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)


One thought on “Work, Energy and the “Satellite Drag Paradox”

  1. Just to elaborate on the drag force calculation for Q. 8. From the information given just below Fig. 3, if the altitude fell by 2 km in 92 days, that corresponds to an average dr/dt = (2000 m / 92 days/ 86400 s/day) = 0.00025 m/s.

    Then from Equation 4 with GM = 3.986E+14, r = (400 km + 6378 km) = 6.778E+6 m, and the ISS mass is 4.2E+5 kg, and orbital speed vc = 7668 m/s, we find the drag force to be FD=0.0598 N, rounded to 0.060 N. Not very much for such a large object moving at high speed – the atmosphere up there must be very thin indeed!

    Also note that Eq. 4 simplifies to dr/dt = -2 (FD/Fg) * vc, where Fg is the gravitational force on the ISS. Or using vc = Sqrt[GM/r] , it can also be written directly as dr/dt = -2 FD /m Sqrt[r^3/GM].


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