Energy Of An Orbiting Satellite

Using Eq. 

$V^2=\frac{GM_E}{R_E+h}$

the kinetic energy of the satellite in a circular orbit with speed v is

$K.E=\frac{1}{2}mv^2$

$=\frac{GM_Em}{2(R_E+h)}$    (1)

Considering gravitational potential energy at infinity to be zero, the potential energy at distance (R +h) from the centre of the earth is

$P.E=-\frac{GM_Em}{R_E+h}$   (2)

The K.E is positive whereas the P.E is negative. However, in magnitude the K.E. is half the P.E, so that the total E is

E = K.E + P.E $=-\frac{GmM_E}{2(R_E+h)}$    (3)

The total energy of an circularly orbiting satellite is thus negative, with the potential energy being negative but twice is magnitude of the positive kinetic energy.

When the orbit of a satellite becomes elliptic, both the K.E. and P.E. vary from point to point. The total energy which remains constant is negative as in the circular orbit case. This is what we expect, since as we have discussed before if the total energy is positive or zero, the object escapes to infinity. Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero.

Example 1

A 400 kg satellite is in a circular orbit of radius $2R_E$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4R_E$? What are the changes in the kinetic and potential energies?

Answer 

Initially,

$E_i=-\frac{GM_Em}{4R_E}$

While finally

$E_f=-\frac{GM_Em}{8R_E}$

The change in the total energy is

$\Delta E=E_f-E_i$

$=\frac{GmM_E}{8R_E}=\left(\frac{GM_E}{R_E^2}\right)\frac{mR_E}{8}$

$\Delta E=\frac{gmR_E}{8}$

$=\frac{9.81 \times 400 \times 6.37 \times 10^6}{8}$

$=3.13 \times 10^9$ J

The kinetic energy is reduced and it mimics ∆E, namely, 

∆K = K$_f$ – K$_i$ = – 3.13 × 10$^9$ J.

The change in potential energy is twice the change in the total energy, namely 

∆V = V$_f$ – V$_i$  = – 6.25 × 10$^9$ J