Energy of an Orbiting Satellite: Overview, Questions, Preparation

Consider a satellite of mass m orbiting Earth (mass M) in a circular orbit of radius r.

K.E. = ½ m*v^2 (1)

Gravitational force of a circular orbit =

GMm/r^2 = mv^2/r

V = under root(GM/r) (2)

Putting 2 in 1, we get:

½*m*(GM/r) = GMm/2r

The gravitational potential energy of a satellite at a distance r from Earth's center is given by: 

P.E.=-GMm/r

The negative sign in the above equation indicates that work must be done to move the satellite to infinity. For a satellite near Earth's surface: P.E.=-gR2mr

The potential energy becomes less negative (increases) as r increases, approaching zero as r→∞ .

The total mechanical energy of the satellite E  is the sum of kinetic and potential energies: 

E=K.E. + P.E.

Substitute the values,

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

By solving the above equation, we get:

E = -GMm/2r

• The total energy is negative, indicating a stable and bound system (the satellite remains in a particular orbit).
• The total energy is equal to the negative of the kinetic energy:E=-K .
• The potential energy is twice the kinetic energy in magnitude but opposite in sign:E=-2K.E.
• As the orbital radius r increases, the total energy becomes less negative, slowly approaching zero at infinity. This is due to the fact that the potential energy increases and the kinetic energy starts decreasing.
• Satellites can also change their orbits by gaining and loosing energies.

The escape energy is the energy required to move the satellite to infinity, where its total energy becomes zero. The binding energy of a satellite is the energy needed to overcome its negative total energy:  

Binding Energy =E=-GMm/2r

This is equal to the kinetic energy K . To escape, the satellite's kinetic energy must be 0.

Thus, the escape speed is: 

E=E(escape) – E(orbit)

0 – (-GMm/2r)

GMm/2r

This shows that the escape speed is 2 times the orbital speed.

Kinetic energy required to escape: 1/2mv^2=E=GMm/2r

V=under root 2GM/a

For a geostationary satellite, the orbital radius is fixed at r≈42300km , with a period of 24 hours. The energy calculations follow the same formulas. For example, with R=6.4×106m,g=9.8m/s2 , and r=4.23×107m  :

K=gR2m2r≈9.8×6.4×1062m2×4.23×107≈4.75×109mJU=-gR2mr≈-9.5×109mJE=-gR2m2r≈-4.75×109mJ

These values highlight the large energy scales involved in geostationary orbits.

For Revision Notes H2