Question

The escape velocity from earth is $ 11.2km/s $ . If a body is to be projected in a direction making an angle $ {45^0} $ to the vertical, then the escape velocity is
(A) $ 11.2 \times 2km/s $
(B) $ 11.2km/s $
(C) $ \dfrac{{11.2}}{{\sqrt 2 }}km/s $
(D) $ 11.2\sqrt 2 km/s $

Answer 1

Hint: Escape velocity (though a velocity) is a constant of a particular planet. Its value is directly proportional to the square root of the mass of the body and inversely related to the square root of the radius of the planet or gravitational body.

Formula used: In this solution we will be using the following formulae;
 $ {v_{esc}} = \sqrt {\dfrac{{2GM}}{R}} $ where $ {v_{esc}} $ is the velocity of escape of a particular planet or gravitational body, $ G $ is the universal gravitational constant, $ M $ is the mass of the gravitational body and $ R $ is the radius of the same body.

Complete Step-by-Step solution
We are told that the escape velocity of the earth is $ 11.2km/s $ . Now, the question is if a body is to be projected in a direction making an angle of $ {45^0} $ to the vertical, then the escape velocity will be what.
First, escape velocity is the velocity required to entirely escape the gravitational pull of a particular planet or star, etc. if a body is fired at escape velocity, both its potential energy and kinetic energy at a farthest point would be zero. Basically, it is to take the object to an infinite distance away from the body.
The escape velocity can generally be given by
 $ {v_{esc}} = \sqrt {\dfrac{{2GM}}{R}} $ where $ G $ is the universal gravitational constant, $ M $ is the mass of the gravitational body and $ R $ is the radius of the same body.
Thus, as seen from the above, there is no such thing as angle of projection, hence the escape velocity will remain $ 11.2km/s $ .
The correct option is B.

Note
For clarity escape velocity is entirely different from the velocity required to place an object in orbit. One major difference is that the velocity required to place an object in orbit is tangential hence depends on the angle of projection of the resultant velocity.