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

Answer 1

Mass of the Earth, M = 6.0 × 10²⁴ kg

Radius of the Earth, R = 6400 km = 6.4 × 10⁶ m

Height of a geostationary satellite from the surface of the Earth,

h = 36000 km = 3.6 × 10⁷ m

Gravitational potential energy due to Earth’s gravity at height h,

`= (-GM)/(R+h)`

= `(6.67xx10^(-11)xx6.0xx10^(24))/(3.6xx10^7+0.64xx10^7)`

`=-(6.67xx6)/(4.24) xx 10^(13-7)`

`=-9.4 xx 10^(6) J/Kg`

Answer 2

Distance of satellite from the centre of earth = R = r + x

= 6400 + 36000 = 42400 km = 4.24 x 10⁷ m

Using potential, `V = -(GM)/R`, we get

V = `-((6.67xx10^(-11))xx(6xx10^(24)))/(4.24xx10^(7))`

= `-9.44 xx 10^6 J kg^(-1)`