Question

A ball is thrown upwards from earth’s surface with a speed $3{{v}_{e}}$, when the ball crosses earth’s gravitational field, then, (${{v}_{e}}$ is escape velocity)

  $\begin{align}
  & A)v'=2{{v}_{e}} \\
 & B)v'=\sqrt{7}{{v}_{e}} \\
 & C)v'=2\sqrt{2}{{v}_{e}} \\
 & D)v'={{v}_{e}} \\
\end{align}$

Answer 1

Hint: To solve this question, we will use the law of conservation of energy. We will calculate the total energy of the ball at earth surface which will be the sum of both kinetic and potential energy. This energy will be equal to the total energy at the limit of earth’s gravitational field. We must know that, at above earth’s gravitational field, the potential energy of every body will be zero. So, we will have energy at both points and solve for $v'$.

Formula used:
$\begin{align}
  & K.E=\dfrac{1}{2}m{{v}^{2}} \\
 & P.E=-\dfrac{GMm}{R} \\
 & {{v}_{e}}=\sqrt{\dfrac{2GM}{R}} \\
\end{align}$

Complete step by step answer:
We will calculate the total energy of the ball at the surface of earth. Total energy will be equal to sum of kinetic and potential energy.
$\begin{align}
  & T.E=K.E+P.E \\
 & T.E=\dfrac{1}{2}m{{\left( 3{{v}_{e}} \right)}^{2}}+\left( -\dfrac{GMm}{R} \right) \\
\end{align}$
There, the velocity of the ball at earth's surface is given as $3{{v}_{e}}$, where ${{v}_{e}}$ is the escape velocity and it is given by the formula,
${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$
Where, G is the universal gravitational constant.
              M is the mass of the earth.
              R is the radius of the earth.
We will substitute this equation in the expression for total energy.
\[\begin{align}
  & T.E=\dfrac{1}{2}m{{\left( 3{{v}_{e}} \right)}^{2}}+\left( -\dfrac{GMm}{R} \right) \\
 & \Rightarrow T.E=\dfrac{9}{2}m{{v}_{e}}^{2}+\left( -\dfrac{GMm}{R} \right)=\dfrac{\left( 8+1 \right)}{2}m{{v}_{e}}^{2}+\left( -\dfrac{GMm}{R} \right) \\
 & \Rightarrow T.E=\dfrac{1}{2}m{{v}_{e}}^{2}+\dfrac{8}{2}m{{v}_{e}}^{2}+\left( -\dfrac{GMm}{R} \right) \\
 & \Rightarrow T.E=\dfrac{1}{2}m{{\left( \sqrt{\dfrac{2GM}{R}} \right)}^{2}}+\left( -\dfrac{GMm}{R} \right)+\dfrac{8}{2}m{{v}_{e}}^{2} \\
 & \Rightarrow T.E=\dfrac{m}{2}\times \dfrac{2GM}{R}+\left( -\dfrac{GMm}{R} \right)+\dfrac{8}{2}m{{v}_{e}}^{2} \\
 & \Rightarrow T.E=\dfrac{GMm}{R}-\dfrac{GMm}{R}+\dfrac{8}{2}m{{v}_{e}}^{2} \\
 & \therefore T.E=4m{{v}_{e}}^{2} \\
\end{align}\]
So, we have got the total energy of the ball at the surface of earth as \[4m{{v}_{e}}^{2}\].
Now, at the above limit of the gravitational field, the gravitational potential energy will become zero. So, total energy will only comprise the kinetic energy of the ball. i.e.
$\begin{align}
  & T.E'=K.E'+P.E' \\
 & T.E'=\dfrac{1}{2}m{{\left( v' \right)}^{2}}+0 \\
 & T.E'=\dfrac{1}{2}m{{\left( v' \right)}^{2}} \\
\end{align}$
According to the law of conservation of energy, both these energies are equal. That is,
$\begin{align}
  & T.E=T.E' \\
 & \Rightarrow 4m{{v}_{e}}^{2}=\dfrac{1}{2}m{{\left( v' \right)}^{2}} \\
 & \Rightarrow {{\left( v' \right)}^{2}}=8{{v}_{e}}^{2} \\
 & \Rightarrow v'=\sqrt{8{{v}_{e}}^{2}} \\
 & \therefore v'=2\sqrt{2}{{v}_{e}} \\
\end{align}$
Therefore, the velocity of the ball when it crosses earth’s gravitational field is found to be $2\sqrt{2}{{v}_{e}}$. So, option C is correct.

Note:
We must know that the gravitational potential energy beyond the earth’s field limit is taken as zero because, the distance beyond that is considered as infinity while calculating the potential energy and anything divided by infinity will be equal to zero. Also, the escape velocity is the minimum velocity needed to escape from earth’s gravitational field.