Q.Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth. Given: $G=6.67\times10^{-11}$ N m^2/kg^2, R = 6400 km, M = $6\times10^{24}$ kg. [Ans (as printed in the book): KE = $40.02\times10^9$ J, PE = $-80.09\times10^9$ J, TE = $-40.07\times10^9$ J, BE = $40.02\times10^9$ J]

For a satellite in circular orbit, KE = GMm/(2r), PE = -GMm/r, so Total Energy E = KE + PE = -GMm/(2r); the Binding Energy is the negative of this, i.e. +GMm/(2r).

Setup: Consider a satellite of mass m moving in a circular orbit of radius r around a planet of mass M (M much greater than m, planet treated as fixed at the centre).

Kinetic Energy of the satellite:

For a stable circular orbit, gravitational force provides exactly the centripetal force needed:

GMm/r^2 = m*v^2/r

Solving for v^2:

v^2 = GM/r

So the kinetic energy is:

KE = (1/2)mv^2 = (1/2)m(GM/r) = GMm / (2r)

Gravitational Potential Energy of the satellite:

The gravitational potential energy of mass m at distance r from mass M (taking PE = 0 at infinite separation) is:

PE = -GMm / r

(the negative sign reflects that gravity is attractive, so work must be done AGAINST gravity to increase r, i.e. PE increases -- becomes less negative -- as r increases, approaching zero at infinity.)

Total Energy of the satellite:

E = KE + PE = GMm/(2r) + (-GMm/r) = GMm/(2r) - GMm/r

E = GMm/(2r) - 2*GMm/(2r) = -GMm/(2r)

So the total mechanical energy of a satellite in a stable circular orbit is:

E = -GMm / (2r)

Note this is exactly half the potential energy (in magnitude) and negative -- the negative sign shows the satellite is in a 'bound' orbit (it cannot escape to infinity on its own; external energy must be supplied).

Binding Energy:

The binding energy of the satellite is defined as the minimum energy that must be supplied to the satellite to just remove it from its orbit to infinity, where its total energy would become zero (a body 'free' of the gravitational field, at rest at infinity, has E = 0).

Binding Energy = 0 - E = -E = -(-GMm/(2r)) = +GMm/(2r)

So the binding energy of an orbiting satellite equals GMm/(2r), the same magnitude as its total energy but positive, since this is the amount of energy that must be ADDED to bring the satellite's total energy up from its negative bound value to zero.

Binding energy of a satellite is the (positive) energy needed to free it from orbit to infinity, equal in magnitude to its total orbital energy.

Binding energy of a satellite in orbit is defined as the minimum energy that must be supplied to the satellite to remove it completely from the gravitational field of the Earth (i.e., to take it from its orbit to infinity, where PE $\to 0$ and, at the boundary case, KE $\to 0$). It equals the magnitude of the total energy of the orbiting satellite (which is negative, since it is gravitationally bound).

Derivation: For a satellite of mass $m$ in a circular orbit of radius $r$ around Earth of mass $M$, gravity supplies the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v^2 = \frac{GM}{r}$

Kinetic energy:

$KE = \frac{1}{2}mv^2 = \frac{GMm}{2r}$

Potential energy (gravitational PE, taking zero at infinity):

$PE = -\frac{GMm}{r}$

Total energy:

$E = KE+PE = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

This negative total energy shows the satellite is bound. The binding energy is the magnitude of this (i.e., the energy required to bring $E$ up to zero, the free/unbound state):

$E_B = |E| = \frac{GMm}{2r}$