Question

Escape velocity of a body from the surface of the earth is $11.2km/s$. Escape velocity, when thrown at an angle of ${45}^o$ from horizontal will be

• A. $11.2km/s$
• B. $22.4km/s$
• C. $\frac{11.2}{\sqrt{2}}km/s$
• D. $11.2\sqrt{2}km/s$

Answer 1

The correct option is A $11.2km/s$

Escape velocity, $v_e=\sqrt{\frac{2GM}{R}}=11.2km/s$

Now if the particle is projected with an angle of ${45}^o$ with horizontal, then it has horizontal and vertical components as $v_1$ and $v_2$ respectively such that $v_1^2+v_2^2=v^2$

Apply conservation of energy at surface and infinity-

$\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+(-\frac{GMm}{R})=0$

$\frac{1}{2}m[v_1^2+v_2^2]=\frac{GMm}{R}$

$\frac{1}{2}m{v}^2=\frac{GMm}{R}$

$⟹v=\sqrt{\frac{2GM}{R}}=11.2km/s$

Thus escape velocity will remain same.