1
$\begingroup$

If a planet becomes lighter, then I believe its orbit will change.

For example, the moon is 1/81 of the mass of the earth. So, if the moon disappeared, then the earth-moon system would lose 1/81th of its mass. How would the orbit of the earth change in that circumstance?

$\endgroup$
  • $\begingroup$ So should we move the Earth to the location of the original Earth-Moon center of mass position, and give it the center of mass' velocity? $\endgroup$ – uhoh Feb 10 at 1:39
  • $\begingroup$ This question has been asked many times, in many forms. The problem is that the Earth cannot instantaneously lose 1/81th of its mass, in either Newtonian mechanics or general relativity. You essentially are asking what the laws of physics say will happen when the laws of physics are violated. Such questions do not make sense. $\endgroup$ – David Hammen Feb 10 at 2:15
  • 1
    $\begingroup$ @DavidHammen you're not a big Space 1999 fan? (humor) youtu.be/Y6BXaGEuqxo?t=2393 lower volume before playing! $\endgroup$ – uhoh Feb 10 at 2:36
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because the question has no basis in science. $\endgroup$ – David Hammen Feb 10 at 2:52
  • 1
    $\begingroup$ Consider the following thought experiment. Suppose we have two identical bowling balls side by side in identical orbits around the Sun, attached together by a very light, strong thread. The balls are small enough that we can neglect their gravitational attraction to each other. The total mass of the object is 2m, where m is the mass of one ball. Do you propose that when we cut the thread, and now have two independently orbiting objects of mass m each, that their orbits will significantly change? $\endgroup$ – Eric Lippert Feb 13 at 23:04
3
$\begingroup$

I will answer the specific case of the Moon disappearing since the more general question, is a bit open-ended.

Of course the Moon cannot just "disappear." However, there are very real circumstances where a planet or other stellar object might be disrupted. The best example I can think of is if a sufficiently large asteroid or moon from another planet somehow make their way into Earth's orbit and collide with the Moon. Such collisions between Solar System objects are rare, but occasionally occur in models of the future evolution of the Solar System (see work by Jacques Laskar).

However, one must be careful to first consider whether that mass is actually lost from the system. For instance, in the Giant Impact Scenario a Mars-sized object is theorized to have slammed into the Earth, but the material never left the system. If a similar collision were to disintegrate the Moon and the mass remained as rocks and dust in orbit around the Earth, the effect on Earth's orbit would be minor. This is because it is the Earth-Moon system that orbits around the Sun. Note that there is a small change since the Earth-Moon barycenter has shifted position, but the effect is tiny so let's ignore it for now.

Instead, let's consider a case where the collision disrupts the Moon, and its material leaves Earth's orbit. If this occurs sufficiently quickly, (faster than the orbital period, i.e., 1 year) then from an orbital dynamics standpoint, the mass is effectively lost instantaneously. The Dutch astrophysicist Adriaan Blaauw, first calculated the relevant physics. Earth's orbit will expand, but by an almost miniscule amount. The reason is that the dynamics of the Earth's orbit around the Sun is dictated by the total mass of the system (i.e. Earth's mass plus the Sun's mass). To lowest order, the new orbit's orbital separation ($a$) will be:

$$ a \approx 1 + \frac{\Delta M}{M} {\rm AU} \approx 1 + \frac{1/81 M_{\oplus}}{M_{\odot}} {\rm AU} \approx 1 + 4\times10^{-8} {\rm AU} $$

So, the orbit will expand by roughly 40 parts in a billion. A quick calculation using Kepler's third law ($a^3\sim P^2$) shows that the length of a year would increase by about 2 seconds. A similar change in the Earth's orbital eccentricity will occur ($\Delta e \sim \frac{\Delta M}{M}$), but since this change is smaller than the precision to which we have measured the Earth's orbital eccentricity, the effect is negligible.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Okay, but I was expecting more of a Keplerian equation based on stable orbits. $\endgroup$ – Tyler Durden Feb 10 at 12:27
  • $\begingroup$ Surely something needs to be said about angular momentum loss? $\endgroup$ – Rob Jeffries Feb 10 at 19:52
  • $\begingroup$ I added something about orbital eccentricity, which should help address angular momentum loss. Or maybe I'm missing something? Happy to add more, @TylerDurden if you can tell me more about what you're looking for. $\endgroup$ – Jeff Feb 11 at 2:42

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