Question

n order better to map the surface features of the Moon, a 311 kg imaging satellite is put into circular orbit around the Moon at an altitude of 133 km. Calculate the satellite's kinetic energy, gravitational potential energy, and total orbital energy. The radius and mass of the Moon are 1740 km and 7.36×1022 kg.

Answer 1

Given

mass of the satellite m = 311 kg , which is in circular orbit around the moon

mass of the moon M = 7.36*10^22 kg

radius of the moon R = 1740 km

the altitude of the satellite is r = 133 km

the orbital speed of the satellite is v = sqrt(G*M/r1)

r1 = R+r = 1740+133 km = 1873*10^3 m

v = sqrt((6.674*10^-11*7.36*10^22)/(1873*10^3)) m/s

v = 1619.43 m/s

as the satellite is in orbit around the moon so it is in rotational motion as well as linear motion

the rotational / angular velocity is v = r*w ==> W = v/r

W = 1619.43/133000 = 0.012176 rad/s

the kientic energy k.e = 0.5*m*v^2 = 0.5*311*1619.43^2 J = 407807073.12 J

gravitational potential energy P.e= G*M*m/r1 = 6.674*10^-11* 7.36*10^22*311/(1873*10^3) J = 815617674.32 J

so the total orbital energy is sum of the p.e and k.e

total energy = 815617674.32+407807073.12 J = 1223424747.44 J