dr/dt \right |\ll v_{c}}, as asserted above.

5.  Calculate the change in the orbital speed \boldsymbol{v_{c}} during that three month period.  (Hint:  \boldsymbol{\Delta v/v_{c}=(?)\Delta r/r}.  Ans: \boldsymbol{\Delta v_{c}=1.13\: \textup{m/s}})

Now let’s calculate how much rocket fuel must be consumed to boost the ISS back to its original altitude of 406.5 km.  To keep things simple, let’s assume that the boost is done in a single short burn, as shown in Figure 4.  A single burn will change a circular orbit to an elliptical one, with perigee \boldsymbol{r_{p}=R_{E}+404.5 \: \textup{km}} at the site of the burn.  If we wish the semi-major axis to be \boldsymbol{a=R_{E}+406.5\: \textup{km}}, then the apogee must be \boldsymbol{r_{a}=R_{E}+408.5\: \textup{km}}, located 180º from the site of the burn.  This results in an eccentricity \boldsymbol{\epsilon \approx 3\times 10^{-4}}, which we will ignore.  (To improve the circularity of the boosted orbit, a two burn strategy is sometimes used.  See Ref. 3.)

Figure 4:  A single burn introduces a slight eccentricity, with the altitude at perigee slightly less that the altitude at apogee.

The change in the ISS’s orbital energy due to the burn is

\large \boldsymbol{E_{f}-E_{i}=\frac{1}{2}m(v_{i}+\Delta v)^{2}-\frac{1}{2}mv_{i}^{2}\simeq mv_{i}\Delta v},

where \boldsymbol{v_{i}} is the speed before the burn, and \boldsymbol{v_{i}+\Delta v} is the speed afterwards.  We can solve for \boldsymbol{\Delta v} using Eqn. 3, or, since the altitude change is so small, set \boldsymbol{\Delta U=mg\Delta a}, where \boldsymbol{g} is the gravitational acceleration at the altitude of the ISS, and \boldsymbol{\Delta a=406.5-404.5=2.0\: \textup{km}} is the change in the mean altitude.  From Eqn. 3, \boldsymbol{\Delta E=\frac{1}{2}\Delta U}.

6.  What is the change of velocity needed to boost the altitude of the ISS by 2.0 km?  (Ans: 1.13 km/s)

ISS orbit boosts are carried out using thrusters on the cargo spacecraft that supply the ISS, such as the Russian Progress modules.  (In the past, the US Space Shuttles were used.)  The Progress thrusters use a fuel (UDMH + NTO) with exhaust velocity 2.7 km/s.

7.  The mass of the ISS is \boldsymbol{4.2\times 10^{5}\: \textup{kg}}.  Calculate the fuel mass needed to boost its altitude by 2.0 km.  (Ans: 176 kg)

8.  Calculate the magnitude of the drag force F_{d} acting on the ISS during November 2016 to February 2017.  You might be surprised by your answer!  (Ans: .06 N)

9.  In ref. 4, astronaut Jeff Williams discusses what happens during a typical orbit boost.  For the maneuver shown in the video, \boldsymbol{\Delta v=2.7\: \textup{m/s}}, and the acceleration was \boldsymbol{1.85\: cm/s^{2}}.  What was the duration of the burn?  (Ans: 146 s)

10.  Does your answer to Question 7 include the fuel required to overcome the drag that lowered the orbit in the first place?  Why is this a small correction?  (Ans: the duration of the reboost is only a few minutes, whereas the orbital period is about 90 minutes.  It took many orbits for the drag force to lower the ISS’s altitude.)

11.  There are at least two minor misstatements in the video of Ref. 4.  Can you spot them?  (Ans: At 1:09, Williams states that because of atmospheric drag, “over a period of time we slow down and our altitude over the Earth decreases.  At 3:04 he states “We’re in weightlessness right now, and there’s no acceleration …”  Actually, he and the ISS are undergoing centripetal acceleration of about \boldsymbol{8.66\: m/s^{2}} towards the Earth.)


After writing a draft of this post, I asked Dr. Philip Blanco – a valued contributor to this blog – to review it for correctness and to offer suggestions for improvement.  He not only did that, but also supplied a numerical simulation of the trajectory of an ISS-like spacecraft immersed in a thin atmosphere.  His solution, shown below (Figure 5), shows that the spacecraft indeed follows a near-circular orbit right up to the point where a catastrophe occurs, whence it plunges toward Earth.  It would be interesting to calculate when that catastrophe occurs.

My thanks to Dr. Blanco for his assistance.  Of course, I bear the full responsibility for any remaining errors.

Figure 5:  An ISS-like spacecraft starts out in a circular orbit of altitude 150 km.  Its trajectory remains near-circular as it loses altitude, until just before it plunges to the Earth.  Courtesy:  Dr. Philip Blanco


 1.  Blake D. Mills, Am. J. Phys. 27, 115 (1959)

2.   Chris Peat,

3.  Robert Frost,

4., courtesy NASA Johnson Space Center

5.  Leon Blitzer, Am. J. Phys. 39, 882 (1971)

Rosetta’s Final Act: Free-Fall to the Surface of Comet 67P

Figure 1:  Artist’s conception of Rosetta approaching comet 67P in September 2016.  Credit:  ESA/Rosetta

The European Space Agency’s historic Rosetta mission is over.  On September 30, 2016, the spacecraft was allowed to fall ballistically (unpowered) to the surface of comet 67P/Churyumov-Gerasimenko, capturing numerous high-resolution photos of the comet’s surface during its 14 hour descent.  The final image, taken from an altitude of 20 meters, is shown in Figure 2.

Comets are presumed to be the primordial building blocks of the solar system, so by observing 67P at close range, Rosetta was investigating the origins of our solar system.

Figure 2:  Rosetta‘s last image of the comet surface, taken from an altitude of 20 m, shows an area of about 1 square meter.  Credit:  ESA/Rosetta

During its 12 year lifespan, Rosetta became the first spacecraft to orbit a comet and observe the chemical processes taking place as it passed through perihelion in August 2015, when the rate of gas and dust evolution peaked.  In November 2014, Rosetta was the first spacecraft to deploy a lander, Philae, to the surface of a comet.  Although Philae functioned only briefly, the two allied space probes made numerous discoveries about the chemical makeup and geology of comets.

Comet 67P has a period of 6.45 years and an aphelion distance of 5.7 AU.  At that distance, Rosetta‘s onboard heating system would have been incapable of keeping the spacecraft warm enough for its instruments to hibernate safely.  Consequently, flight directors opted to end the mission by crash-landing the spacecraft onto the comet, gathering as much data as possible while its instruments were still functioning.  Today, 67P is 4.3 AU from the Sun.  As it heads toward aphelion in November, 2018, it carries the remains of Rosetta and Philae, the two pioneering spacecraft that doggedly chased it for over a decade.

In this blog post, we present a set of student-ready exercises based on Rosetta‘s final maneuver and descent to the surface of 67P.  These exercises were inspired by Dr. Philip Blanco’s recent letter (Ref. 1) to the editor of The Physics Teacher.  Both Philip and I are deeply indebted to Pablo Muñoz, flight dynamics engineer at the ESA’s European Space Operations Centre in Darmstadt, Germany.  Pablo supplied indispensible data on the position and velocity of the spacecraft before and after its final thruster burn, plus the mass of Rosetta and the amount of fuel consumed during that final burn.  Without his input, none of these exercises could have been written.

To make the exercises suitable for introductory students, I have modified the details slightly, as indicated below.   To fully appreciate the complexity of Rosetta‘s dance with its cometary partner, follow the links in Ref. 4 and 5 to the beautiful ESA videos of the mission.  Have fun!

A.  Problems

In these exercises, \boldsymbol{ GM=666.15\mathrm{ m^{3}/s^{2}}}, where
\boldsymbol{ M} is the mass of the comet (\boldsymbol{ M\simeq 1.0\times 10^{13}} \textup{ kg}), and model the comet as a homogeneous sphere of radius 2.00 km.

Prior to its plunge to the comet surface, Rosetta was placed in an unpowered elliptical orbit with apocenter (farthest distance to the comet’s CM) \boldsymbol{r_{a}=23.44}\textup{ km} and pericenter \boldsymbol{r_{p}=13.58}\textup{ km}.  This is shown nicely in Ref. 4.

1.  Calculate the specific energy \boldsymbol{ E/m} (where \boldsymbol{m} is Rosetta‘s mass) and the orbital period of the spacecraft.  (Ans: \boldsymbol{E/m=-1.800\times 10^{-2}}\textup{ J/kg}\boldsymbol{ T=6.13\times 10^{5}}\textup{ s}\simeq 1\textup{ week})

2.  What was the speed \boldsymbol{ v_{0}} of the spacecraft at apocenter? Compare this to the speed of a spacecraft in low Earth orbit.  Why the big difference?  (Ans: \boldsymbol{ v_{0}=0.1444}\textup{ m/s}\boldsymbol{M_{67P}/M_{Earth}\sim 10^{-12}})

The final maneuver that sent Rosetta diving to the comet surface occurred at a true anomaly of 208º, i.e., 28º beyond apocenter (0º corresponds to pericenter).  (See Ref. 4.)  To streamline the following questions, assume instead that the maneuver took place exactly at apocenter.  This alters the results modestly.

3.   Immediately after the burn, Rosetta‘s velocity was \boldsymbol{ \vec{v}{_{0}}'=-0.0211\hat{i}-0.3255\hat{j}}\textup{ m/s}, where \boldsymbol{ \hat{i}} and \boldsymbol{ \hat{j}} refer to the radial and azimuthal directions (Ref. 2) at apocenter.  See Figure 3.  Show that the burn inserted the spacecraft into an unbounded hyperbolic trajectory.  (Ans: \boldsymbol{ {E}'/m=2.478\times 10^{-2}}\textup{ J/kg}>0)

Figure 3:  Spacecraft velocity immediately after its final burn at apocenter

4.  Note that \boldsymbol{ {\vec{v}}'_{0}} is not perfectly radial, i.e., directed toward the comet’s CM.  Convince someone that the spacecraft will still strike the comet.  This question and the solution below were suggested by Philip Blanco.          (Ans:  Compare the angle subtended by the comet at apocenter to the angle between the velocity vector \boldsymbol{{\vec{v}}'_{0}} and the radial direction.)

5.  With what speed \boldsymbol{ v} will the spacecraft strike the surface?  Compare this to the surface escape speed.  (Ans: \boldsymbol{ v=0.846}\textup{ m/s}\boldsymbol{ v_{esc}=0.816}\textup{ m/s})

6.  Find the spacecraft’s velocity \boldsymbol{ \vec{v}=v_{x}\hat{i}+\vec{v}_{y}\hat{j}} just before it strikes the surface.  Hint: first conserve angular momentum.  (Ans: \boldsymbol{ \vec{v}=-0.809\hat{i}-0.247\hat{j}}\textup{ m/s})

7.  Calculate the magnitude of the change in velocity caused by the thruster burn.  The mass of the spacecraft at this time was 1422.5 kg, and the burn consumed 0.2 kg of rocket fuel.  What was the effective exhaust velocity of the fuel (a mixture of monomethyl hydrazine (MMH) and nitrogen tetroxide (\large \mathrm{N_{2}O_{4}} or NOX))?  (Ans: \boldsymbol{ \Delta v_{0}=.365}\textup{ m/s}\boldsymbol{ v_{ex}=2.6}\textup{ km/s}.  This is substantially less than the “published” value of 3.35 km/s (Ref. 3) for MMH/NOX because a fuel leak forced flight engineers to operate the thrusters at lower than optimal pressure.)

8.  In September 2016, comet 67P has a rotation period of 12.055 hours.  This is less than its period in 2014 (12.4 hours), when Philae was released, due to outgassing from the comet as it passed through perihelion.   Assume that the comet’s rotation axis was perpendicular to the plane of the spacecraft’s motion (Figure 3).  Rosetta‘s velocity \boldsymbol{ \vec{v}_{0}{}'} was selected so that its impact velocity relative to the rotating surface was nearly vertical, i.e., perpendicular to the impact surface.  (Hence, an observer standing on the surface at the impact point would have seen the spacecraft descending vertically.)  Show that this is roughly consistent with your calculated value of \boldsymbol{v_{x}} in Question 6.  (Note: because the comet has such an irregular shape, the local vertical direction was significantly different from the radial direction.) (Ans:\boldsymbol{ v_{rot}=-\omega R=-29.0\textup{ cm/s} \approx v_{x}=-24.7\textup{ cm/s}})

The following question is suitable for users of Physics from Planet Earth who have studied Section 12.9: Application: the “Slingshot Effect” Revisited, or who have seen a comparable treatment elsewhere.

9.  Suppose Rosetta‘s final maneuver had been incorrectly executed, and the spacecraft just missed striking the comet, instead grazing its surface as it flew by.

a.  From Question 3, we know that Rosetta‘s speed after the burn was 0.3262 m/s at apocenter.  In the comet’s reference frame, what will be its speed after it has flown past the comet and is very far from it?  (Ans: \boldsymbol{ v_{\infty }=0.2226\textup{ m/s}})

b.  By what angle Θ will the spacecraft be deflected after its close fly-by of the comet?  (Ans:  \boldsymbol{ \tan (\Theta /2)=1.769}\boldsymbol{ \Theta =121^{\circ}}.  See Eqn. 12.12 of PPE.)


B.  Sphere of Influence of Comet 67P

Here is an additional “space-themed” topic that is quite suitable for discussion in an introductory mechanics course.  As far as I know, it is not discussed in any currently available textbook (including PPE).  I thank Philip Blanco for suggesting it.

So far, we have ignored the influence of the Sun on the motion of Rosetta about comet 67P.  We might justify this by invoking Einstein’s Principle of Equivalence (see Section 7.5 of PPE) which states that in a free-fall reference frame, the local effects of gravity disappear.  Comet 67P’s heliocentric orbit is free-fall motion about the Sun, so in the comet’s reference frame, we can ignore the Sun’s gravitational force on the comet (which is obvious).  However, there is a small but important complication when we consider the motion of the spacecraft in the same reference frame.  Since the Sun’s gravitational force varies inversely with the square of the distance, the spacecraft’s acceleration toward the Sun varies as the distance between it and the Sun changes along its orbit.  Let \boldsymbol{ R} be the distance from the Sun to the comet, and \boldsymbol{ r} be the distance from the comet’s CM to the spacecraft.  See Figure 4.  When the spacecraft is at position a(b) in the figure, its distance from the Sun is a maximum (minimum), and its acceleration due to the Sun is

\LARGE \boldsymbol{a=\frac{GM_{Sun}}{(R\pm r)^{^{2}}}\simeq \frac{GM_{Sun}}{R^{2}}\left ( 1\mp \frac{2r}{R} \right )}.


Figure 4:  R is the distance from the Sun to the comet CM, and r is the distance from the comet CM to the spacecraft.  At a and b, the craft is farthest and nearest to the Sun.

The acceleration of the comet toward the Sun is \boldsymbol{ {a}'=GM_{Sun}/R^{2}}, so the maximum difference between the spacecraft’s acceleration and that of the comet is \boldsymbol{ \Delta a_{max}=2GM_{Sun}\, r/R^{3}}.  How close is the spacecraft’s motion to an unperturbed Keplerian orbit about the comet?  The answer depends on the ratio between \boldsymbol{ \Delta a_{max}} and the centripetal acceleration of the spacecraft toward the comet, \boldsymbol{ a_{c}=GM/r^{2}}, where \boldsymbol{ M} is the comet mass.  The smaller this ratio is, the smaller the influence of the Sun, and the better the motion can be described as an unperturbed Keplerian orbit about the comet.

Now let’s ask the opposite question.  Imagine the spacecraft is far enough from the comet that its motion is better described as a heliocentric orbit perturbed slightly by the comet.  How close is its motion to a pure heliocentric orbit?  The answer depends on the ratio between the acceleration \boldsymbol{ a_{c}} toward the comet and the acceleration toward the Sun \boldsymbol{ {a}'}:

\LARGE\boldsymbol{ \frac{a_{c}}{{a}'}=\frac{M}{M_{Sun}}\frac{R^{2}}{r^{2}}}.


The smaller this ratio is, the smaller the influence of the comet, and the better the motion can be described as a Keplerian orbit about the Sun.  The “sphere of influence” of the comet is the region within which the motion of the spacecraft is only weakly perturbed by the Sun.  By convention, its radius \boldsymbol{ r_{soi}} is found by equating the ratios in Eqns. 1 and 2, and solving for \boldsymbol{ r\textup{ }\: (=r_{soi})}:



\LARGE\boldsymbol{ \frac{r^{5}}{R^{5}}=\frac{1}{2}\left ( \frac{M}{M_{Sun}} \right )^{2}}.

Ignoring the factor \boldsymbol{ (1/2)^{1/5}=0.87\approx 1}, we obtain the radius of the sphere of influence:

\LARGE\boldsymbol{ r_{soi}=R\left ( \frac{M}{M_{Sun}} \right )^{\frac{2}{5}}}.


10.  a.  Philae was released from Rosetta when comet 67P was about 3 AU from the Sun, from an initial distance 22.5 km from the comet’s CM.  Show that the spacecraft was well within the comet’s sphere of influence at that time.  (Ans:  \boldsymbol{ r_{soi}=54\textup{ km}})

      b.  At perihelion (August 13, 2015), 67P was 1.24 AU from the Sun.  To escape possible damage from the comet’s efflux, Rosetta was removed to a distance of 330 km.  Was it in an unpowered orbit about 67P at that time?  (Ans: No, \boldsymbol{ r_{soi}=22\textup{ km}}.  See Ref. 5, fast-forward to August 2015.)

C. References

1a.  “Modeling Rosetta’s final descent,” Philip Blanco, The Physics Teacher 54, 516 (2016)

2a.  If students are familiar with polar coordinates, substitute \boldsymbol{ \hat{r}} and \boldsymbol{ \hat{\theta}} for \boldsymbol{ \hat{i}} and \boldsymbol{ \hat{j}}.




Gravitational Radiation 3: GW151226, Encore from LIGO

Figure 1: A filtered, reconstructed image of the gravitational wave detected by LIGO on December 26, 2015 (UTC).  The signal was generated from the merger of two black holes, of approximately 14.2 and 7.7 solar masses, roughly 1.4 billion light years from  Earth.  Figure taken from Abbott et al, Ref. 3.

LIGO (Laser Interferometer Gravitational-Wave Observatory) consists of two gargantuan Michelson-like interferometers, with arms 4 km long, located 3000 km apart in Livingston LA and Hanford WA.  Its first observing session began on September 12, 2015.  Incredibly, just two days later, it captured the fleeting signal (designated GW150914) produced by the merger of two inspiraling black holes more than 1 billion light years from Earth.  (Ref. 1)  That groundbreaking discovery, announced on February 11, 2016, ended a 60 year race to directly detect gravitational waves, and came 100 years after the phenomenon was first predicted by Albert Einstein.

On Christmas Day (in the USA), LIGO scientists detected a second event, GW151226 (Ref. 3), proving that the earlier discovery was not a fluke, and that black hole mergers are frequent events.  (Ref. 4,5)  Since January, LIGO’s sensitivity has been improved, and it will begin a new observing run in Autumn, 2016.  This time, the two US detectors will be joined by VIRGO, a similar instrument with arms 3 km long, located near Pisa, Italy.  The addition of VIRGO will greatly assist in localizing – by triangulation – the sources of gravitational radiation.  In the near future, when LIGO and VIRGO reach their full design sensitivities, they may be capable of detecting one or more black hole collisions daily!  Indeed, the 21st Century seems destined to become the era of gravitational wave astronomy.

In this brief post, we will use the mathematical treatment outlined in Ref. 2 to analyze the published data given in Ref. 3 for GW151226.  Our educational goal is to add to the collection of homework exercises related to gravitational waves that are suitable for first year physics students.  In this way, we hope to inspire instructors to include this exciting topic in their mechanics syllabi.


Analyzing GW151226

In Ref. 2, the chirp mass of the binary black hole system was defined as \large \mathfrak{M}=\frac{(m_{1}m_{2})^{3/5}}{(m_{1}+m_{2})^{1/5}}, where \large m_{1} and \large m_{2} are the masses of the black holes.  This quantity is important because it can be found directly from the time dependence of the gravitational wave’s (GW) frequency (Eqn. 10 of Ref. 2):

\large \mathfrak{M}=\frac{c^{3}}{G}\left ( \frac{5}{96}\: \pi ^{-8/3}f^{-11/3}\: \frac{df}{dt} \right ) .


Following Ref. 2, define \large A=\frac{c^{5}}{G^{5/3}}\: \frac{5}{96}\: \pi ^{-8/3}=5.45\times 10^{56} (SI units), and integrate Eqn. 1 over the time interval \large \Delta t=t_{2}-t_{1} to obtain

\large \mathfrak{M}^{5/3}\Delta t=A\int_{t_{1}}^{t^{_{2}}}f^{-11/3}df=-\frac{3}{8}Af^{-8/3}\mid _{t_{1}}^{t_{2}} .


1.  Referring to Fig. 1 above, at 0.80 s before the two black holes coalesce, the frequency of the GW was 39.2 Hz.  Later, 0.40 s before the merger, the frequency was 50.3 Hz. (Ref. 6)  Calculate the chirp mass and express your answer in terms of \large M_{Sun}.  (Ans: \large 9.7\: M_{Sun}.)

2.  Show that the total mass \large M=m_{1}+m_{2} of the binary system before coalescence was at least \large 22\: M_{Sun}.  (Hint: Let \large m_{2}=\alpha m_{1}.  Express \large M\large \mathfrak{M}, and \large M/\mathfrak{M} in terms of \large m_{1} and \large \alpha.  Show that the ratio is a minimum when \large \alpha=1.)

3.  The GW frequency at the moment of coalescence of the two black holes was 420 Hz.  (See Fig. 1.)  Recall from Ref. 2 that the orbital frequency of the binary system is half the frequency of the GW.   Use the total mass given in Ref. 3, \large M=22\: M_{Sun}, to find the distance between the bodies just before they merged.  (Ans: 120 km)

4.  Assume that one black hole was twice as massive as the other, i.e., \large m_{2}=2m_{1}, which is about what the LIGO team concluded.  If the orbits were circular, what were their radii just before coalescence?  What were their speeds? Ignore relativity.   (Ans: 40 and 80 km; \large v_{1}=0.35\, c)


1.  B. P. Abbott et al, “Observation of Gravitational Waves from a Binary Black Hole Merger,”, Phys. Rev. Lett. 116, 061102 (2016)

2.  this blog:  “Gravitational Radiation 2: The Chirp Heard Round the World”

3.  B. P. Abbott et al, “Observation of Gravitational Waves from a 22-Solar-Mass Binary Black Hole Coalescence,”, Phys. Rev. Lett. 116, 241103 (2016)

4.  A. Cho, “LIGO Detects Another Black Hole Crash,” Science 352, 1374, 17 June 2016

5.  D. Castelvecchi, “LIGO Sees a Second Black Hole Crash,” Nature 534, 448, 23 June 2016

6.  We thank Jonah Kanner at the LIGO Open Science Center for providing these numbers, which were calculated using the tutorial on the Center’s webpage:  The numbers obtained from the tutorial differ slightly from those reported in Ref. 3 because the analysis is not identical to the one used for the paper.  (In Ref. 3, \large \mathfrak{M}=8.9\pm 0.3\: M_{Sun}.)  The LIGO Open Science Center is a service of LIGO Laboratory and the LIGO Scientific Collaboration.    LIGO is funded by the U.S. National Science Foundation.

Proxima Centauri b: An Earth-like Neighbor in our Galactic backyard

proxima-centauri-b (II)
Figure 1:  Artist’s conception of sunset on Proxima Centauri b.  The binary star Alpha Centauri is also visible.  Inset:  Proxima Centauri is the Sun’s nearest neighbor, 1.294 pc (4.224 ly) away.  Image credit:  Pale Red Dot/ESO

In August (2016), astronomers at the European Southern Observatory in Chile announced the discovery of an Earth-sized exoplanet orbiting the red dwarf Proxima Centauri, the nearest star to our Sun (Ref. 1, 2).  The planet, called Proxima Centauri b, was detected by the radial velocity method:  the orbiting planet causes the star to execute a much smaller orbit of its own (as required by momentum conservation), and light from the star is Doppler-shifted due to its radial motion relative to Earth.  The discovery of Proxima Centauri b is exciting because its orbit lies within the “Goldilocks,” or temperate, zone of the star, where the planet’s surface temperature would allow water to exist in liquid form – a likely prerequisite for life.  Red dwarfs are the most common stars in our galaxy, so if this particular red dwarf harbors a life-supporting planet, it is likely that the Milky Way, which contains over 100 billion stars, is brimming  with life.

In this post, we will use the data reported in Reference 1, plus the strategy presented in Section 8.12 of PPE, to calculate the mass and orbital radius of the exoplanet Proxima Centauri b.  Our goal is to help students share the excitement of this new discovery through their understanding of introductory mechanics.


A.  Calculating the exoplanet’s mass and orbital radius

In the following, we will assume \large m\ll M, where \large m and \large M are the masses of the planet and star, respectively.  We will also assume that their orbits are circular, which is consistent with observations, and begin by treating the simplest case where the orbits are viewed edge-on from Earth.  (Later, we’ll relax this restriction.)  Radial velocity measurements of Proxima Centauri collected over the past 16 years are compiled in Figure 2.  Careful analysis indicates that the period \large T of the star’s orbit is 11.186 ± .002 d, and its orbital speed \large v is 1.38 ± 0.02 m/s.

Figure 2:  Radial velocity measurements of Proxima Centauri, compiled over 16 years of observations.  Note the sinusoidal shape of the best-fit curve.

Kepler’s 3rd law states that

\large \frac{T^{2}}{r^{3}}=\frac{4\pi ^{2}}{GM},

where \large r is the orbital radius of the planet.  Observations of the star’s luminosity and color indicate that its mass \large M=0.120\pm .015\, M_{Sun}.




1.  Find the planetary orbit radius \large r (in AU).  Hint: use  \large T=11.186/365.25=.03063\, \mathrm{yr} and \large r^{3}\propto MT^{2} to compare this orbit to Earth’s orbit about the Sun.  (Ans: \large 0.048 \: \mathrm{AU} = 7.24\times 10^{9}\: \mathrm{m}.

2.  The orbital radius of the star \large R can be found from its period and radial velocity \large v\large vT=2\pi R.  Using  \ 1\: \mathrm{d}=8.64\times 10^{4}\: \boldsymbol{\mathrm{s}}, find \large R.  (Ans: \large 2.12\times 10^{5}\: \mathrm{m})

3.  The planet and star co-orbit their center of mass.  Find the mass of the exoplanet.  (Ans:   \large m=MR/r=2.93\times 10^{-5}\, M=6.99\times 10^{24}\, \mathrm{kg}=1.18\, m_{Earth})

If the orbit is not viewed edge-on, but is inclined to the line of sight (LOS), the measured radial speed of the star is less than the actual orbital speed \large v used in the above analysis:  \large v_{meas}=vsin(i), where i, the angle of inclination, is unmeasurable.  See Figure 3.  In this case, \large 2\pi R=Tv=Tv_{meas}/sin(i)).

Figure 3:  In the general case, the orbits of the star and planet are not viewed edge-on (i = 90 deg) from Earth.  The measured speed is less than the actual speed by a factor of sin(i).

4.  Prove that, in the general case, the radial velocity measurements are only sufficient to determine \large msin(i).  Is the exoplanet mass you found in Question 3 the maximum, or minimum, mass of the planet?

5.  Because red dwarfs are much cooler than the Sun, their temperate zones are much closer to them than the Earth is to the Sun.  In addition, red dwarfs are less massive than the Sun.  Explain why both of these factors favor the detection of small habitable exoplanets.  (See Ref. 2 for a brief discussion.)

6.  Proxima Centauri subtends an angle of 1.02 ± .08 mas (milliarcseconds) when viewed from Earth.  (See PPE, Section 1.7)  Find its radius \large R_{s}.  (Ans: \large 1.0\times 10^{8}\: \mathrm{m}=0.14\, R_{\mathrm{Sun}})

If Proxima centauri b transits, or passes in front of, its star, as viewed from Earth, it will intercept some of the starlight and allow its radius to be determined.  Additionally, if the planet has an atmosphere, it will absorb certain wavelengths of light preferentially, enabling astronomers to detect the presence of atmospheric gases such as water, oxygen, carbon dioxide, and methane.  Methane (CH4), in particular, is often associated with the presence of life.

The probability of transit is found from the geometric argument outlined in Figure 4.  In the figure, the horizontal dotted line is the LOS from Earth, and the vertical line represents the plane perpendicular to the LOS passing through the star’s center.  For an orbit to be transiting, its normal (shown as a red line for each of the two orbits depicted) must lie within \large \pm \theta _{max}=\pm R_{s}/r of the vertical plane, where \large R_{s} is the star’s radius.  But the normal to the orbit does not need to lie in the plane of the figure: any transiting orbit rotated by any angle (0 – 2π) about the LOS would remain a transiting orbit.  The total solid angle available for transiting orbits is therefore \large 2\pi \cdot 2\theta _{max}=4\pi R_{s}/r, whereas the total solid angle associated with all possible orientations is \large 4\pi.  Therefore, the probability of a transiting orbit is \large R_{s}/r.  (Check: if \large r=R_{s}, then the probability equals 1.)  For a fuller discussion, see .

Figure 4:  In a transiting orbit, the planet passes in front of the star, as seen from Earth.  For this to happen, the radius of the planet’s orbit r must be less than the stellar radius Rs.

7.  How probable is it that Proxima Centauri b transits its star?  (Ans: 1.5%)

B.  How Warm is Proxima Centauri b?

The following questions are only suitable for students who have studied blackbody radiation in a course of thermal physics.

8.  The luminosity of a star (its total radiated power) is proportional to its surface area and the fourth power of its surface temperature: \large L_{star}\propto R_{star}^{2}T_{star}^{4}.  Proxima Centauri’s surface temperature is about 3050 K, whereas the Sun’s is 5780 K.  Show that \large L_{Proxima}\simeq 0.0015\, L_{Sun}.  This indicates that the temperate zone lies much closer than 1 AU to the star.

9.  The fraction of the star’s radiation that is intercepted by a planet is equal to the solid angle subtended by the planet divided by 4π:  \large \pi R_{p}^{2}/4\pi r^{2}.  The absorbed radiation warms the planet, which reaches a steady state temperature \large T_{p} when the blackbody radiation emitted by the planet equals the radiation absorbed from the star:
\large 4\pi R_{p}^{2}\sigma _{B}T_{p}^{4}=L_{Proxima}\pi R_{p}^{2}/4\pi r^{2}, where  \large \sigma _{B} is called the Stefan-Boltzmann constant.  (Its value is not needed to answer this question.)  Use this information to compare the surface temperature of Proxima Centauri b to that of the Earth, assuming both planets are perfect black bodies (or have the same albedo).  (Ans:   \large T_{p}/T_{Earth}=0.90)


1.    G.  Anglada-Escudé et al, Nature 536, 437 (25 August, 2016)

2.   A. P. Hatzes, Nature 536, 408 (25 August, 2016)




Inserting Juno into Orbit around Jupiter

Juno at Jupiter
Figure 1: Simulated view of Juno’s main engine firing during the craft’s insertion into orbit about Jupiter, on July 4, 2016.  Screenshot is from JPL’s Eyes on the Solar System application.

After a 5-year interplanetary journey, which included a gravity assist from Earth, NASA’s spacecraft Juno was successfully inserted into orbit about  Jupiter on July 4, 2016.  Juno is NASA’s most remote spacecraft to be entirely solar-powered; its three 8.9 meter long solar panels are the largest of any on NASA’s deep-space probes.  It houses 9 primary instruments enabling a close-up study of the planet’s gravitational field, its massive magnetic field and auroras, the chemical composition (including water) of its atmosphere, and providing detailed color images.  During the next 20 months, Juno will slip under the planet’s intense radiation belts, approaching within 5000 km of its cloud cover, and execute approximately 40 tight orbits of period 14 days before burning up in the planet’s atmosphere.  Astronomers believe that Jupiter was the first planet to form in the Solar System, and so it must have played a critical role in the formation of the remaining planets.   Juno’s mission is to learn how Jupiter was formed, to help us understand how Earth came to be.

Shortly before Juno‘s orbit insertion, Dr. Philip Blanco of Grossmont College mined NASA’s press releases and other on-line tools for solid numerical data on the spacecraft’s past and planned trajectory, to make the orbit insertion maneuver a “teachable moment” for introductory physics students.  Using his data, he kindly wrote a “guest post” (presented below) for use by physics instructors in their introductory mechanics classes.   I edited it just a bit, but all credit for the real work belongs to Phil.  We hope you will find it useful!


Eyes on Juno for Jupiter Orbit Insertion

by Philip Blanco, Grossmont College

The Juno spacecraft has reached Jupiter after an epic journey that began with launch from Cape Canaveral in August, 2011.  At the end of May, 2016, when Jupiter’s gravity dominated its motion, mission planners began preparations for the Jupiter Orbit Insertion (JOI) phase of the mission.  (See this melodramatic NASA video.)  In this post, we’ll see how mission information provided by NASA’s Eyes on the Solar System online application can be used along with some introductory mechanics to calculate the  velocity change required to put the spacecraft into its first “capture” orbit.

NASA provided some dynamical information about JOI in a Press Kit.  However, non-metric units and vague statements about speed and distance appearing in the kit made it frustrating for this educator to extract “hard numbers.”  (An alternate source with more details is Spaceflight 101.)  Fortunately, NASA provides a much better resource for our purposes: JPL’s Eyes on the Solar System uses real mission data to simulate solar system and spacecraft motions, in the past, (predicted) present, and near-future.  Eyes gets its data from SPICE format files produced by the mission teams themselves.  Once downloaded (PC and Mac versions are both available), select the Advanced mode and find your favorite mission to “fly along” with.  After first switching to metric units (under Visual Controls on the right of the screen), I found an option under Cool Tools that allows the user to display relative distance and speed between any two selected objects.  I chose  Jupiter and Juno, and noted that the distance shown in km appears to be from Jupiter’s cloud tops, not its center, so I added Jupiter’s equatorial radius to the distances displayed by the tool.

A.  Juno’s Approach Energy and Speed

I used the following physical characteristics of Jupiter: \large GM=1.267\times 10^{8} \mathrm{\: km^{3}/s^{2}}, and equatorial radius \large R=71492\mathrm{ \; km} (used here because the initial approach and subsequent capture orbits pass over the poles with perijove – the closest distance from Jupiter’s center –  above the equator).

For simplicity I analyzed Juno‘s motion relative to Jupiter starting at noon on July 3, 2016.  Although one could say that the Sun’s gravitational pull can be neglected at this point, it is more accurate to say that we can neglect the difference between Juno‘s and Jupiter’s freefall acceleration toward the Sun, i.e., we can treat the Jovian frame of reference as inertial for calculating Juno‘s subsequent motion relative to the planet.

Approximating Jupiter as a spherical body, and ignoring all other influences, Juno‘s orbital mechanical energy per unit mass ε is conserved during its unpowered approach to perijove, and is expressed as:

\large \varepsilon =\frac{1}{2}v^{2}-\frac{GM}{r}.


Juno‘s incoming speed and distance relative to Jupiter’s center on 2016 July 3.5 were \large r=1.40\times 10^{6}\mathrm{\; km} and \large v=14.49\mathrm{\; km/s}, for which \large \varepsilon _{in}=14.5\mathrm{\; MJ/kg}.  The fact that \large \varepsilon >0 confirms that Juno was on an (unbound) hyperbolic trajectory which would continue around and away from Jupiter unless ε is corrected to a negative value consistent with a bound elliptical orbit.  (Although this value of \large \varepsilon _{in} looks huge, in fact it is tiny compared to the kinetic energy per unit mass needed for escape from the Jovian surface:  \large \frac{1}{2}v_{esc}^{2}=\sqrt{GM/R}=1.77\mathrm{\; GJ/kg}.)

Rearranging Eqn. 1, Juno‘s uncorrected incoming perijove speed is

\large v_{p,in}=\sqrt{2\varepsilon _{in}+\frac{2GM}{r_{p}}}.


NASA’s Press Kit and Eyes give different values for the perijove distance \large r_{p}, which I derived from their stated altitudes above the Jovian cloud tops.  Let’s take \large r_{p}=76000\pm 200\mathrm{\; km}, or 1.06 Jupiter radii, for which Eqn. 2 yields \large v_{p,in}=58.02\mathrm{\; km/s}.  Compare this to the local escape speed \large v_{esc}=\sqrt{2GM/r_{p}}=57.74\mathrm{\; km/s}.  To enter a capture orbit, the speed must be reduced by at least 0.26 km/s.

B.  Required Speed at Perijove for Capture Orbit

NASA’s Press Kit states that, after firing its main engine, Juno‘s path will be converted from an unbound hyperbolic trajectory to an elliptical “capture” orbit with period \ T=53.5\boldsymbol{\mathrm{\; Earth\: days}}=4620\mathrm{\; ks}.  What instantaneous change of speed (a.k.a. specific impulse) at perijove is required to achieve this new orbit?

Kepler’s 3rd law states that \large T^{2}/a^{3}=4\pi ^{2}/GM, where a is the orbital semi-major axis.  Solving for a, we obtain \large a=4.09\times 10^{6}\mathrm{\; km}, or about 57.2 Jovian radii.  The orbital energy per unit mass of a body in elliptical orbit is given by (See PPE, Eqn. 10.13)

\large \varepsilon _{capture}=-\frac{GM}{2a},


which for the values given above yields \large \varepsilon _{capture}=-1.548\times 10^{7}\mathrm{\; J/kg}, negative for the bound orbit.  Substituting \large \varepsilon _{capture} and \large r_{p}=76000\mathrm{\; km} into Eqn. 2 gives the perijove speed for this orbit: \large  v_{p,capture}=57.48\mathrm{\; km/s}, just below the local escape speed given above.

Therefore, Juno‘s required change in speed is

\large \Delta v=v_{p,capture}-v_{p,in}=-0.54\mathrm{\; km/s},

very close to the figure of 541 m/s stated in advance by NASA!  The figure below shows the incoming hyperbolic path and the capture orbit predicted by our simple model.

Figure 2
Figure 2:  Incoming hyperbolic trajectory of Juno (blue solid line, projected as blue dashed line) and elliptical capture orbit (green) for values given in text.  (a) Full capture orbit. (b) Close-up of region around perijove.  The distance scale is numbered in units of Jupiter’s equatorial radius.


Questions for students

Instructors might ask students to perform some or all of the calculations given above.   Here are a few additional questions related to the Juno mission.

1.  Juno‘s main engine is a British-built  Leros 1b  which burns a mixture of hydrazine (N2H4) and nitrogen tetroxide (N2O4).  A rocket engine’s efficiency is characterized by its specific impulse \large I_{sp}, which is (different from the earlier use of this term) the ratio between the rocket thrust (see PPE, section 6.14) and the weight (on Earth) of fuel consumed per second:

\large F_{thrust}=I_{sp}g\, \dot{m}=v_{ex}\dot{m}

where \large g=9.8\mathrm{\: m/s^{2}}.  The Leros 1b is advertised as having a nominal thrust of 635 N and a specific impulse of 317 s.

a.  What is the exhaust velocity?  (Ans: 3.1 km/s)

b.  What mass of fuel is consumed per second (\large \dot{m})?   How much fuel was consumed in the 35 minute JOI burn?    (Ans: 0.2 kg/s, 430 kg)

c.  Verify that your answers are consistent with the value of  \large \Delta v calculated above.

2.  Following two 53.5 day capture orbits, Juno will execute a Period Reduction Maneuver at perijove to enter its “science” orbit of period 13.965 days = 1206.6 ks.  Assuming that this maneuver does not change the perijove distance, calculate the required \large \Delta v.  (Ans: 400 m/s.  This answer is higher than NASA’s “published” value of 350 m/s.  But the uncertainty in the perijove distance (± 200 km) propagates to an uncertainty in the perijove speed of nearly 100 m/s.  Students might be asked to calculate this uncertainty.)

Gravitational Radiation 2: The Chirp Heard Round the World

Figure 1: Gravitational wave GW150914 “chirp” detected in Hanford WA and Livingston LA. The bright arc in each panel plots the signal’s frequency vs. time, while the brightness registers its intensity. The two wiggly lines are the strain measured at each detector. Image courtesy Caltech/MIT/LIGO Laboratory





A.  The Discovery

Sometime, maybe 1.5 billion years ago, two large collapsed stars began a long, slow mating dance.  Wedded by gravity, they patiently circled their center of mass, inching together imperceptibly (perhaps a few meters each year, like the Hulse-Taylor binary), all the while quickening their speed.  As they approached one another – over millions of years – they shed vast amounts of energy via gravitational radiation, and seconds before the end of their dance, their gyration increased feverishly as their relative speed approached half that of light.  Then, in a split second, they embraced to form a single black hole, emitting a final burst of gravitational radiation that briefly outshone the light emitted by the entire visible universe!   1.3 billion years later, on September 15, 2014, this blast of energy reached planet Earth, where it was detected by two gargantuan, souped-up Michelson interferometers, each with arms 4 km long, collectively called the Laser Interferometer Gravitational Wave Observatory (LIGO).  (See Fig. 2.)

The passing gravitational wave differentially changed the length of each detector’s arms by about one-thousandth the diameter of an atomic nucleus, inducing an unmistakable signal 24 times greater than background noise.  The age of gravitational-wave observational astronomy had begun.

Figure 2: LIGO Livingston, courtesy Caltech/MIT/LIGO Laboratory

The LIGO Collaboration first reported its discovery in the February 12, 2016 issue of Physical Review Letters.  (See Ref. 1.)  Their main conclusions, summarized in Section II of that paper, are described clearly and simply, and best of all, they can be understood by students having an introductory-level comprehension of classical mechanics.  (Because of its groundbreaking significance, the PRL paper is available open access — no subscription or institutional affiliation is necessary to download it.)  Two other recently published papers will surely be helpful to educators wishing to discuss the LIGO discovery in their classrooms.  The first (Ref. 2), by Lior Burko, describes an introductory lab exercise using data taken from the LIGO Open Science Center ( ).  It is clearly written and discusses the LIGO results in more detail than the present post.  The second (Ref. 3), by Louis Rubbo et al., illustrates how to extract astrophysical information from LIGO-like (simulated) data, and a separate link offers a complete worksheet and teacher’s guide suitable for use by first-year physics or astronomy students ( ).

  1.   When the two bodies merged, they shed about 3 solar masses of energy in less than 0.1 s.  Show that the average power emitted during this fleeting time interval was greater than the total electromagnetic radiation emitted by the entire visible universe.  (The luminosity of the Milky Way galaxy is roughly \large 10^{10}L_{Sun}, where \large L_{Sun}=3.9\times 10^{26} \textup{W} is the luminosity of the Sun.  There are about \large 10^{11} galaxies in the observable universe.)


B.     A Quick Review of Gravitational Wave Physics

Our goal is to introduce LIGO’s exciting discovery into the introductory mechanics curriculum, in a way that supports and complements the fundamental topics of the course.  For the intended first-year audience, the discussion must rest primarily on Newtonian physics, and concepts from general relativity must be strictly limited: no tensors allowed!  In the previous post, we introduced an expression (Eqn. 1) for the gravitational luminosity of two equal-mass stars in circular orbit, and later generalized (Eqn. 3) that expression to treat binary systems with unequal masses in elliptical orbit:

\large L=\frac{32}{5}\frac{G}{c^{^{5}}}\mu ^{2}a^{4}\omega_{orb}^{6}f(\varepsilon )


where \large \mu =m_{1}m_{2}/(m_{1}+m_{2}) is the system’s reduced mass, and \large f(\varepsilon ) is a correction associated with the orbit eccentricity ε.  In the following discussion, we’ll use Eqn. 7 rather than Eqn. 1 to conform to the notation commonly found in the literature.  Eqn. 7 – plus a lot of tedious algebra – is all we need to extract physical information from the LIGO signal.  That signal was produced by two massive in-spiraling bodies in the split second before they merged to form a single entity, and by that time the energy drain due to gravitational radiation had circularized their orbits (\large \varepsilon =0\large f(\varepsilon )=1).  As usual,  the total orbital energy of the binary system is half the potential energy (see PPE, section 10.6): \large E=-Gm_{1}m_{2}/2a.  Setting \large dE/dt=-L, and proceeding just as we did in the previous post to derive Eqn. 4, we find

\large \frac{1}{T}\frac{dT}{dt}=-\frac{96}{5}\frac{G^{3}}{a^{4}c^{5}}\frac{\mu ^{2}(m_{1}+m_{2})^{3}}{m_{1}m_{2}} .


For spectroscopic binaries (PPE section 8.11), where we can measure a radial velocity curve for each star, \large m_{1} and \large m_{2} can be determined using Kepler’s 3rd law and momentum conservation.  In the LIGO case, we have no radial velocity curve, but the detected signal allows us to measure the orbital period T and its rate of change \large dT/dt.  This will be sufficient to draw conclusions about the mass and nature of the system.  In particular, we can understand why the two merging bodies were most likely black holes.

C.  Extracting Physical Information from the LIGO Signal

Like a radial velocity curve, the period (or frequency) of the LIGO signal is directly related to the orbital period (or frequency) of the binary system.  In the LIGO case, however, the period of the detected gravitational wave is half that of the orbital period (\large T_{gw}=T_{orb}/2), so the wave frequency is twice that of the orbit: \large f_{gw}=2f_{orb}, where we have added the subscripts “gw” and “orb” to distinguish between wave and orbit.  These relationships are easy to understand in the case of two identical bodies (Figure 3).

Figure 3: The location of two orbiting bodies at the start of an orbit, and half a period later. The gravitational effects are the same at the two times.

When the two bodies execute half an orbit, they exchange their original positions, and the gravitational effects seen far away  are the same as at the start of the orbit.  Therefore, the radiated wave is periodic with half the period of the orbit.

We cannot determine the orbital radius a from the detected signal, so let’s use Kepler’s 3rd law to eliminate a from Eqn. 8:

\large a^{4}=\left ( \frac{T_{orb}^{2}}{4\pi ^{2}}G(m_{1}+m_{2}) \right )^{4/3}

and, after MUCH messy algebra (which might be left as a student exercise), we obtain

\large \frac{1}{T_{orb}}\frac{dT_{orb}}{dt}=-\frac{96}{5c^{5}}G^{5/3}\mathfrak{M}^{5/3}(2\pi)^{8/3}T_{orb}^{-8/3}


where \large \mathfrak{M}=\frac{(m_{1}m_{2})^{3/5}}{(m_{1}+m_{2})^{1/5}} is called the chirp mass.  Finally, using

\large \frac{1}{T_{orb}}\frac{dT_{orb}}{dt}=-\frac{1}{f_{orb}}\frac{df_{orb}}{dt}

and \large f_{gw}=2/T_{orb}, we obtain

\large \frac{1}{f_{gw}}\frac{df_{gw}}{dt}=\frac{96}{5c^{5}}G^{5/3}\mathfrak{M}^{5/3}\pi ^{8/3}f_{gw}^{8/3} .

Solving for the chirp mass,

\large \mathfrak{M}=\frac{c^{3}}{G}\left ( \frac{5}{96}\pi ^{-8/3}f^{-11/3}\frac{df}{dt} \right )^{3/5}


where we have now dropped the subscript “gw” to adopt the notation used in the literature.

Figure 4 is taken from Ref. 1.  It shows a calculated waveform derived from the actual detected signals shown in Fig. 1 at the beginning of this post.  According to Ref. 1, the signal increases in frequency from 35 to 150 Hz in the time interval 0.250 to 0.425 s immediately before the two bodies coalesce.  Let’s use these numbers to calculate the chirp mass.

Abbott Fig. 2
Figure 4: (top) LIGO signal reconstructed from data shown in Fig. 1; (bottom) Black curve is separation of the two bodies in units of the sum of their Schwarzchild radii. Green curve is their relative speed in units of c

2.     Let\large A=\frac{c^{5}}{G^{5/3}}\, \frac{5}{96}\, \pi^{-8/3}\simeq 5.45\times 10^{56} (in SI units), and rewrite Eqn. 10 as

\large \mathfrak{M}^{5/3}=Af^{-11/3}df/dt .

Next, integrate both sides over the time interval \large \Delta t=t_{2}-t_{1}=0.425-0.250=0.175\: s,

\large \mathfrak{M}^{5/3}\Delta t=A\int_{t_{1}}^{t_{2}}f^{-11/3}df=-\frac{3}{8}A\left( f^{-8/3} \right )_{t_{1}}^{t_{2}} .

Finally, plug in the LIGO numbers to show \large \mathfrak{M}\simeq 30M_{Sun}.

3.     Use Eqn. 10 to show that the total mass of the system \large M=m_{1}+m_{2} must be greater than about \large 70M_{Sun}.  (Hint: let \large m_{2}=\alpha m_{1} and derive expressions for \large \mathfrak{M} and in terms of \large \alpha and \large m_{1}.  Then minimize M.)

4.     Just before the two bodies merged, their orbital frequency was about 75 Hz.  Assuming \large M=70M_{Sun}, estimate the separation a between the bodies at this time.  (Ans: \large a\simeq 350\: \textup{km})

Your answers to questions 3 and 4 should show that the two bodies must have been highly compact objects, either black holes or neutron stars.  (Compare your answer to question 4 to the radius of the Sun.)  Since a neutron star has a mass of about \large 1.4M_{Sun}, they could not both have been neutron stars.  Could one of them have been a neutron star?

5.     Assuming one of the merging bodies was a neutron star, what was the mass of the other?  (Hint: let \large m_{1}=1.4M_{Sun}\large m_{2}=\alpha m_{1}, and derive an expression for \large \mathfrak{M} in terms of \large \alpha and \large m_{1}.  Solve for \large \alpha , noting that \large \alpha \gg 1.)  (Ans: \large \alpha \simeq 2160, so \large m_{2}\simeq 3\times 10^{3}\, M_{Sun} .)

The effective radius of a black hole of mass m is given by its Schwarzchild radius \large R_{s}=2Gm/c^{2}.  General relativity is needed to understand this properly, but a crude qualitative explanation follows from Newtonian reasoning: \large R_{s} is the distance from a point source m at which the escape speed equals c.  (Nothing, including light, can escape from a body whose radius is less than \large R_{s}, so the body is “black.”)

6.     Calculate \large R_{s} for the mass \large m_{2} found in Question 5.  Assuming the bodies merge when their separation is equal to \large R_{s}, calculate the orbital period \large T_{orb} immediately before merging.  Then find the frequency of the gravitational wave emitted at that time.  (Don’t forget the factor of 2.)  (Ans:  \large R_{s}\simeq 9\times 10^{3}\: \textup{km}\large T_{orb}\simeq 0.26 \: \textup{s}\large f_{gw}\simeq 7.6 \: \textup{Hz} .)

7.     So why did the LIGO team rule out a black hole – neutron star merger?


 1.   B. P. Abbott et al.,  “Observation of Gravitational Waves from a Binary Black Hole Merger,” Phys. Rev. Lett. 116, 061102 (2016)

2.   Lior Burko, “Gravitational Wave Detection in the Introductory Lab,” arXiv:1602.04666 [physics.ed-ph]

3.   Louis J. Rubbo et al., “Hands-on Gravitational Wave Astronomy: Extracting astrophysical information from simulated signals,” Am. J. Phys. 75, 597 (2007) and accompanying teacher’s guide: