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Physics

Potential Energy of an Object at a Height

A satellite o...

Question

A satellite of mass m is just placed over the surface of earth. In this position mechanical energy of satellite is $E_{1}$ Now it starts orbiting round the earth in a circular path at height $h =$ radius of earth. In this position, kinetic energy, potential energy and total mechanical energy of satellite are $K_{2}, U_{2}$ and $E_{2}$ respectively.Then

U_{2} = \frac{E_{1}}{2}

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E_{2} = \frac{E_{1}}{4}

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K_{2} = - E_{2}

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K_{2} = - \frac{U_{2}}{2}

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Solution

The correct option is D $K_{2} = - \frac{U_{2}}{2}$
Since initially at rest,

$E_{1} = - \frac{G M m}{2 R}$

Also,

$U_{2} = - \frac{G M m}{(2 R)} = - \frac{G M m}{2 R}$

Thus, $U_{2} = E_{1}$

$K_{2} = \frac{G M m}{2 (2 R)} = \frac{G M m}{4 R}$

$K_{2} = - \frac{U_{2}}{2}$

Answer is option D.

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The correct option is D K2=−U22Since initially at rest,E1=−GMm2RAlso, U2=−GMm(2R)=−GMm2RThus, U2=E1K2=GMm2(2R)=GMm4RK2=−U22Answer is option D.

The correct option is D $K_{2} = - \frac{U_{2}}{2}$
Since initially at rest,

$E_{1} = - \frac{G M m}{2 R}$

Also,

$U_{2} = - \frac{G M m}{(2 R)} = - \frac{G M m}{2 R}$

Thus, $U_{2} = E_{1}$

$K_{2} = \frac{G M m}{2 (2 R)} = \frac{G M m}{4 R}$

$K_{2} = - \frac{U_{2}}{2}$

Answer is option D.