A satellite in a circular orbit around the earth with a radius 1.011 times the m

Question

A satellite in a circular orbit around the earth with a radius 1.011 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 68.0 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 373.0 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.

Total Work = 4.633 x 10^7 J

CALCULATE THE AMOUNT OF WORK CONVERTED TO HEAT?!

Explanation / Answer

r = 1.011 * Re = 1.011 * 6.37 * 10^6 m m = 68.0 kg v = 373.0 m/s the total work done by gravity on the satellite frament is K = (1/2)mv^2 the gravitational potential energy of the satellite is U = (GMe * m/r) where G = 6.67 * 10^-11 Nm^2/kg^2 and Me = 5.98 * 10^24 kg the amount of work converted to heat is W = K - U

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