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(A) The orbital speed $v_0$ of a satellite at a distance $r$ from the center of the earth is given by $v_0 = \sqrt{\frac{GM}{r}}$.For a body near the

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$4$

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(A) The orbital speed $v_0$ of a satellite at a distance $r$ from the center of the earth is given by $v_0 = \sqrt{\frac{GM}{r}}$.For a body near the earth's surface,$r_1 = R = 6400 \ km$,so $v_1 = \sqrt{\frac{GM}{R}} = 8 \ km s^{-1}$.For a body at a height $h = 19,200 \ km$,the distance from the center is $r_2 = R + h = 6400 + 19,200 = 25,600 \ km$.The orbital speed at this height is $v_2 = \sqrt{\frac{GM}{r_2}}$.Taking the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{R}{r_2}} = \sqrt{\frac{6400}{25600}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$v_2 = \frac{v_1}{2} = \frac{8 \ km s^{-1}}{2} = 4 \ km s^{-1}$.

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