Is it really necessary for the orbital plane of a satellite to pass through the center of mass of the celestial body around which it's orbiting? Does the answer depend on, whether the celestial body is a sphere of radially symmetric density or it's irregularly shaped (for example, a comet or an asteroid)? If there exist cases where the orbital plane doesn't pass through the center of mass, how is Kepler's First Law defined for such cases, since the center of mass of the larger body doesn't lie on the orbital plane of the satellite which contains the foci of the elliptical orbit?

Please Note: The answers for the question Why do satellites orbit around the centre of a planet? did not clarify my above mentioned doubts.


It is absolutely false that the orbit of a satellite should always be a curve in a plane. And it is also false that the direction of the force should always point towards the center of mass of the attracting body.

As far as the first statement, it should be noticed that in the case of central forces the existence of a plane where the orbit is confined to stay is ensured by the conservation of angular momentum. If the attracting body has a spherical distribution of its mass, a mathematical theorem ensures that the resulting force is central and the situation is equivalent to concentrate the whole mass of the body in its center of mass. However celestial bodies can have shapes deviating from the spherical one and/or a non spherical mass distribution. As a consequence, real interaction energies, in particular at small distances, can be far from being spherical. A general description of anisotropic gravitational interaction, like that felt by low Earth orbit satellites, can be described by a multipole expansion. In such a case, angular momentum conservation does not hold and there is no planar orbit and no Kepler's first law.

A part the angular dependence of the potential energy, it is also interesting to notice that in general a non-spherical mass distribution does not attract in the direction of the center of mass. This can be easily verified examining simple examples of non spherical mass distributions.

  • $\begingroup$ Let's say we have a point mass $m_0$ present in the vicinity of a giant arbitrarily-shaped planet of mass $M$ and volume $V$. The net gravitational force acting on the point mass due to the planet is given by $\int_V Gm_0 dm \frac{\mathbf{r}}{r^3}$ (where $\mathbf{r}$ is the position vector from $m_0$ to the infinitesimal mass $dm$ : origin is located at $m_0$). Is there a special name for this point : $\int_V dm \frac{\mathbf{r}}{Mr^3}$ ? (Similar to how we call $\int_V dm \frac{\mathbf{r}}{M}$ as the COM of the planet) $\endgroup$
    – Ajay Mohan
    Nov 27 '19 at 8:35
  • $\begingroup$ @AjayMohan The point you refer to depends on the position of the test mass $m_0$. It is only if the large object has a mass distribution that is sherically symmetric that the point you refer to coincides with the center of symmetry (which is perforce the center of mass) wherever the test mass $m_0$ is. Otherwise, it does depend on $m_0$'s position ! $\endgroup$
    – Alfred
    Nov 27 '19 at 9:02
  • $\begingroup$ @Alfred Yes, you're right. I made a mistake. Thanks for correcting me. Upon imagining a dumbbell shaped planet, I can visually see how the direction of the net force on the test mass doesn't point towards a common point irrespective of its position. $\endgroup$
    – Ajay Mohan
    Nov 27 '19 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.