Question

In order better to map the surface features of the Moon, a 393 kg imaging satellite is put into circular orbit around the Moon at an altitude of 149 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

a)

orbital velocity of satellite

v^2 = GM / (R +h)

v^2 = 6.67* 10^-11* 7.36* 10^22 / ( 1740* 10^3 + 149*10^3)

v = 1612.08 m/s

kE of satellite

kE = 0.5* m v^ = 0.5* 393* 1612.08 ^2 = 5.107* 10^8 J

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b)

gravitational pE

pE = - GM m / ( R + h)

pE = - 6.67* 10^-11* 7.36* 10^22* 393 / ( 1740* 10^3 + 149* 10^3)

pE = - 10.21* 10^8 J

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c)

total orbital energy

U = pE + kE

U = - 5.103* 10^8 J

Note: neglect negative sigb in order to have magnitude only

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