What is the ratio of the kinetic energies of the moon and the Earth?

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Considering only the revolution of the Earth and the moon around their common center of mass;
what is the ratio of the kinetic energies of the moon and the Earth?
If you like use Me for the mass of the Earth, Mm for the mass of the moon and R for the distance between their centers.

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(Ω)Mistress Bekki

In the center of mass frame, their momenta must be equal and opposite.
Kinetic energy is:
KE = p^2 / 2m
So if magnitude of momentum is equal, kinetic energy is inversely proportional to mass.
So KE(moon) / KE(earth) = m(earth) / m(moon)
D–Holy overkill batman! Good job anyway though. :p

Terry R

They must be equal. KE= 1/2 mv^2. The higher moon velocity compensates for its lesser mass. You may compare this to using Jupiter as a slingshot for a spacecraft. The energy gained by the spacecraft in slinging it faster matches exactly energy lost by Jupiter (however so immeasurably inconsequential to that planet). Remember, KISS- Keep It Simple [something].

(Ω) Dr D

Since we’re only considering revolution about their common center of mass, I presume we’re ignoring revolution around the sun (?).
The gravitational force on each of them is the same
This force is constant and always pointing to each other, therefore both will move in circular orbits, and both will always remain colinear with their common center of rotation.
Let ρ = distance of center of rotation from the earth.
The centripetal acceleration of the moon is
G*Me/R^2 = vm^2 / (R – ρ)
The centripetal acceleration of the earth is
G*Mm/R^2 = ve^2 / ρ
Also their angular velocities about the center are equal
ve/ρ = vm/(R-ρ) = C
G*Me/R^2 = vm* C
G*Mm/R^2 = ve* C
Me*ve / Mm*vm = 1
Me*ve^2 / Mm*vm^2 = Mm/Me
The ratio of the KEs is the inverse ratio of the masses.


Even if you allow for the revolution of both bodies around the sun, the ratio of their kinetic energies will still be in proportion to their masses. However, the additional kinetic energy in the rotation of the earth about its own axis will tend to skew the ratio slightly toward Earth, which has both a disproportionately higher moment of inertia because of its larger radius and a greater angular velocity.


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