I'm reading An Introduction to Mechanics by Kleppner and Kolenkow. In the chapter on angular momentum, a (beautiful!) example is given by discussing Kepler's second law of planetary motion.

The law states that the when a planet is orbiting the sun, the area swept by its radius to the sun for a given time length is constant.

The authors prove Kepler's law using conservation of angular momentum. To my understanding, they show that the angular momentum of a given planet about the sun is conserved, by showing that the torque acting on it is 0. This is done by showing that the gravity force exerted on the planet by the sun is parallel to the position vector to the planet from the sun.

Here's what's confusing to me: it seems to me that given planet A in the solar system - in addition to the force of gravity from the sun, additional (gravity) forces act on it. These are the forces that act on A from other planets in the system.

So why can we say that the total torque on planet A is 0? If we can't say that, we can't rely on conservation of angular momentum.


3 Answers 3


Deriving Kepler's laws in this manner is somewhat of an idealization that works because the perturbing effects of other planets are relatively small. If you consider more than just one planet and the sun, you are getting off into an $N$-body problem, which is an absolute monster of a problem. So even though the story is more complex than one planet and the sun, the conclusions from considering the problem from that standpoint work well enough to allow one to analytically derive Kepler's laws.

As noted in the comments below, the reality of the situation is that the presence of other gravitating bodies does alter the orbit of a given planet, even if it is by a small amount. For example the great astronomer, Le Verrier among others, discovered Neptune by noticing perturbations in the orbit of Uranus.

  • $\begingroup$ Thank you. Two questions: how can we know that the accumulating effects of other planets on planet A don't sum up to be significant? It seems to me that simply ignoring them could lead to very wrong conclusions. Am I wrong? $\endgroup$
    – Aviv Cohn
    Commented Mar 18 at 0:35
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    $\begingroup$ @AvivCohn You can prove the second law independently of the first law, all you need for the second law is the conservation of angular momentum. This one of the reasons that the second law is often the first law proved in many textbooks. Proof of the first law requires that one know the specific form of the attractive force, i.e. Newton's universal law of gravitation. $\endgroup$ Commented Mar 18 at 1:45
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    $\begingroup$ @AvivCohn We know that "the accumulating effects of other planets on planet A don't sum up to be significant" because Kepler originally derived his laws empirically by looking at actual data of planetary observations collected by Tycho Brahe over many years. Planets do slightly perturb each other's orbits and in fact the existence of Neptune was predicted by noticing the perturbations of Uranus. $\endgroup$
    – KDP
    Commented Mar 18 at 2:00
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    $\begingroup$ @AlbertusMagnus Yes $\endgroup$
    – KDP
    Commented Mar 18 at 2:04
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    $\begingroup$ @AvivCohn Have a look at the Nice Model. The effects of the other planets don't add up to anything observable during a human lifetime, but they do accumulate over the life of the Solar System. If you plan to integrate the equations of motion for a billion years or so, you need to be careful that numerical errors don't overwhelm the small effects that you care about. $\endgroup$ Commented Mar 18 at 20:33

You're not wrong, strictly speaking we should consider the force of gravity from the other planets. However, the Sun is by a large margin the most massive body in the solar system, hence to first order we can neglect the other planets.

Consider for example the motion of the Earth. The relevant parameters are:

$M_{Sun} \sim 2\times 10^{30} kg$

$M_{Earth} \sim 6\times 10^{24} kg$

$r_{Sun-Earth} \sim 1 AU = 1.5 \times 10^{11} m$

The gravitational force between the Earth and the Sun is therefore, from Newton's law of universal gravitation, about $3.6 \times 10^{22} N$.

Meanwhile, the second-largest body in the solar system is Jupiter. The relevant parameters are:

$M_{Jupiter} \sim 2\times 10^{27} kg$

$M_{Earth} \sim 6\times 10^{24} kg$

$r_{Jupiter-Earth} \sim 5.88 \times 10^{11} m$ (at closest approach - sometimes Jupiter is on the opposite side of the Sun)

Again applying Newton's law of universal gravitation, we get a gravitational force of $2.3 \times 10^{18} N$. That's four orders of magnitude less than the gravitational force of the Sun. It's why we can neglect Jupiter (and all other planets) when analyzing the motion of the Earth.

Just for fun, if you work through the math, the gravitational force between the Earth and the Moon is actually larger than that between Earth and Jupiter - but you still don't have to take the Moon into account, because that force is still two orders of magnitude less than the Sun's gravitational force.

Edit: As pointed out in the comments, the above is only to first order, since over the long term a force four orders of magnitude weaker will still cause noticeable effects. Nonetheless, these other planets/moons are sometimes ahead of the Earth and sometimes behind it, hence their effect tends to cancel. You can take them into account anyway, but the complexity of the calculation you'll have to do increases dramatically.

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    $\begingroup$ Another thing about the Moon is that half the month it's ahead of the Earth, half behind. So its effects mostly cancel out over time. $\endgroup$
    – Barmar
    Commented Mar 18 at 14:21
  • $\begingroup$ Good answer. However, it's an exaggeration to say "we can neglect Jupiter (and all other planets) when analyzing the motion of the Earth". Jupiter does in fact have a significant influence on the other bodies in the solar system, cf PM 2Ring's answer. Even a force 4 orders of magnitude weaker than the central attraction can destabilise orbits, cause precession etc.. In these cases, a single orbital revolution still looks almost Keplerian, but it's not exactly closed anymore and over many years can become completely different. $\endgroup$ Commented Mar 19 at 11:51
  • $\begingroup$ I'm sure the influence of the moon is quite palpable -- after all, the common barycenter is 4600 km or so from the center of the Earth, which means that the Earth jiggles quite a lot (thousands of kilometers back and forth each month!) on its orbit around the Sun. It's just that the angular momentum of the entire Moon-Earth-system is constant (sans other planets' influence): The angular momentum of the Earth oscillates around a constant. $\endgroup$ Commented Mar 19 at 18:11
  • $\begingroup$ And, by the way, the influence of other planets should cancel itself out in the long run as well. They are, after all, not permanently leading or trailing Earth but pass it or are passed. $\endgroup$ Commented Mar 19 at 18:15
  • $\begingroup$ @Peter Those influences don't quite cancel out, so the Earth's orbital plane (the ecliptic) slowly precesses. This is distinct from the precession of the equatorial plane. en.wikipedia.org/wiki/Axial_precession $\endgroup$
    – PM 2Ring
    Commented Mar 20 at 4:07

Yes, the trajectory of a planet is affected by the gravity of the other planets, but the effect on the specific angular momentum $h$ is relatively small.

We can get Solar System trajectory data from JPL's Horizons system. The JPL Development Ephemeris is computed by integrating the equations of motion of the major Solar System bodies (including individual perturbations from 343 asteroids), relative to the barycentre of the Solar System. I have more info about Horizons and the JPL DE in this answer.

Horizons doesn't directly supply specific angular momentum data, but we can easily calculate it by taking the cross product of the position and velocity vectors.

Here's a 1 year plot of $h$ for Earth (relative to the Sun), using a 1 day time step. (Data points are plotted at 0:00 TDB, which is quite close to 0:00 UTC). The main variation in Earth's $h$ is caused by the Moon, so it has a period of ~1 month. The maximum relative variation from the mean is only ~$1/2200$.

Angular momentum, Earth

We can also get the angular momentum of the Earth-Moon barycentre (EMB). The relative variation here is much smaller. It's mostly caused by Jupiter, and to a lesser extent, by Venus. Now the maximum relative variation from the mean has dropped to ~$1/90000$.

Angular momentum, EMB

Here's my Sage / Python plotting script, running on the SageMathCell server. The script can also plot radial distance or angular velocity. Any major body can be used as the observation center, but of course you'll get rather odd looking angular velocity or angular momentum graphs if the target isn't orbiting the center, or vice versa. ;)

The plots for Jupiter and Saturn are rather interesting because those two planets have quite a strong effect on each other.

Please see the Horizons manual for details on specifying bodies and times.

  • $\begingroup$ Because the Moon's mass is relatively large, its orbit is quite complex, with perturbations on various time scales. I have info & graphs here: astronomy.stackexchange.com/a/55112/16685 $\endgroup$
    – PM 2Ring
    Commented Mar 20 at 4:03

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