Question

If a rocket is fired with a velocity, $V=2\sqrt{gR}$ near the Earth's surface and goes upwards, its speed in the inter-stellar space is

• A. $4\sqrt{gR}$
• B. $\sqrt{2gR}$
• C. $\sqrt{gR}$
• D. $\sqrt{4gR}$

Answer 1

Answer: (B)

$\sqrt{2gR}$

$V=\sqrt{gR}$

As Earth's escape velocity=$\sqrt{\frac{2GM}{R}}=V_{esc}$

So, we can say that energy required to escape earth's gravitational field is

$\frac{1}{2}m(v_{esc}{)}^2$

$K{E}_i=\frac{1}{2}m\times \frac{2GM}{R}$

$K{E}_i=\frac{GMm}{R}$

Now, energy given is $\frac{1}{2}m\times 4\times gR$

$⟹2gmR$

So, $E_{remaining}=\frac{1}{2}m{v}^2=\frac{GMm}{R}+2mR\left(\frac{GM}{R^2}\right)$

as, $g=\frac{GM}{R^2}$

$\frac{V^2}{2}=\frac{2GM}{R}-\frac{GM}{R}=\frac{GM}{R}$

$V=\sqrt{\frac{2GM}{R}}$ or $V=\sqrt{2gR}$