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  • Can someone help me with this, 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

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

  • Option 1)

    \frac{{mg_0 R^2 }}{{2\left( {R + H} \right)}}

  • Option 2)

    - \frac{{mg_0 R^2 }}{{2\left( {R + H} \right)}}

  • Option 3)

    \frac{{2mg_0 R^2 }}{{R + H}}

  • Option 4)

    - \frac{{2mg_0 R^2 }}{{R + H}}

 

Answers (1)

best_answer

 

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)}

 

 


Option 1)

\frac{{mg_0 R^2 }}{{2\left( {R + H} \right)}}

This option is incorrect

Option 2)

- \frac{{mg_0 R^2 }}{{2\left( {R + H} \right)}}

this option is correct.

Option 3)

\frac{{2mg_0 R^2 }}{{R + H}}

This option is incorrect.

Option 4)

- \frac{{2mg_0 R^2 }}{{R + H}}

this option is incorrect.

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Plabita

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