A satellite in a circular orbit around the earth with a radius 1.013 times the m
ID: 1474090 • Letter: A
Question
A satellite in a circular orbit around the earth with a radius 1.013 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 79.0 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. Find the total work done by gravity on the satellite fragment. RE 6.37·103 km; Mass of the earth= ME 5.98·1024 kg. Calculate the amount of that work converted to heat.
Explanation / Answer
Work done by Gravity = m*g*h
Where
m = 79.0 kg
h = Re * (1.013 - 1) m
h = 6370 * 10^3 * (0.013) m
h = 82823 m
Work done = 79.0 * 9.8 *82823 J
Work done = 6.41 * 10^7 J
Final Kinetic Energy = 1/2 * m * v^2
K.E fin = 1/2 * 79.0 * 371^2
K.E fin = 5.436 * 10^6 J
Amount of work converted to heat = 6.41 * 10^7 - 5.436 * 10^6 J
Amount of work converted to heat = 5.86 * 10^7 J
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