Question

Derive the expression for escape velocity of a projectile from Earth.

Answer 1

Hint: To solve this question we will understand about escape velocity. Escape velocity is the minimum velocity requisite by a body to be projected to overcome the gravitational pull of earth. It is the minimum velocity required by an object to escape the gravitational field that is, escape the land without falling back, ignoring the losses due to the atmosphere. By using this concept we can easily approach our answer.

Formula used:
$V = \sqrt {\dfrac{{2GM}}{R}} $
Where,
$V = $ Escape velocity
$G = $ Gravitational constant is $6.67408 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
$M = $ Mass of the planet
$R = $ Radius from the centre of gravity
Another expression for the escape velocity particularly useful at the surface on the body is ${V_{esc}} = g{R^2}$
Where $g$ is the acceleration due to gravity of earth. Therefore, escape velocity is also given by ${V_{esc}} = \sqrt {2gR} $ It is indicated in $m/s$ and escape velocity of earth is $11,200m/s$

Complete step-by-step solution:
Projectile motion is a type of motion, which is experienced through the object or by the particle which near the earth’s surface is projected, and under the action of gravity only it moves along a curved path (air resistance effects are assumed negligible).
The minimum speed required for the free non-propelled object which can escape from the massive body’s gravitational influence to achieve the infinite distance is known as escape velocity. Escape velocity is the function of the body’s mass and the distance to the center of the body’s mass. From the earth’s surface, the escape velocity is about 11,186 m/s.
Here is the derivation for the escape velocity of a projectile from Earth.
When it overcomes the gravitational binding energy due to the earth then the particles escape from the earth. It has zero energy in infinity.
Gravitational Binding Energy= \[\dfrac{{ - GMm}}{R}\]
With initial velocity ${v_{\sec }}$ it is projected so it escapes.
$ \Rightarrow KE = \dfrac{1}{2}m{v_{esc}}^2$ (KE=initial) (m: mass of body)
From the conservation of energy: -
$ - \dfrac{{GMm}}{R} + \dfrac{1}{2}m{v_{esc}}^2 = 0$
E=0 at infinity.$ \Rightarrow \dfrac{{GMm}}{R} = \dfrac{1}{2}m{v_{esc}}^2 \Rightarrow {v_{\sec }} = \sqrt {\dfrac{{2GM}}{R}} $
Also, as g, $g = \dfrac{{GM}}{{{R^2}}}$ we get,
$ \Rightarrow {v_{\sec }} = \sqrt {(\dfrac{{2GM}}{{{R^2}}}}. R) = \sqrt {2gR} $ (Req).
Hence, this is the derivation for the escape velocity of a projectile from the earth.

Note: We have seen rockets going into space leaving the earth. We also noticed that a very huge kick-start is required to leave the surface of the earth because of the strong gravitational field. This is because of the escape velocity.