If we transport rocks from Earth to Moon, enough rocks to double the mass of the Moon?

Question

what will be the new period of the Moon?
Right now T = 28 days.


No rockets, just use long ropes to pull the rocks.

Answer 1

I'm going to do this under the approximation that the Earth is vastly more massive than the Moon.

M = Earth mass
m = Moon mass
Ro = current distance between the centers of the Earth and Moon
R1 = new distance
wo = current angular frequency = 2*pi/(28 days)
w1 = new angular frequency

a) Because it's all held together by gravity, the gravitational force provides the centripetal force:
GMm/Ro^2 = mRowo^2
=> wo^2 = (GM)/Ro^3 ; and later,
w1^2 = (G(M-m))/R1^3

b) Because angular momentum will be conserved in this process:
mwo*Ro^2 = L
= (2m)w1*R1^2

b) implies that:
w1/wo = (Ro/R1)^2 / 2
2*w1/wo = (Ro/R1)^2
Ro/R1 = sqrt(2*w1/wo)

a) implies that:
(w1/wo)^2 = (Ro/R1)^3 * ((M-m)/M)

(w1/wo) = sqrt( (sqrt(2*w1/wo))^3 * (1-m/M))
w1/wo = (w1/wo)^(3/4) * (1/2)^(1/4) * sqrt(1-m/M)
(w1/wo)^(1/4) = (1/2)^(1/4) * (1-m/M)^(1/2)
w1/wo = (1/2) * (1-m/M)^2

So
T1/To = wo/w1 = 2/(1-m/M)^2

from my reference, m/M = 0.0123, so:
2/(1-0.0123)^2 = 2/(.9877)^2 = 2.1015

Therefore, T1 = 2.1 * 28 days = 58.84 days.
(I guess if I were completely consistent in ignoring the ratio m/M, I would get a straight factor of 2. But I believe that if I did it even more carefully, it would only be a 2nd-order effect. So taking the finite ratio into account here is a reasonable compromise.)

Answer 2

supastremph is only conserving graviational potential energy! But even gravitional plus kinetic energy will not be conserved because you're doing work on the system big time.

I'm not quite sure nealjking did it right because I don't see any distances to the center of mass being used in the centripetal acceleration and angular momentum equations he uses. The problem is solved by simultaneously equating angular momentum about the center of mass of the present to that of the new earth-moon systems, and equating centripetal acceleration (again about the center of mass) with gravitational acceleration in both cases. Details are left to the reader.

As for des0ne, why is it that some people have such a hard time with the hypothetical? The rope assumption was necessary to avoid the need to consider the angular momentum of the rocket exhaust (an intractible dilema).

Oops. I've neglected the angular momentum of the earth due to it's own rotation. this is a significant contribution to the whole earth-moon system. How this changes the answer depends on where the material is removed from. If from the poles, the earth's own angular momentum will stay the same. If from the equator, angular momentum will be incrementally removed from each load moved.

Answer 3

Because you are pulling the rocks up to the moon with ropes, you can view this as central force motion, hence the angular momentum of the earth moon system will remain constant. Using conservation of energy, you can determine the new earth-moon distance, r2, GMm/r1 = G(M-m)(2m)/r2.

Then use the conservation of angular momentum to find the new period using this new geometry.

Answer 4

long ropes!! hahahaha
I have no answer sorry, thought that last part was funny though
thanks for the 2pts!