Question

An object is projected horizontally with speed $\frac{1}{2}\sqrt{\frac{GM}{R}}$, from a point at height $3R$ [where $R$ is radius and $M$ is mass of earth, then object will].

• A. Fall back on surface of earth by following parabolic path
• B. Fall back on surface of earth by following hyperbolic path
• C. Start rotating around earth in a circular orbit
• D. Escape from gravitational field of earth

Answer 1

Answer: (C)

At height $3R$, i.e at distance $4R$ from the centre of the earth,
$V_{\text{orbital }}=\sqrt{\frac{GM}{r}}$
Here, $r=4R\Rightarrow V_0=\sqrt{\frac{GM}{4R}}=\frac{1}{2}\sqrt{\frac{GM}{R}}$
Thus, an object taken to a height $3R$ if projected horizontally with speed $\frac{1}{2}\sqrt{\frac{GM}{R}}$,
will start rotating around earth in a circular orbit.