Celestia Forums

How distant can a moon be from the Earth? I know that if the moon has significant mass, it works with the Earh's gravity to keep it orbit so:

How distant can a moon of negligible mass be?
How distant can a moon of a third of The Moon's (Luna's) mass be?
How distant can a moon of Luna's mass be?
How distant can a moon of half the Earth's mass be?
How distant can a moon the same mass as Earth be? (But then it's not really a moon any more)

Michael Kilderry

You'd need to know the size of the primary's Hill Sphere.

http://open-encyclopedia.com/Hill_sphere
http://www.asterism.org/tutorials/tut22-1.htm

Note that in the formula to calculate it (shown in a more understandable format in the second link) the "minor body" is actually the Earth, and the "major body" is the sun, and the moon's mass is considered negligible. The calculation probably doesn't apply is the mass of the third object is similar to the mass of the "minor body".

So it actually depends on three things - the masses of the planet and star, and the distance between the planet and the star.

I am still a little confused on how you would work out the hill sphere. How would you work it out on a calculator?

Michael Kilderry

Michael Kilderry wrote:I am still a little confused on how you would work out the hill sphere. How would you work it out on a calculator?

Michael Kilderry

To start with, you need to know the numbers you're plugging in for the bodies involved. Earth is 5.9742e24 kg, the Sun is 1.9891e30 kg. Distance is 1.496e11 metres.

First step is to multiply the Sun's mass by 3. Then press the 1/X button to invert that. Then multiply that number by the Earth's mass. This gives you the m/3M term in the brackets in this equation.

Then, (I'm assuming you have a reasonably scientific calculator here), press the "x to the power y" button and then enter (1/3), then press "equals" to get the final answer. Or just press the "cube root" button if your calculator has that.

Then multiply that result by the distance.

Or, if you have Excel, then use this formula in a cell to figure it out:

=a*((m/(3*M))^(1/3))

(where a, m, and M are the cells that have the orbital distance in metres, the mass of the Earth, and the mass of the Sun respectively).

EDIT: Corrected the order and equations.

Evil Dr Ganymede wrote:First step is to multiply the Sun's mass by 3. Then press the 1/X button to invert that. Then multiply that number by the Earth's mass. This gives you the m/3M term in the brackets in this equation.

Then multiply that number by the distance.

Finally, (I'm assuming you have a reasonably scientific calculator here), press the "x to the power y" button and then enter (1/3), then press "equals" to get the final answer. Or just press the "cube root" button if your calculator has that.

Or, if you have Excel, then use this formula in a cell to figure it out:

=(a*(m/(3*M)))^(1/3)

(where a, m, and M are the cells that have the orbital distance in metres, the mass of the Earth, and the mass of the Sun respectively).

You have to do the cube root before you multiply by a. So the calculator instructions should be

First step is to multiply the Sun's mass by 3. Then press the 1/X button to invert that. Then multiply that number by the Earth's mass.

Then press the "x to the power y" button and then enter (1/3), then press "equals" to get the final answer. Or just press the "cube root" button if your calculator has that.

Finally multiply that number by the distance.

or for Excel:
=a*(m/(3*M))^(1/3)

Other interesting Hill stuff:

This formula works for objects that orbit in the same plane as the priamary and secondary. But an object in a retrograde orbit can orbit well past the Hill radius as defined by this formula, in a weird triangluar orbit.

No object can exist beyond the Moon's orbit in a prograde orbit, not because of the Hill radius, but because the Moon's influence will either eject it, or cause it to collide with the Earth or the Moon.

Mars' moons, Phobos and Deimos have a hill sphere that is below their surfaces. Similar to the question posed in one of your links, "can an astronaut orbit the space shuttle?", the answer is no, nothing can orbit Phobos or Deimos.

Oops. You're right, Tony. Sorry! I've corrected it now.

I thought the number I ended up with was a little funny, oh well, I'll just have to do the math again. I've never heard of triangular orbits before!

Interesting question: How far away does an object orbiting the Earth beyond the Moon have to be from the Moon to avoid being flung into a different orbit?

Would an object be able to orbit our Moon?

Michael Kilderry

Michael Kilderry wrote:I thought the number I ended up with was a little funny, oh well, I'll just have to do the math again. I've never heard of triangular orbits before!

Interesting question: How far away does an object orbiting the Earth beyond the Moon have to be from the Moon to avoid being flung into a different orbit?

Would an object be able to orbit our Moon?

Michael Kilderry

Yes, objects can orbit the Moon. The Apollo Command Modules orbited the Moon, as did the Clemintine spacecraft, and most recently, ESA's SMART-1. But no natural objects orbit the Moon. The Moon used to be much closer to the Earth. That would have made its Hill sphere very small at the time, so nothing could form there.

In a prograde orbit, an object would have to be beyond Earth's Hill sphere to avoid the Moon disrupting its orbit. So nothing can orbit the Earth in the same direction as the Moon beyond the Moon's orbit.

But in a retrograde orbit, things can orbit beyond the Moon, and at very close range too.

Triangular orbits... Well kind of...

I think the orbit looks more egg shaped rather than triangular, would an orbit like this be able to last long periods of time?

Michael Kilderry

Michael Kilderry wrote:I think the orbit looks more egg shaped rather than triangular, would an orbit like this be able to last long periods of time?

Michael Kilderry

Surprisingly it is pretty stable. I don't know about millions of years, but in my simulator it lasts hundreds of years and would probably last thousands or longer if I ran it long enough.

I've heard of something called a quasi-satellite. It seesm to be a moon orbiting around a planet in a bit of a strangely shaped orbit but it's really orbiting around the Sun close to the planet.

Do you think the sattelite could be one of these?

Michael Kilderry

Michael Kilderry wrote:I've heard of something called a quasi-satellite. It seesm to be a moon orbiting around a planet in a bit of a strangely shaped orbit but it's really orbiting around the Sun close to the planet.

Do you think the sattelite could be one of these?

Michael Kilderry

A quasi-satellite has a 1:1 resonance with Earth. Cruithine is one, and I forget the name of the other one that has a strange horseshoe type orbit. They are not always near Earth though. They often are on the opposite side of the Sun, 2 AU away.

tony873004 wrote:No object can exist beyond the Moon's orbit in a prograde orbit, not because of the Hill radius, but because the Moon's influence will either eject it, or cause it to collide with the Earth or the Moon.

Tony,

even if the object is in ressonance with the Moon?

I've been reading back on this topic just now, and I think the resonance question is an interesting one. Would an earth-orbiting object's orbit be stable if it was resonanced with the moon, even if it was prograde and orbited beyond the moon's distance?

There is a location beyond the Moon called the L2 point where an object orbits at the same speed as the Moon, because of the way thw gravity of the Moon and the Earth add together; this is not a stable location, and an object placed there would wander away after a few years.


Presumably that means that the Earth can't have two large moons; that isn't very fair, as Jupiter has four.

L2 has angular motion behind the Moon. However, since the Earth/Moon relationship changes, L2 moves, and appears as a circling orbit behind the Moon.

d.m.f.