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 = 55.0 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 371.0 m/s.

(a ) Find the total work done by gravity on the satellite fragment. The radius of the Earth = 6.37 times 10^3 km; The mass of the Earth = 5.98 times 10^24 kg.

(b) The work done by gravity must equal the change of potential energy of the satellite fragment. Use the general formula for Gravitational Potential Energy, NOTe: PE = mgh, is only valid for small distances above the surface.

(c) Calculate the amount of that work that is converted in to heat.

Explanation / Answer

Use conservation of energy to find the work done.

Work done = change in potential energy
W = G*m1*m2/(1.017*r) - G*m1*m2/r
W = (G*m1*m2*/r)*((1/1.013)-1)
W = (6.67300

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