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.

Answers (1)
D Devendra Khairwa

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

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