Question

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 × 1024 kg, radius = 6400 km. Question 8.22 Gravitation

Answer 1

Mass of the Earth, M = 6.0 x 10^24 kg
Radius of the Earth, R =6400 km => 6.4 x 10^6 m
Height of a geostationary satellite from the surface of the Earth,
h =36000 km=> 3.6 x 10^7 m
Now
Gravitational potential energy due to Earth's gravity at height h
= -GM/(R+h)

= -6.67×10-¹¹ ×6×10²⁴
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3.6×10^7+0.64×10^7

=> -9.94 × 10^6 J/Kg .

Answer 2

here,
radius of earth ( Re ) = 6400 km =6400*10³ m
mass of earth ( Me ) = 6 * 10²⁴ kg
height of geostationary satellite (h) = 36000km = 36000*10³m

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∴ gravitational potential  V = - GMe/( Re + h) 
                                              = -6.67×10⁻¹¹ * 6*10²⁴/(6400+36000)*10³
                                              =40.02*10¹³/42400*10³
                                              =-9.43 × 10⁶ j/kg