Question

A satellite in a circular orbit around the earth with a radius 1.015 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 89.0 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 359.0 m/s. Find the total work done by gravity on the satellite fragment. R_E 6.37·10³ km; Mass of the earth= M_E 5.98·10²⁴ kg.

Calculate the amount of that work converted to heat.

Answer 1

A)

total work done by gravity is U1-U2

U1 is the potential energy of the system when the fragment is at 1.015*Re

U2 is the PE of the system when the fragment is at Re

then U1 = -G*m*M/(1.015*RE) = -(6.67*10^-11*89*5.98*10^24)/(1.015*6.37*10^6) = 549*10^7 J

U2 = -G*m*M/RE = -1.015*549*10^7 = -557.28*10^7 J

W_gravity = (-549+557.28)*10^7 = 8.28*10^7

B) amount of that work converted into heat is (0.5*m*v^2)- 8.28*10^7 J
v be the speed of the fragnment at the surface of the earth

V = sqrt(2*g*h)

h = 1.015*RE - RE = 0.015*RE = 0.015*6.37*10^6 = 9.55*10^4 m

then v= sqrt(2*9.81*9.55*10^4) = 1369.2 m/s

then amount of that work converted into heat is (0.5*m*v^2)- 8.28*10^7 J = 0.5*89*1369.2^2 - 8.28*10^7 = 623749.5 = 6.23*10^5 J