Question #eb544

1 Answer
Nov 2, 2016

Answer:

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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