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Q: 8.19 A satellite orbits the earth at a height of $\small 400\hspace {1mm}km$ above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite $\small =200\hspace {1mm}kg$ ; ! mass of the earth $\small =6.0 \times10^2^4\hspace {1mm}kg$ ; radius of the earth $\small =6.4 \times10^6\hspace {1mm}m$ ; $\small G=6.67 \times10^-^1^1\hspace {1mm}Nm^2 kg^-^2$ .

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The total energy of the satellite at height h is given by :

$=\ \frac{1}{2}mv^2\ +\ \left ( \frac{-GM_em}{R_e\ +\ h} \right )$

We know that the orbital speed of the satellite is :

$v\ =\ \sqrt{\left ( \frac{GM_e}{R_e\ +\ h} \right )}$

Thus the total energy becomes :

$=\ \frac{1}{2}m\times \left ( \frac{GM_e}{R_e\ +\ h} \right )\ +\ \left ( \frac{-GM_em}{R_e\ +\ h} \right )$

or $=\ - \frac{1}{2} \left ( \frac{GM_em}{R_e\ +\ h} \right )$

Thus the required energy is negative of the total energy :

$E_{req}\ =\ \frac{1}{2} \left ( \frac{GM_em}{R_e\ +\ h} \right )$

or $=\ \frac{1}{2} \left ( \frac{6.67\times 10^{-11}\times 6 \times 10^{24}\times 200}{6.4\times 10^6\ +\ 0.4\times 10^6} \right )$

or $=\ 5.9\times 10^9\ J$

Posted by

Devendra Khairwa

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