Question

The escape velocity for a body projected vertically upwards from the surface of the earth is 11 km/s. If the body is projected at an angle of ${45^ \circ }$ with the vertical, the escape velocity will be
A. $11\sqrt 2$ km/s
B. $22$ km/s
C. $11$ km/s
D. $\dfrac{{11}}{{\sqrt 2 }}$ km/s

Answer 1

To find the escape velocity of an object, it is important for the object to have zero gravitational force acting on the body such that the sum of kinetic energy and the potential energy is equal to zero.

Complete step by step solution:
Any object on the surface of the Earth is bound to the surface due to the gravitational pull exerted by the force on all objects present on the Earth’s surface. At any point above the Earth’s surface, there is presence of gravitational field, due to which at any point, the body will have a finite gravitational energy.

If a body is oriented towards space, and is ejected from the surface with a high velocity called the escape velocity, the body breaks free of the gravitational force and can escape the gravitational field of the Earth. This happens because at this velocity, the sum total of the gravitational potential energy and the kinetic energy due to the velocity is equal to zero.

At escape velocity,

$P + K = 0$

Potential energy of a body in the gravitational field at a height h from the Earth is –

$P = - \dfrac{{GMm}}{r}$

where r = h + R (radius of Earth)
Kinetic energy of the body,

$K = \dfrac{1}{2}m{v^2}$

Substituting, we get –
$\dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{r} = 0$
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{GMm}}{r}$
$\Rightarrow \dfrac{1}{2}{v^2} = \dfrac{{GM}}{r}$
$\therefore v = \sqrt {\dfrac{{2GM}}{r}}$

At this velocity, the sum of energies is zero and hence, the body can escape the surface of the Earth. Here, it can be observed that the velocity depends only on the magnitude of the velocity and not the direction of the projectile.
By the above equation, substituting the values of gravitational constant G, mass of Earth M and the radius of Earth r, we obtain the standard value of escape velocity of approximately 11 km/sec.
Hence, even if the angle is projected at an angle of ${45^ \circ }$ also, the magnitude must be equal to 11 km/sec in order to achieve the escape velocity.

Hence, the correct option is Option C.

Note: This is the reason that it is referred to, generally, as escape “speed” and not escape “velocity” because the direction is not important and only the magnitude. Hence, only the scalar quantity of speed is used, in general terms.