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  • I have a doubt, kindly clarify. 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

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

  • Option 1)

    gx

  • Option 2)

    \frac{gR}{R-x}

  • Option 3)

    \frac{gR^{2}}{R+x}

  • Option 4)

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

 

Answers (1)

As we learnt in 

For a satellite

centripetal force = Gravitational force

\therefore \; \; \frac{mv_{0}^{2}}{(R+x)}=\frac{GMm}{(R+x)^{2}}

or\; \; \; v_{0}^{2}=\frac{GM}{(R+x)}=\frac{gR^{2}}{(R+x)}\; \; \; \; \; \; \; \left [ \because \; \; g=\frac{GM}{R^{2}} \right ]

or\; \; \; v_{0}=\sqrt{\frac{gR^{2}}{R+x}}

 

 

Orbital velocity of satellite -

V=\sqrt{\frac{GM}{r}}

r=R+h

r\rightarrow Position of satellite from the centre of earth

V\rightarrow Orbital velocity

- wherein

The velocity required to put the satellite into its orbit around the earth.

 \frac{mv_{o}^2}{\left ( R+x \right )} = \frac{Gmm}{\left ( R+x \right )^2}

v_{0}^2 = \frac{Gm}{\left ( R+x \right )}

= \frac{gR^2}{\left ( R+x \right )}

v_{0} = \sqrt{\frac{gR^2}{\left ( R+x \right )}}

 


Option 1)

gx

Incorrect

Option 2)

\frac{gR}{R-x}

Incorrect

Option 3)

\frac{gR^{2}}{R+x}

Incorrect

Option 4)

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

Correct

Posted by

Sabhrant Ambastha

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