As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0×10^{24} kg, radius = 6400 km

Asked by Pragya Singh | 1 year ago | 127

Given:

Radius of the Earth, R = 6400 km = 0.64 × 10^{7 }m

Mass of Earth, M = 6 x 10^{24 }kg

Height of the geostationary satellite from earth’s surface,

h = 36000 km = 3.6 x 10^{7} m

Therefore, gravitational potential at height ‘h’ on the geostationary satellite due to the earth’s gravity:

G_{P }=\( \dfrac{-GM}{R+h}\)

\(⇒ GP = \dfrac{-(6.67\times 10^{-11})\times (6\times 10^{24})}{(0.64\times 10^{7})+(3.6\times 10^{7})}\)

\(⇒ GP = \dfrac{-40.02\times 10^{13}}{4.24\times 10^{7}}\)

Therefore, the gravitational potential due to Earth’s gravity on a geostationary satellite orbiting earth is -9.439 × 10^{6} J/Kg

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