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From Newton鈥檚 law of universal gravitation, it can be concluded that the force acting among two objects shows a direct relation with the masses of objects. Also, it varies inversely with the variation in the square of the distance between them.

The magnitude of the gravitational force is given by:

$F_{\mathrm{G}}=\frac{G{m}_1{m}_2}{r^2}$

Here, $G$ is the universal gravitational constant, $m_1$ and $m_2$are the masses of the objects and r is the distance between the objects.

This force is attractive and acts along the line joining the two objects.

• The mass of the Earth is, $m_1=5.97脳{10}^{24}\mathrm{kg}$.
• The mass of the Moon is, $m_2=7.35脳{10}^{22}\mathrm{kg}$.
• The distance between the centers of the Earth and the Moon is, $d=3.84脳{10}^8\mathrm{m}$.

Let a spacecraft of mass m travelling directly from the Earth to the Moon experience zero net force at some point P which is at a distance x from the Earth.

The gravitational force on the spacecraft due to the Earth is:

$F_1=G\frac{m{m}_1}{x^2}$

The direction of this force is along the line joining the centre of the Earth and the spacecraft, acting towards the Earth.

The gravitational force on the spacecraft due to the Moon is:

$F_2=G\frac{m{m}_2}{{(d-x)}^2}$

The direction of this force is along the line joining the centre of the Moon and the spacecraft, acting towards the Moon.

The net force on the spacecraft at point P will be zero when the gravitational force on the spacecraft due to the Earth will be equal to the gravitational force on the spacecraft due to the Moon, i.e.,

$\begin{array}{c}{F}_1=F_2\\ G\frac{m{m}_1}{x^2}=G\frac{m{m}_2}{{(d-x)}^2}\end{array}$

On rearranging the above expression, you will get the expression for distance xas:

$\begin{array}{c}{m}_1{\left(d-x\right)}^2=m_2{x}^2\\ \sqrt{m_1}\left(d-x\right)=\sqrt{m_2}x\\ \sqrt{m_1}d=\sqrt{m_2}x+\sqrt{m_1}x\\ x=d\frac{\sqrt{m_1}}{\sqrt{m_1}+\sqrt{m_2}}\end{array}$

On putting the given values, you will get:

$\begin{array}{c}x=\left(3.84脳{10}^8\mathrm{m}\right)\frac{\sqrt{5.97脳{10}^{24}\mathrm{kg}}}{\sqrt{7.35脳{10}^{22}\mathrm{kg}}+\sqrt{5.97脳{10}^{24}\mathrm{kg}}}\\ =3.46脳{10}^8\mathrm{m}\end{array}$

Thus, the point P is at a distance $3.46脳{10}^8\mathrm{m}$from the Earth.