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Satellites

Potential ene...

Question

Potential energy of a satellite having mass ‘m’ and rotating at a height of $6.4 \times 10^{6}$ from the earth centre is

$- 0.5 m g R_{e}$

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$- m g R_{e}$

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$- 2 m g R_{e}$

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$4 m g R_{e}$

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Solution

The correct option is A
$- 0.5 m g R_{e}$

Potential energy $= - \frac{G M m}{r} = - \frac{G M m}{R_{e} + h} = - \frac{G M m}{2 R_{e}} [A s h = R_{e} (g i v e n)]$

$∴$ Potential energy $= - \frac{g R_{e}^{2} m}{2 R_{e}} = - 0.5 m g R_{e} [A s G M = g R^{2}]$

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Similar questions

3.6 \times 10^{6}

= 500

= 6 \times 10^{24}

= 6.4 \times 10^{6}

= 6.67 \times 10^{- 11}

m^{2}

k g^{- 2}

A satellite of mass $m$ is revolving around the earth at height $R$ (radius of the earth) from the earth's surface. Its potential energy will be:

m

R_{e}

R_{e} =

1 \times 10^{6}

35 \times 10^{6}

= M = 6 \times 10^{3 x} k g

= R = 6.4 \times 10^{6} m

6.67 \times 10^{- 11} {N m}^{2} {K g}^{- 2}

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