Question

Question #eb544

Answer 1

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(M_pxxm_s)/R_O^2`
   where, `M_p 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=(m_sv^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(M_pxxm_s)/R_O^2=(m_sv^2)/R_O`
`=>G(M_p)/R_O=v^2`
`=>v^2prop1/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.