Mumbai University > EXTC > Sem 8 > Satellite Communication and Networks

Marks: 10 M

Year: Dec 2012

• For a geostationary orbit, the law of conservation of energy is valid at all points of orbit when a satellite is in motion.

• That means, the sum of kinetic and potential energy of the satellite always remains constant and value of this constant is

  $U = \frac{- G m_1 m_2}{2a}= \frac{-G m_1 m_2}{2r}$

  where, $m_1$ is the mass of earth

  $m_2$ is the mass of satellite

  a is the distance of satellite from the earth but since it is geostationary satellite a = r.

• The kinetic energy of satellite is given by:

  $KE = \frac{1}{2} m_2 v^2$

• Potential energy of a satellite at a distance r from the center of earth is given by:

$$PE=- \frac{Gm_1 m_2}{r}$$

• Therefore we get:

$$U=KE+PE \\ So, \frac{- G m_1 m_2}{2r}= \frac{1}{2} m_2 v^2-\frac{Gm_1 m_2}{r} \\ \frac{- G m_1 m_2}{2r}= \frac{m_2 v^2 r-2Gm_1 m_2}{2r} \\ \frac{- G m_1 m_2}{2r}= \frac{m_2 (v^2 r-2Gm_1 )}{2r} \\ - G m_1= (v^2 r-2Gm_1 ) \\ v^2= \frac{- G m_1}{r}+ \frac{2Gm_1}{r} \\ v=\sqrt{(Gm_1 (\frac{1}{r})} \\ Gm_1=constant= μ \\ μ=3.986 ×10^{14} m^3/s^2 \\ v=\sqrt{μ(\frac{1}{r}) } \\ ∴v= \sqrt{\frac{μ}{r}}$$

• Thus, the significance of the equation given is to find out the velocity of the satellite.