# Question #eb544

Nov 2, 2016

Satellite moves faster in orbit when it is close to the planet it orbits, and slower when it is farther away.

#### Explanation:

The satellite experiences two forces while in orbit

1. Force due to gravity ${F}_{g}$ of the planet
${F}_{g} = G \frac{{M}_{p} \times {m}_{s}}{R} _ {O}^{2}$
where, ${M}_{p} \mathmr{and} {m}_{s}$ are mass of the planet and mass of the satellite respectively; $G$ is Universal gravitational constant and ${R}_{O}$ is the radius of the orbit measured from the center of the planet.
We also know that ${R}_{O} = {R}_{p} + h$, where ${R}_{p}$ is the radius of planet and $h$ is the height of the satellite above planet's surface.
2. Net centrifugal force ${F}_{C}$ due to its circular motion
${F}_{C} = \frac{{m}_{s} {v}^{2}}{R} _ O$
where $v$ is the velocity of the satellite.

As the satellite-planet system is in equilibrium, equating both forces we get
$G \frac{{M}_{p} \times {m}_{s}}{R} _ {O}^{2} = \frac{{m}_{s} {v}^{2}}{R} _ O$
$\implies G \frac{{M}_{p}}{R} _ O = {v}^{2}$
$\implies {v}^{2} \propto \frac{1}{R} _ O$

From above equation it is evident that when a satellite is moved to a larger radius/higher from the planet’s surface, ${R}_{O}$ increases. To keep the system in equilibrium, the velocity must decrease.

Heights of satellites above earth and their velocities.