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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
?

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
3 years ago
576-1396_1.PNG
Arun
25750 Points
3 years ago

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 )}}

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