Question 249: Gravitation

Question

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is____________ $	imes 10^{10} \mathrm{~J}$.
(Mass of earth $=6 	imes 10^{24} \mathrm{~kg}$, Radius of earth $=6.4 	imes 10^6 \mathrm{~m}$, Gravitational constant $=6.67 	imes 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$ )

Accepted Answer

The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following formula:
$ \mathrm{KE} = \frac{1}{2} m v^2 = \frac{1}{2} m \frac{GM_e}{r} = \frac{GM_e m}{2r} $
For this satellite, the radius $ r $ is the sum of the Earth's radius $ R_E $ and the height $ h $ above the Earth's surface:
$ r = R_E + h $
Given:
Mass of the satellite, $ m = 1000 \, 	ext{kg} $
Mass of the Earth, $ M_e = 6 	imes 10^{24} \, 	ext{kg} $
Radius of the Earth, $ R_E = 6.4 	imes 10^6 \, 	ext{m} $
Height of orbit above the Earth's surface, $ h = 270 \, 	ext{km} = 2.7 	imes 10^5 \, 	ext{m} $
Gravitational constant, $ G = 6.67 	imes 10^{-11} \, 	ext{Nm}^2 \, 	ext{kg}^{-2} $
Substitute these values into the equation:
$ \mathrm{KE} = \frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes 1000}{2(6.4 	imes 10^6 + 2.7 	imes 10^5)} $
$ = \frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes 1000}{2 	imes 6.67 	imes 10^6} $
$ = 3 	imes 10^{10} \, 	ext{J} $
Hence, the kinetic energy of the satellite in its orbit is $ 3 	imes 10^{10} \, 	ext{J} $.

Gravitation

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