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(C) At height $h = 3R$,the distance from the center of the Earth is $r = R + h = R + 3R = 4R$.The orbital velocity $V_o$ required for a circular orbit

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Start rotating around the Earth in a circular orbit

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(C) At height $h = 3R$,the distance from the center of the Earth is $r = R + h = R + 3R = 4R$.The orbital velocity $V_o$ required for a circular orbit at a distance $r$ from the center of the Earth is given by $V_o = \sqrt{\frac{GM}{r}}$.Substituting $r = 4R$ into the formula,we get:$V_o = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}}$.Since the object is projected horizontally with exactly this speed,it will maintain a circular orbit at a distance of $4R$ from the center of the Earth.Therefore,the correct option is $(c)$.

An object is projected horizontally with speed $\frac{1}{2} \sqrt{\frac{GM}{R}}$ from a point at height $3R$ (where $R$ is the radius and $M$ is the mass of the Earth). The object will:

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