Question

The distance between the earth and the moon is about $3.8\times {{10}^{5}}km$. At what points will the net gravitational force of the earth moon-system be zero? [Given the mass of the earth is 81 times the moon’s mass]. (Hint: Assume a body of unit mass at the null point).

Answer 1

Hint: This problem can be solved by finding out the individual gravitational forces on a unit mass at a specific point by the earth and the moon. At this point the gravitational forces due to both of them will cancel each other out. Since, gravitational force is an attractive force, the point has to lie in between the earth and the moon, so that the individual gravitational forces can act in opposite directions and cancel each other out.

Formula used:
${{F}_{g}}=\dfrac{GMm}{{{R}^{2}}}$
where ${{F}_{g}}$ is the magnitude of the gravitational force exerted by a body of mass $M$ on a body of mass $m$ where $R$ is the separation between their centers and G is the universal gravitational constant equal to $6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-1}}$.
This force attracts the mass $m$ towards the center of mass $M$.
When a body is in static equilibrium, summation of all forces $\overrightarrow{F}$ on the body will be zero
$\sum{\overrightarrow{F}}=0$

Complete step by step answer:
This problem can be solved by finding out the specific point in between the earth and the moon where the attractive gravitational forces exerted by the moon and the earth on a body of unit mass cancel each other out.
Since, gravitational force is an attractive force, the point has to lie in between the earth and the moon, so that the individual gravitational forces can act in opposite directions and cancel each other out.
The expression for gravitational force is given by
${{F}_{g}}=\dfrac{GMm}{{{R}^{2}}}$ --(1)
where ${{F}_{g}}$ is the magnitude of the gravitational force exerted by a body of mass $M$on a body of mass $m$where $R$is the separation between their centers and G is the universal gravitational constant equal to $6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-1}}$.
This force attracts the mass $m$ towards the center of mass $M$.
Now, let us analyze the question.
Let the mass of the earth be ${{M}_{E}}$.
Let the mass of the moon be ${{M}_{M}}$.
The total distance between the earth and the moon be $L$.
The distance of the required point form the earth on the same line joining the centers of the earth and the moon be $l$.
Therefore, the distance of this point from the moon will be $L-l$.
Now, we will consider a body of unit mass at this point. That is
$\text{Mass of body = 1kg}$
Now, using (1), the gravitational force exerted by the earth $\left( {{F}_{E}} \right)$ on the body will be,
${{F}_{E}}=\dfrac{G{{M}_{E}}\times 1}{{{l}^{2}}}=\dfrac{G{{M}_{E}}}{{{l}^{2}}}$ --(2)
This force attracts the body towards the center of the earth.
Now, using (1), the gravitational force exerted by the moon $\left( {{F}_{M}} \right)$ on the body will be,
${{F}_{M}}=\dfrac{G{{M}_{M}}\times 1}{{{\left( L-l \right)}^{2}}}=\dfrac{G{{M}_{M}}}{{{\left( L-l \right)}^{2}}}$ --(3)
This force attracts the body towards the center of the moon.
Now, since at this point, the two forces will oppose each other in direction and have the same magnitude, they will cancel each other out.
Hence, now we will equate their magnitudes.
Therefore, using (2) and (3)
${{F}_{E}}={{F}_{M}}$
$\Rightarrow \dfrac{G{{M}_{E}}}{{{l}^{2}}}=\dfrac{G{{M}_{M}}}{{{\left( L-l \right)}^{2}}}$
$\Rightarrow \dfrac{{{M}_{E}}}{{{l}^{2}}}=\dfrac{{{M}_{M}}}{{{\left( L-l \right)}^{2}}}$ --(4)
Now, according to the problem, the mass of the earth is 81 times the mass of the moon. Hence,
${{M}_{E}}=81{{M}_{M}}$ ---(5)
Also, the distance between the earth and the moon is given to be $3.8\times {{10}^{5}}km$.
Hence,
$L=3.8\times {{10}^{5}}km$--(6)
Putting (5) in (4), we get,
$\dfrac{81{{M}_{M}}}{{{l}^{2}}}=\dfrac{{{M}_{M}}}{{{\left( L-l \right)}^{2}}}$
$\therefore 81={{\left( \dfrac{l}{L-l} \right)}^{2}}$
Square rooting both sides we get,
$\sqrt{81}=\sqrt{{{\left( \dfrac{l}{L-l} \right)}^{2}}}$
$\Rightarrow 9=\dfrac{l}{L-l}$
$\Rightarrow 9\left( L-l \right)=l$
$\Rightarrow 9L-9l=l$
$\Rightarrow 9L=l+9l=10l$
$\Rightarrow \dfrac{9}{10}L=l$
Putting (6) in the above equation, we get
$l=\dfrac{9}{10}\times 3.8\times {{10}^{5}}km=3.42\times {{10}^{5}}km$
Hence, the required point lies at a distance of $3.42\times {{10}^{5}}km$from the earth on the line joining the centers of the moon and the earth.

Note: During the calculation when we are square rooting, we only take the positive root and not the negative one since it will give us the value of $l$ greater than $L$, that is the point will not be between the earth and the moon anymore. It will still be a point where the magnitudes of the forces by the moon and the earth will be the same, however they will have the same direction and thus add up. Thus, there will be a net attractive force on a body here. Thus, here the gravitational force will not be zero.
The same answer could be arrived upon by taking a body of any mass since the mass of the body will anyway cancel out during the calculations. However, assuming a unit mass avoids the introduction of an unnecessary variable in the calculation and makes it cleaner.