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If the moon were shrunk to having a radius of one meter, without changing its distance (of the apogee) from the earth, what would be the slowest it could be slowed down to and still orbit the earth, albeit in a highly elliptical orbit? It shouldn't hit the earth. The speed at the apogee is what is sought. The apogee must be at 400,000 km from the earth, same as the moon's center is now.

The moon orbits the earth at close to one kilometer per second in a fairly round orbit. So the answer must be considerably less than that speed.

Someone better at math than me might be able to calculate it from the equations on this page: https://en.wikipedia.org/wiki/Vis-viva_equation.

The lunar distance is approximately 400,000 km says https://en.wikipedia.org/wiki/Lunar_distance_(astronomy)

The radius of the earth is approximately 6400 km says Wikipedia: "A globally-average value is usually considered to be 6,371 kilometre" https://en.wikipedia.org/wiki/Earth_radius.

The nearest to the surface of the earth that a one meter radius lump of moon rock could fly and stay in orbit, I estimate to be 100 km.

Matthew Christopher Bartsh
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The size of a satellite does not matter for the period/velocity of its orbit. In fact, as long as the mass is much lower than the planet's, the satellite's mass doesn't matter either. So the moon would orbit at about the same speed if it was shrunk to 1 meter.

qazwsx
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By making the Moon’s orbit highly elliptical - in other words, by making the semi-major axis $a$ sufficiently large - you can make the Moon’s velocity at apogee as small as you like, and its orbital period as large as you like. Since you can stretch the orbit in this way while keeping the perigee (the closest point to the Earth) the same as the current Earth-Moon distance, the size of the Moon is irrelevant.

(This, of course, treats the Earth-Moon as a two body system and ignores the effect of the Sun and other bodies in the Solar System)

Note: this answer was written in response to the original question, not the edited version which adds the constraint that apogee is unchanged.

gandalf61
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Okay, so the answer is a fairly straight forward application of the Vis-viva equation.

So, since you want perigee to be (100+6400) km and apogee to be 400000, hence the semi major axis becomes a = 1/2 * ( 6500 + 400000)= 203250 km

You want the velocity at apogee. so the distance to satellite, r = 400000 km

Put these values of a and r in the equation,

and you get v = 178.51 m/sec

Hence, you would need to slow moon down to about 178 m/sec at its apogee to have its orbit pass at a distance of 100 km to earth's surface .

This is of course under the assumptions of that equation which assumes there is negligible air friction at 100 km, but that may not be a valid assumption in this case.

silverrahul
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Is what you're asking this? "What would be the orbital velocity at apogee of a point mass that has the moon's apogee but just barely misses the Earth at perigee?"

Then you can use the vis viva equation from the page you linked. The semi-major axis $a$ would be half of the sum of the lunar apogee and the Earth's radius (including your 100 km). Put in the apogee for $r$ to get your answer.

The equation is for "small" mass satellites so it would probably not be entirely accurate. I'm not sure where to go for a better approximation.

Kristoffer Sjöö
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