"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." is the important part of this closed question:If you could shrink the moon to one meter radius, how much could you slow the moon down by and have it continue to orbit the earth?
"Here's how to do the apogee speed calculation in Google, using a perigee altitude of 400 km = 6778 km perigee radius and μ⊕ from en.wikipedia.org/wiki/Standard_gravitational_parameter sqrt((3.9860044E14 m^3/s^2)*(2/(406700 km)-1/(206739 km)))" says PM 2Ring.
I pasted the numbers into the Google search box and Google returned "sqrt(((3.9860044E14 (m^3)) / (s^2)) * ((2 / (406 700 km)) - (1 / (206 739 km)))) = 179.255041 m/s" which I think is convenient and is impressive on Google's part. The answer is believable although I had guessed about 50 m/s. Now all I need to do is figure out what the numbers mean.
So could someone explain the logic behind this calculation? I think the equations on this page have something to do with it, especially the Vis viva equation: https://en.wikipedia.org/wiki/Vis-viva_equation which seems to be:
For any Keplerian orbit (elliptic, parabolic, hyperbolic, or radial), the vis-viva equation is as follows: $$\displaystyle v^{2}=GM\left({2 \over r}-{1 \over a}\right)$$ where:
v is the relative speed of the two bodies
r is the distance between the two bodies
a is the length of the semi-major axis (a > 0 for ellipses, a = ∞ or 1/a = 0 for parabolas, and a < 0 for hyperbolas)
G is the gravitational constant
M is the mass of the central body
The product of GM can also be expressed as the standard gravitational parameter using the Greek letter μ.
