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(C) The orbital velocity of a satellite at a height $h$ above the Earth's surface is given by $V_{0} = \sqrt{\frac{GM}{R+h}}$.The escape velocity from

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$R$

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(C) The orbital velocity of a satellite at a height $h$ above the Earth's surface is given by $V_{0} = \sqrt{\frac{GM}{R+h}}$.The escape velocity from the Earth's surface is given by $V_{e} = \sqrt{\frac{2GM}{R}}$.According to the problem,the orbital speed is half the escape velocity: $V_{0} = \frac{V_{e}}{2}$.Substituting the expressions: $\sqrt{\frac{GM}{R+h}} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.Squaring both sides: $\frac{GM}{R+h} = \frac{1}{4} 	imes \frac{2GM}{R}$.Simplifying the equation: $\frac{1}{R+h} = \frac{1}{2R}$.Therefore,$R+h = 2R$,which gives $h = R$.

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth,the height of the satellite above the surface of the earth is

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