Gravitation - Result Question 54
57. The escape velocity from earth is $11.2 km / s$. If a body is to be projected in a direction making an angle $45^{\circ}$ to the vertical, then the escape velocity is
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(a) $11.2 \times 2 km / s$
(b) $11.2 km / s$
(c) $11.2 / \sqrt{2} km / s$
(d) $11.2 \sqrt{2} km / s$
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Answer:
Correct Answer: 57. (b)
Solution:
- (b) Escape velocity does not depend on the angle of projection.
The value of escape velocity is given by
$ V_e=\sqrt{\frac{2 G M}{R}} $
Here, $M=$ mass of earth, $G=$ gravitational constant $R=$ radius of earth
$ \begin{aligned} V_e & =\sqrt{\frac{2 G M}{R^{2}}} \\ & =\sqrt{\frac{2 \times G}{R^{2}} \times \frac{4}{3} \pi R^{3} \rho} \quad(\because M=\frac{4}{3} \pi R^{3} \rho) \\ V_e & =R \sqrt{\frac{8 \pi G \rho}{3}} \end{aligned} $
Thus, value of escape velocity depends upon the mass $M$, density $\rho$ and radius $R$ of earth/planet from which the body is projected and independent of angle of projection.
