A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth's surface, is

Answers (1)

Gravitational potential energy at height 'h' -

U_{h}=-\frac{GMm}{R+h}

U_{h}=-\frac{gR^{2}m}{R+h}

U_{h}\rightarrow Potential energy at height h

R\rightarrow Radius of earth

- wherein

U_{h}=-\frac{mgR}{1+\frac{h}{R}}

 

 total energy of satelliteis equal to half of its potential energy.

\therefore E= \frac{-G m_{s}.Me}{2(R+h)}

\because g.=\frac{GMe}{R_{2}}  or  GMe = g_{0}R^{2}

E =\frac{ g_{0} m_{s}.R^{2}}{2(R+h)}

 

 

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