Question #eb544

1 Answer
A08
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(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.

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