Question

A small block of superdense material has a mass of $3\times {{10}^{24}}kg$. It is situated at a height h (much smaller than the earth’s radius) from where it falls on the earth’s surface. Find its speed when its height from the earth’s surface has reduced to h/2. The mass of the earth is $6\times {{10}^{24}}kg$.

Answer 1

Hint:Use the law of conservation of mechanical energy and equate the initial and the final mechanical energy of the system. Then use the law of conservation of momentum because the net force on the system is zero.
Formula used:
$U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}$
$K=\dfrac{1}{2}m{{v}^{2}}$
$P=mv$

Complete answer:
Let us use the law of conservation of mechanical energy. The mechanical energy of a system is equal to the sum of the potential energy and kinetic energy of the system.
Therefore, the sum of the potential energy and the kinetic energy of the system when the block is at a height h and the sum of potential energy and the kinetic energy of the system when the block is at height h/2.
The potential energy of a system of two bodies is given as $U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}$, where ${{m}_{1}}$ and ${{m}_{2}}$ are the masses of the bodies. r is the distance between the centres of the two bodies.
The kinetic energy of a body is given as $K=\dfrac{1}{2}m{{v}^{2}}$, m and v are the mass and the speed of the body.
The potential energy of the given system when the block is at height h is ${{U}_{1}}=-\dfrac{G{{m}_{b}}{{m}_{e}}}{R+h}$.
At this time, the kinetic energy of both the bodies is zero.
The potential energy of the given system when the block is at height h/2 is ${{U}_{2}}=-\dfrac{G{{m}_{b}}{{m}_{e}}}{R+\dfrac{h}{2}}$.
At this time, the kinetic energies of both the bodies are ${{K}_{e}}=\dfrac{1}{2}{{m}_{e}}v_{e}^{2}$ and ${{K}_{b}}=\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$.
Therefore, the kinetic energy of the system is ${{K}_{2}}={{K}_{e}}+{{K}_{b}}$
 We know that ${{U}_{1}}+{{K}_{1}}={{U}_{2}}+{{K}_{2}}$.
$\Rightarrow -\dfrac{G{{m}_{b}}{{m}_{e}}}{R+h}+0=-\dfrac{G{{m}_{b}}{{m}_{e}}}{R+\dfrac{h}{2}}+{{K}_{e}}+{{K}_{b}}$
$\Rightarrow -\dfrac{G{{m}_{b}}{{m}_{e}}}{R+h}+0=-\dfrac{G{{m}_{b}}{{m}_{e}}}{R+\dfrac{h}{2}}+\dfrac{1}{2}{{m}_{e}}v_{e}^{2}+\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$ ….. (i).
The net force on the system of the earth and the block is zero. When the net force on a system is zero, the momentum of the system is conserved.
The momentum of a body is as P = mv.
The initial momentum of the system is zero. i.e. ${{P}_{i}}=0$
The final momentum of the system is ${{P}_{f}}={{P}_{e}}+{{P}_{b}}={{m}_{e}}{{v}_{e}}+{{m}_{b}}{{v}_{b}}$.
But, ${{P}_{i}}={{P}_{f}}$.
$\Rightarrow {{m}_{e}}{{v}_{e}}+{{m}_{b}}{{v}_{b}}=0$.
$\Rightarrow {{m}_{e}}{{v}_{e}}=-{{m}_{b}}{{v}_{b}}$.
$\Rightarrow {{v}_{e}}=-\dfrac{{{m}_{b}}{{v}_{b}}}{{{m}_{e}}}$.
If we only consider only the speed of the earth then,
 $\Rightarrow {{v}_{e}}=\dfrac{{{m}_{b}}{{v}_{b}}}{{{m}_{e}}}$
Substitute this value in (i).
$\Rightarrow -\dfrac{G{{m}_{b}}{{m}_{e}}}{R+h}=-\dfrac{G{{m}_{b}}{{m}_{e}}}{R+\dfrac{h}{2}}+\dfrac{1}{2}{{m}_{e}}{{\left( \dfrac{{{m}_{b}}{{v}_{b}}}{{{m}_{e}}} \right)}^{2}}+\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$
$\Rightarrow -\dfrac{G{{m}_{b}}{{m}_{e}}}{R+h}+\dfrac{G{{m}_{b}}{{m}_{e}}}{R+\dfrac{h}{2}}=\dfrac{m_{b}^{2}}{2{{m}_{e}}}v_{b}^{2}+\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$
$\Rightarrow -\dfrac{G{{m}_{b}}{{m}_{e}}}{R+h}+\dfrac{2G{{m}_{b}}{{m}_{e}}}{2R+h}=\left( \dfrac{{{m}_{b}}}{{{m}_{e}}}+1 \right)\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$
$\Rightarrow G{{m}_{b}}{{m}_{e}}\left( \dfrac{-1}{R+h}+\dfrac{2}{2R+h} \right)=\left( \dfrac{{{m}_{b}}}{{{m}_{e}}}+1 \right)\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$
$\Rightarrow G{{m}_{b}}{{m}_{e}}\left( \dfrac{-2R-h+2R+2h}{(R+h)(2R+h)} \right)=\left( \dfrac{{{m}_{b}}+{{m}_{e}}}{{{m}_{e}}} \right)\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$
$\Rightarrow G{{m}_{b}}{{m}_{e}}\left( \dfrac{h}{2{{R}^{2}}+3Rh+{{h}^{2}}} \right)=\left( \dfrac{{{m}_{b}}+{{m}_{e}}}{{{m}_{e}}} \right)\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$.
It is given that h <<<< R.
$\Rightarrow 2{{R}^{2}}+3Rh+{{h}^{2}}\approx 2{{R}^{2}}$
$\Rightarrow G{{m}_{b}}{{m}_{e}}\left( \dfrac{h}{2{{R}^{2}}} \right)=\left( \dfrac{{{m}_{b}}+{{m}_{e}}}{{{m}_{e}}} \right)\dfrac{1}{2}{{m}_{b}}v_{b}^{2}$
$\Rightarrow \dfrac{G{{m}_{e}}}{{{R}^{2}}}h=\left( \dfrac{{{m}_{b}}+{{m}_{e}}}{{{m}_{e}}} \right)v_{b}^{2}$
We know that $\dfrac{G{{m}_{e}}}{{{R}^{2}}}=g$, i.e. acceleration due to gravity of earth.
$\Rightarrow gh=\left( \dfrac{{{m}_{b}}}{{{m}_{e}}}+1 \right)v_{b}^{2}$
$\Rightarrow gh=\left( \dfrac{{{m}_{b}}}{{{m}_{e}}}+1 \right)v_{b}^{2}$
Substitute the masses of the earth and the block.
$\Rightarrow gh=\left( \dfrac{3\times {{10}^{24}}}{6\times {{10}^{24}}}+1 \right)v_{b}^{2}$
$\Rightarrow gh=\left( \dfrac{1}{2}+1 \right)v_{b}^{2}$
$\Rightarrow gh=\dfrac{3}{2}v_{b}^{2}$
$\Rightarrow v_{b}^{2}=\dfrac{2}{3}gh$
$\Rightarrow {{v}_{b}}=\sqrt{\dfrac{2gh}{3}}$.
This means that the speed of the block at the height h/2 is $\sqrt{\dfrac{2gh}{3}}$.

Note:
Usually when we get these kinds of questions, we only consider the kinetic energy of the block while calculating the mechanical energy of the system. This is because in these cases the mass of the block is much less than the mass of earth. Due to this reason, the displacement of earth is neglected. Therefore, it is considered to be at rest.
However, in the given question, the mass of the block is comparable to the mass of earth. Therefore, the movement of earth must be taken into consideration.