Gravitation Question 348

Question: A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Options:

A) $ [\frac{gR^{2}}{R+x}] $

B) $ [\frac{gR^{2}}{R-x}] $

C) $ [gx] $

D) $ [{{\left( \frac{gR^{2}}{R+x} \right)}^{1/2}}] $

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Answer:

Correct Answer: D

Solution:

  • Gravitational force provides the necessary centripetal force.

    $ [\therefore \frac{mv^{2}}{\left( R+x \right)}=\frac{GmM}{{{\left( R+x \right)}^{2}}}also,,g=\frac{GM}{R^{2}}] $

    $ [\therefore \frac{mv^{2}}{\left( R+x \right)}=m\left( \frac{GM}{R^{2}} \right)\frac{R^{2}}{{{\left( R+x \right)}^{2}}}\frac{n!}{r!\left( n-r \right)!}] $

    $ [\therefore \frac{mv^{2}}{\left( R+x \right)}=mg\frac{R^{2}}{{{\left( R+x \right)}^{2}}}] $

    $ [\therefore v^{2}=\frac{gR^{2}}{R+x}\Rightarrow v={{\left( \frac{gR^{2}}{R+x} \right)}^{1/2}}] $



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