Question

A 425 kg satellite is in a circular orbit at an altitude of 400 km above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface, where it hits the ground with a speed of 2.00 km/s. How much energy was transformed to internal energy by means of friction?
J

Answer 1

Using Energy conservation:

KEi + PEi + Wf = KEf + PEf

KEi = Initial kinetic energy of satellite = (1/2)*m*Vi^2

Vi = Initial speed of satellite in circular orbit = sqrt (G*M/R)

R = distance between satellite and center of earth = 6.371*10^6 m + 400*10^3 m = 6.771*10^6 m

M = mass of earth = 5.98*10^24 kg, So

KEi = (1/2)*m*(sqrt (GM/R))^2

KEi = G*m*M/(2R)

PEi = Initial potential energy at the given altitude = -G*m*M/R

KEf = (1/2)*m*Vf^2

Vf = final speed of satellite = 2.00 km/sec = 2000 m/sec

m = mass of satellite = 425 kg

PEf = final potential energy at the surface of earth = -G*m*M/Re

Re = Radius of earth = 6.371*10^6 m

Wf = Work-done by friction = energy transformed into friction

So,

Wf = (KEf - KEi) + (PEf - PEi)

Wf = (1/2)*m*Vf^2 - G*m*M/(2R) + [(-G*m*M/Re) - (-G*m*M/R)]

Wf = (1/2)*m*Vf^2 + G*m*M/(2R) - G*m*M/Re

Using known values:

Wf = (1/2)*425*2000^2 + 6.67*10^-11*425*5.98*10^24/(2*6.771*10^6) - 6.67*10^-11*425*5.98*10^24/(6.371*10^6)

Wf = -1.32*10^10 J (-ve sign means energy is lost)

energy transformed is 1.32*10^10 J

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