Question

Derive an expression for the energy of satellite.

Answer 1

The total energy of the satellite is the sum of its kinetic energy and gravitational potential energy. The potential energy of the satellite is,

U = `-("GM"_"s""M"_"E")/(("R"_"E" + "h"))` ......(1)

Here M_s – mass of the satellite, M_E – mass of the Earth, R_E – radius of the Earth.

The Kinetic energy of the satellite is

K.E = `1/2"M"_"s""v"^2` .............(2)

Here v is the orbital speed of the satellite and is equal to

v = `sqrt("GM"_"E"/(("R"_"E" + "h")))`

Substituting the value of v in (2) the kinetic energy of the satellite becomes,

K.E = `1/2("GM"_"E""M"_"s")/(("R"_"E" + "h"))`

Therefore the total energy of the satellite is

E = `1/2("GM"_"E""M"_"s")/(("R"_"E" + "h")) - ("GM"_"s""M"_"E")/(("R"_"E" + "h"))`

E = `-("GM"_"s""M"_"E")/(2("R"_"E" + "h"))`

The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.

As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at a large distance.