Question #eb544

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Question #eb544

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Answer 1 · 2016-11-02T22:15:37

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

Force due to gravity $F_{g}$ $F_g$ of the planet
$F_{g} = G \frac{M_{p} \times m_{s}}{R_{O}^{2}}$ $F_g=G(M_pxxm_s)/R_O^2$
where, $M_{p} and m_{s}$ $M_p and m_s$ are mass of the planet and mass of the satellite respectively; $G$ $G$ is Universal gravitational constant and $R_{O}$ $R_O$ is the radius of the orbit measured from the center of the planet.
We also know that $R_{O} = R_{p} + h$ $R_O=R_p+h$ , where $R_{p}$ $R_p$ is the radius of planet and $h$ $h$ is the height of the satellite above planet's surface.
Net centrifugal force $F_{C}$ $F_C$ due to its circular motion
$F_{C} = \frac{m_{s} v^{2}}{R_{O}}$ $F_C=(m_sv^2)/R_O$
where $v$ $v$ is the velocity of the satellite.

As the satellite-planet system is in equilibrium, equating both forces we get
$G \frac{M_{p} \times m_{s}}{R_{O}^{2}} = \frac{m_{s} v^{2}}{R_{O}}$ $G(M_pxxm_s)/R_O^2=(m_sv^2)/R_O$
$\Rightarrow G \frac{M_{p}}{R_{O}} = v^{2}$ $=>G(M_p)/R_O=v^2$
$\Rightarrow v^{2} \propto \frac{1}{R_{O}}$ $=>v^2prop1/R_O$

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

Heights of satellites above earth and their velocities.

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Question #eb544

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