Question

A 3450-kg spacecraft is in a circular orbit 1630 km above the surface of Mars. How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 3560 km above the surface?

Answer 1

orbital speed v= sqrt(GM/(R+h))

M = mass of mars = 6.39*10^23 kg

R = radius of mars = 3.39*10^6 m

kinetic energy K = (1/2)*m*v^2 = (1/2)*G*M*m/(R+h)

gravitational potential energy U = -G*M*m/(R+h)

total energy E = K + U = -G*M*m/(2*(R+h))

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at height h1 = 1630 km = 1630000 m

E1 = -6.67*10^-11*6.39*10^23/(2*(3.39*10^6+1630000))

E1 = -4245149.4 J

at height h2 = 3560 km = 3560000 m

E2 = -6.67*10^-11*6.39*10^23/(2*(3.39*10^6+3560000))

E2 = -3066280.57 J

work done = E2 - E1 = 1178868.83 J