A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of 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 xx 10^(24) kg, radius = km, G = 6.67 xx 10^(-11)Nm^(-2)//kg^(-2).

A geostationary satellite orbits the earth at a height of nearly 36,00

1.	A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of 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 xx 10^(24) kg, radius = km, G = 6.67 xx 10^(-11)Nm^(-2)//kg^(-2).
Answer» Solution :Gravitational potential at height hfrom the surface of earth is `V = -(GM)/((R+h)) = (-6.67 xx 10^(-11) xx (6 xx 10^(24)))/((6.4 xx 10^(6) + 36 xx 10^(6))) = -9.4 xx 10^(6) J//kg`.