A satellite in a circular orbit around the earth with a radius 1.017 times the m
ID: 1475870 • Letter: A
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 = 66.0 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 357.0 m/s.
a)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.
b)Calculate the amount of that work converted to heat.
Explanation / Answer
radius of earth is R =6.371x10^6 m
Mass of earth M =5.98x10^24 kg
Mass of fragmen tid m =66 kg
G =6.67x10^-11 N.m^2/kg^2
R1 =1.017R
R2 = R
initial potential energy U1 = -GMm/R1^2
Final potential energy U2 =-GMm/R2^2
Work done by gravity = U1 -U2
W = GMm/R [ 1- (1/1.017) ]
W = (6.67x10-11x5.98x1024x66)/(6.371x106)[ 1- (1/1.017) ]
W =69.1 MJ
(b) v =357 m/s
Kinetic energy =(1/2)mv^2 = (1/2)(66*357*357)
K = 4.206 MJ
Energy lost to heat is Q = W - K = 69.1 -4.206
Q = 64.894 MJ
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