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.

Gravitational potential energy is the energy possessed by a body at some point due to its mass and its power to attract object due to its Gravity

We know Gravitational potential at a point is given by the relation


V = -GM/R


Where V is the potential of a point at a distance R from a Body of mass M, G is universal Gravitational constant


Now R is the Distance from the center of the body, so in case of a satellite at some height h from earth surface, the distance of earth’s satellite from the centre of earth R will be equal to sum of earth’s radius r and height from the surface of earth h


R = r + h


The situation has been shown in figure



We are given Radius of earth


r = 6400 Km


the height of satellite from Earth’s Surface


h = 36,000 km


so the net distance of point where the satellite is located to the center of the earth is


R = r + h = 6400 Km + 36,000 km


= 42400 Km = 4.24 × 107 m


Mass of earth


M = 6.0 × 1024 kg


Value of universal gravitational Constant


G = 6.67 × 10-11 Nm2Kg-2


Now putting the values in above equation we get potential at the point as



So the potential due to earth’s gravity at the site of this satellite is


-9.43 × 106 Jkg-1


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