Studying Haumea: A Dwarf Planet with Two Moons and a Ring

Figure 1:  Artist’s rendition of Haumea and its recently discovered ring, which has a radius of approximately 2300 km.  Haumea’s two moons, Ha’iaka and Namaka, are too far from the planet to be visible in this image.  Image credit: IAA-CSIC/UHU

A.  Introduction

Haumea is a Pluto-sized dwarf planet lying beyond Neptune within the Kuiper Belt (Fig. 1).  Discovered in 2004, it is the fifth largest dwarf planet known, and is unique among them because it hosts a ring in addition to its two moons.  It is also one of the fastest spinning objects in the solar system, completing one revolution in less than 4 hours.  The ring was discovered in 2017 by Ortiz et al (Ref. 1) when Haumea passed between Earth and a distant star, temporarily blocking its light from reaching Earth.  This stellar occultation was detected at nine observatories in central and eastern Europe, and the light curves recorded at these sites were used to determine the ring’s radius and opacity in addition to the dimensions of the dwarf planet itself.  The Haumea system offers a wealth of opportunity for illustrating basic physics concepts to first year college students.  Lying in the outer reaches of the solar system, its physical properties are likely to be unknown a priori (in contrast to Saturn, for example), making it an ideal topic for study or testing.  In the following, we offer a simple treatment of the stellar occultation data, plus an assortment of problems suitable for homework exercises, quizzes or exams in an introductory physics course.  We hope you will find these materials useful in your teaching.

B.  The Size and Shape of Haumea

The occultation occurred on January 21, 2017.  Figure 2 shows the light curves recorded at the nine observatories participating in the study.  The large dip centered near 200 s is due to occultation by the dwarf planet, and the narrower, weaker dip near 50 s is due to the ring.  Figure 3 shows the relative motion of Haumea and the occulted star as observed at each observatory.  The blue lines are the occultation chords, i.e., the apparent motion of the star as the planet passed in front of it, as seen at each observatory.  Note that the length of the chord observed at Konkoly (in Hungary) is nearly equal to the diameter b-b′ of Haumea.

To use the light curves to determine the chord lengths, we need to know the proper speed of Haumea relative to Earth.  The proper speed is the component of the relative velocity perpendicular to the line of sight from Earth to the dwarf planet.  Using the online orbital calculator on JPL’s Horizons website (Ref. 2), it is straightforward (but tedious) to find the proper speed at the time of occultation: v_{\perp }=14.367\textup{ km/s}.

Figure 2:  Light curves of the occultation recorded at nine European observatories.  The blue lines are the best square well fits to the data.  From Ref. 1

1.  The ingress/egress times recorded at Konkoly are 3:08:20.3 and 3:10:17.4 UTC.      Find the length of the Konkoly occultation chord (in km) shown in Fig. 3.  (Ans:  1680 km)

In the following questions, assume that the shape of Haumea is a triaxial ellipsoid with semi-axes a > b > c.

Picture 2
Figure 3:  The apparent path of the star behind Haumea, as seen from each observatory.  a, b, and c are the  semi-axes of the triaxial ellipsoid model of Haumea.  The dot-dash line passes through the center of the planet.  Adapted from Ref. 1.


2.  From direct measurements on Fig. 3, the “diameter” b-b′ is 2% greater than the Konkoly chord.  Calculate the semi-axis b of the dwarf planet.  (Ans: 857 km.  Compare to the published value 852 ± 4 km)

3.  Haumea rotates rapidly around its shortest axis, labeled c in Fig. 3.  At the time of the occultation, the brightness of the planet observed from Earth was at its minimum value.  Its maximum brightness is 34% greater than this minimum.  Find the length of the semi-axis a.  (Ans: 1150 km.  Compare to the published value 1161 ± 30 km.)

4.  The length of the rotation axis c = 513 ± 16 km, found from analysis of the brightness vs. time curve.  The volume of a triaxial ellipsoid is V=\frac{4}{3}\pi abc.  Using the dimensions found above, calculate the radius R of a sphere having the same volume.  (Ans: 794 km.  The published value is 797.5 ± 5.5 km.)

5.  Phil Plait, who writes the Bad Astronomy blog (Ref. 3), likened the shape of Haumea to a “flattened potato.”  Why is the planet flattened?  (See PPE, section 8.14.)

C.  The Ring and Moons of Haumea

The ring surrounding Haumea is in the plane of its equator.  According to the online Extended Data, Table 2 of Ref. 1, occultation by the ring was seen at Ondrejov (Czech Republic) at 3:06:48.4 UTC, while the center of the occultation by the dwarf planet occurred at 3:09:20.7 UTC.  (See Figs. 2 and 4.)

Haumea Figure 4
Figure 4:  Haumea’s ring as viewed from Earth.  The occultation path observed at Ondrejov is the dot-dash line.  Adapted from Ref. 1

6.  Estimate the radius of the ring.  (Ans: 2188\textup{ km}.  The published value is 2287_{+75}^{-45}\textup{ km}.)

In the following questions, assume that Haumea is a spherically symmetric body with the radius R given in Question 4 above.

7.  Haumea’s largest moon, Ha’iaka, has a near-circular orbit in the equatorial plane of the planet (Ref. 4).  Its period is P=49.462\pm .083\textup{ d}, and its semi-major axis is a=49,880\pm 200\textup{ km}.  Find the mass of Haumea.  (Ans: 4.02\times 10^{21}\textup{ kg})

8.  A second moon, Namaka, orbits Haumea with period {P}'=18.278\pm .008\textup{ d}.  Find the semi-major axis of its orbit.  (Ans: {a}'=25,680\textup{ km})

9.  Calculate the average density of Haumea, and compare it to that of Pluto (Ref. 5), another dwarf planet lying within the Kuiper Belt.  Pluto has a radius of 1151 km and a mass of 1.309\times 10^{22}\textup{ kg}.  (Ans: \rho _{H}=1.885\times 10^{3}\mathrm{\: kg/m^{3}}, \rho _{P}=2.049\times 10^{3}\mathrm{\: kg/m^{3}}.  Nearly the same!)

10.  Show that Haumea’s moons lie well beyond its Roche limit. (See PPE, section 8.16.) (Ans: R_{Roche}=1950\textup{ kg}.)

Note:  The Roche limit derived in PPE, Eqn. 8.22, assumes that the density of the main body and its satellites are the same.  If Haumea has a rocky core and an icy crust, as suspected from its albedo, and if its ring and moons were formed by impact, then they are likely to have a density closer to that of ice, \rho\approx 1000\mathrm{\: kg/m^{3}}.  As shown in the discussion preceding Eqn. 8.22 in PPE, this lower density would increase the magnitude of the Roche limit.  Ortiz et al assume a satellite density of 500\mathrm{\: kg/m^{3}} to obtain a Roche limit substantially larger than the ring radius.

D.  Surface Gravity on Haumea

In the following exercises, assume once again that Haumea is a spherically symmetric body of radius 797.5 km and density 1.885\times 10^{3}\mathrm{\: kg/m^{3}}, and that its period of rotation is 3.9155 ± .0001 hr.

11.  What is the gravitational acceleration g on the surface of Haumea?  What is the escape velocity from the surface?  (Ans: 0.42\mathrm{\: m/s^{2}}, 820 m/s)

12.  Picture a tiny object lying on the surface of the spinning planet.  Draw a free body diagram for the object.  At what location(s) is the normal force acting on it a maximum?  Where is the normal force a minimum?  (Ans: maximum at the poles, minimum at the equator)

13.  If the dwarf planet is a gravitationally bound “rubble pile,” it will begin to break apart when the minimum normal force on a surface object approaches zero.  (See PPE, problem 8.15.)  What is the minimum period of rotation for Haumea to remain stable?  (Ans: 2.41 hr)

14.  A simple pendulum of length 1 m is placed on the surface of Haumea.  What is its period of oscillation if it is located (a) at the north pole; (b) on the equator?  (Ans: (a) 9.70 s  (b) 12.3 s)

15.  Imagine that the pendulum is located at latitude \theta and is not swinging.  Upon close inspection, you find that it is not hanging vertically, but is deflected by an angle \phi.  Derive an equation showing the dependence of \phi on the latitude and on Haumea’s angular velocity \omega.  Is the pendulum bob deflected toward the pole or toward the equator?  Justify this with a free body diagram.  What is the deflection angle when \theta =45^{\circ} degrees?

(Ans: \large \boldsymbol{tan\phi =\frac{\omega ^{2}Rsin2\theta }{2\left ( g-\omega ^{2}Rcos^{2}\theta \right )}}, toward the equator, \large \phi (45^{\circ})=13.1^{\circ})

E. References

1.  J. L. Ortiz et al, Nature 550, 219 (October 12, 2017)



4.  D. Ragozzine and M. E. Brown, Astro. J. 137, 4766 (June 2009)

5.  See, e.g.,



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