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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 ?

Saurabh Koranglekar
one year ago
Arun
25763 Points
one year ago

For a satellite

centripetal force = Gravitational force

$\dpi{100} \therefore \; \; \frac{mv_{0}^{2}}{(R+x)}=\frac{GMm}{(R+x)^{2}}$

$\dpi{100} or\; \; \; v_{0}^{2}=\frac{GM}{(R+x)}=\frac{gR^{2}}{(R+x)}\; \; \; \; \; \; \; \left [ \because \; \; g=\frac{GM}{R^{2}} \right ]$

$\dpi{100} or\; \; \; v_{0}=\sqrt{\frac{gR^{2}}{R+x}}$

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.

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

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

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

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