A satellite in a circular orbit around the earth with a radius 1.017 times the m

Question

A satellite in a circular orbit around the earth with a radius 1.017 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 72 kg) is ejected in the backwards direction so that it is stationary with respect to the earth and falls directly to the ground. Its speed just before it hits the ground is 369 m/s. Find the total work done by gravity on the satellite fragment. RE 6.37 x 103 km; Mass of the earth= ME 5.98 x 10^24 kg.

I've calculated the gravitational potential energy using U=-G*M*m / R^2 but I still can not seem to get the correct answer.

Explanation / Answer

Fg = G*M*m/ R^2 Fg = 6.67*10^-11 * 5.98*10^24 * 72 / (1.017*6.37*10^6)^2 ; Note the exponent of the RE is 6 because we want to convert to meters Fg = 684.6 N Ep = U = Fg* d, where d is the distance the fragment will travel. Note that this distance is not 1.017*6.37*10^6m - this is the distance to the center of the earth. The distance d is only d = 0.017*6.67300*10^6m =108300 (rounded to the nearest hundred) U = Fg*d = 684.6*108300 = 74 142 180J We can also do a quick check Ep = mgh = 72*9.8*108300 = 76 416 480J. The latter result is slightly larger because the formula uses constant acceleration. The first method accounts for diminishing acceleration at higher altitudes. This however can be used a sanity check.

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