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(C) The escape velocity from the surface of the Earth is given by $v_e = \sqrt{2gR}$.The speed of the satellite in its orbit is given as $v_o = \frac{

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(C) The escape velocity from the surface of the Earth is given by $v_e = \sqrt{2gR}$.The speed of the satellite in its orbit is given as $v_o = \frac{v_e}{4} = \frac{\sqrt{2gR}}{4}$.For a satellite in a circular orbit at a distance $x$ from the center of the Earth,the centripetal force is provided by the gravitational force:$\frac{GMm}{x^2} = \frac{mv_o^2}{x}$.Substituting $v_o^2 = \frac{2gR}{16} = \frac{gR}{8}$ and $GM = gR^2$:$\frac{gR^2}{x^2} = \frac{gR/8}{x}$.Solving for $x$:$x = \frac{gR^2}{gR/8} = 8R$.The height $h$ of the satellite above the Earth's surface is $h = x - R = 8R - R = 7R$.

An artificial satellite is moving around the Earth in a circular orbit with a speed equal to one-fourth the escape speed of a body from the surface of the Earth. The height of the satellite above the Earth's surface is ............ ($R$ is the radius of the Earth). (in $R$)

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