Question

With what velocity should a body be thrown up so that it rises to a height equal to the radius of the earth? [ $g = 10m/{s^2}$ at the surface]
A) $8000m/s$
B) $6400m/s$
C) $1600m/s$
D) $1000m/s$

Answer 1

Take the mass of the earth, $M$ and mass of the body, $m$ and the velocity of projection from the earth’s surface, $v$ and radius of the earth is $R$. Kinetic energy on the earth’s surface is $\dfrac{1}{2}m{v^2}$ and potential energy on the earth’s surface is $- \dfrac{{GMm}}{R}$. Final kinetic energy is $0$ at final height and potential energy is $- \dfrac{{GMm}}{{2R}}$. Equate both the total energies.

Formula used:
Kinetic energy on the earth’s surface is $\dfrac{1}{2}m{v^2}$ and potential energy on the earth’s surface is $- \dfrac{{GMm}}{R}$.

Complete step by step solution:
Initial distance of the body from the center of the earth is equal to the radius of the earth.
Let, the radius of the earth is $R$.
The final distance of the body from the center of the earth is $\left( {R + R} \right) = 2R$ (according to the question, it reaches a height equal to the radius of the earth).
Let, the mass of the earth is $M$ and mass of the body is $m$ and the velocity of projection from the earth’s surface is $v$.
Hence, its kinetic energy on the earth’s surface is $\dfrac{1}{2}m{v^2}$.
At a height $R$ above the earth’s surface, the body stops momentarily and then falls.
Hence, at that height $R$, kinetic energy is $0$.
Now, the potential energy of a body of mass $m$ due to a mass $M$ (mass of earth), at a distance $R$ (radius of the earth) from the center of the earth is $- \dfrac{{GMm}}{R}$ .
So, potential energy on the earth’s surface is $- \dfrac{{GMm}}{R}$ .
So, the total energy of the body on the earth’s surface is $\dfrac{1}{2}m{v^2} + \left( { - \dfrac{{GMm}}{R}} \right) = \dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{R}$ .
Again, potential energy at height $R$ is $- \dfrac{{GMm}}{{2R}}$ and its total energy at that height is $\left( {0 - \dfrac{{GMm}}{{2R}}} \right) = \dfrac{{GMm}}{{2R}}$
So, from the law of conservation of energy,
$\dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{R} = - \dfrac{{GMm}}{{2R}}$
or, $\dfrac{1}{2}m{v^2} = \dfrac{{GMm}}{R} - \dfrac{{GMm}}{{2R}} = \dfrac{{GMm}}{{2R}}$
or, $\dfrac{1}{2}m{v^2} = \dfrac{{GM}}{{{R^2}}} \times \dfrac{{mR}}{2} = g \times \dfrac{{mR}}{2} = \dfrac{1}{2}mgR$
or, ${v^2} = gR$
or, $v = \sqrt {gR} = \sqrt {10 \times 64 \times {{10}^7}} = \sqrt {64 \times {{10}^8}} = 8 \times {10^4}cm/s$ [Radius of the earth is $6400km$ ]
or, $v = 8000m/s$
So, the required velocity is $8000m/s$.
Option (A) is correct answer.

Note:
$g$ is the gravitational field intensity. Let, the mass of a body be $M$ and at any point at a distance $R$ from the center of the body, the gravitational field intensity due to the body is to be calculated.
If a body of unit mass is kept at that point, then from the law gravitation, we know that the force of gravitation is $\dfrac{{GM \times 1}}{{{R^2}}} = \dfrac{{GM}}{{{R^2}}}$. So, the gravitational field intensity at that point is $g = \dfrac{{GM}}{{{R^2}}}$.