Question

The escape velocity from earth is 11.2 km/s . If a body is to be projected in a direction making an angle of 45⁰ to the vertical , then find out the escape velocity.
A) $11.2 \times 2 km/s$
B) $ 11.2 km/s$
C) $ 11.2 \times \dfrac{1}{\sqrt{2}} km/s$
D) $ 11.2 \sqrt{2} km/s$

Answer 1

Hint
The lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object is called escape velocity.According to the question we can know that the escape velocity from earth is 11.2 km/s. And we will describe that if there is any change of velocity by changing the angle of the particle.

Complete answer:
The formula of escape velocity is
\[{V_e} = \sqrt {2gR} = \sqrt {\dfrac{{2GM}}{R}} \]
As we can see, angle is not mentioned in formula.
Here, $G$=gravitational force, $R$= radius
Escape velocity does not depend on the angle of projection. Escape velocity will remain the same.Hence, escape velocity is 11.2km/s .
Option (B) is correct.

Note
An object that has this velocity at the earth's surface will totally escape the earth's gravitational field ignoring the losses due to the atmosphere. For example, a spacecraft leaving the surface of Earth needs to go at 7 miles per second, or around 25,000 miles per hour to leave without falling back to the surface.