PART A. A satellite in a circular orbit around the earth with a radius 1.019 tim

Question

PART A. A satellite in a circular orbit around the earth with a radius 1.019 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 56.0 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 367.0 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.

PART B. Calculate the amount of that work converted to heat.

Explanation / Answer

A)

total work done by gravity is U1-U2

U1 is the potential energy of the system when the fragment is at 1.019*Re

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

then U1 = -G*m*M/(1.019*RE) = -(6.67*10^-11*56*5.98*10^24)/(1.019*6.37*10^6) = 344.11*10^7 J

U2 = -G*m*M/RE = 1.019*344.11*10^7 = 350.65*10^7 J

W_gravity = (344.11-350.65)*10^7 = -6.54*10^7

negative sign represents potential energy is decreased

B) amount of that work converted into heat is 6.54*10^7 J

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.