Question

If the body is projected with a velocity v greater than the escape velocity ${{v}_{e}}$from the surface of the earth, find its speed in interstellar space?

Answer 1

Hint: Escape velocity is the velocity with which an object has to be projected upwards from the surface of a planet in order to escape the gravitational attraction of the planet and move freely in space. So if a body is given a velocity $v$ which is greater than the escape velocity of the planet, then the object will use a part of its velocity as kinetic energy to break away from the gravitational field and the rest of the velocity is used for its movement in outer space.

Complete Step-by-Step solution:
Consider a planet of mass M and radius R. The gravitational potential energy of a body of mass m on the surface of the planet is given by,
$P.E=-\dfrac{GMm}{R}$
So the particle projected upwards should have kinetic energy that is greater or equal to the gravitational potential energy of the mass on the planet in order to escape to outer space. So, we can write
$\dfrac{GMm}{R}=\dfrac{1}{2}m{{v}^{2}}$
$\therefore {{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$ …. Equation(1)
Where ${{v}_{e}}$ is the escape velocity of the planet. G is the gravitational constant.
So when a body of mass is given a velocity $v$ which is greater than the escape velocity ${{v}_{e}}$, we can find the speed of the body in outer space by using the energy conservation theorem.
If the final velocity of the particle is ${{v}_{f}}$ in outer space. The kinetic applied to the body on the surface of the earth will be used to escape the gravitational field of earth, which is possible with the escape velocity and the rest of the kinetic is used to travel in outer space. So, we can write,
$K.E(initial)=K.E(escape)+K.E(final)$
$\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}m{{v}_{e}}^{2}+\dfrac{1}{2}m{{v}_{f}}^{2}$
$\Rightarrow {{v}_{f}}=\sqrt{{{v}^{2}}-{{v}_{e}}^{2}}$
From equation (1) we can substitute ${{v}_{e}}$ in the above equation,
$\therefore {{v}_{f}}=\sqrt{{{v}^{2}}-\dfrac{2GM}{R}}$
So the velocity of the body in outer space is ${{v}_{f}}=\sqrt{{{v}^{2}}-\dfrac{2GM}{R}}$.

Note: In this problem, we are considering that there is no air drag and there is no field other than the gravitational field of the earth.
If the velocity applied to the body is equal to the escape velocity, then the body will escape the gravitational force of earth but after that, it will remain stationary since all of its energy has been spent to tackle the gravitational potential energy of the planet.