Question

A geostationary satellite orbits the earth at a height of nearly 36000 km from the surface of the earth. What is the potential due to earth's gravity at the site of the satellite? (Take the potential energy at infinity to be 0) Mass of the earth equal to 6.0 × 10²⁴ kg; Radius = 6400 km.

Answer 1

gravitational potential(v) is the amount of workdone in slowly carrying a unit mass from infinity to a particular point.

so, V= -GM/R+h

so, V = (6.67*10^-11)*(6.0 × 10²⁴)/42400*10^3

V=-9.4*10^6 J/kg

Answer 2

Hii dear,

◆ Answer: -9.4×10^6 J/kg.

◆ Explaination -
# Given -
M = 6 × 10^24 kg
R = 6400 km = 6.4 × 10^6 m
h = 36000 km = 3.6 × 10^7 m

# Solution-
We know that, the gravitational potential energy due to Earth’s gravity at height h is given by-
GPE = -GM/(R+h)
GPE = –6.67×10^-11×6×10^24 / (3.6×10^7+0.64×10^7)
GPE = -9.4×10^6 J/kg.

Gravitational potential energy at height of geostationary satellite is -9.4×10^6 J/kg.

Hope this was useful.