Newtons Laws EX 40

A satellite on its way to the Moon is attracted towards the Earth with a gravitational force of $F_{1}$. The force gradually decreases as the distance, $x$, between the Earth and satellite increases. At the same time the Moon attracts the satellite with a gravitational force $F_{2}$ (which increases as the distance to the Moon decreases.) These two forces oppose one other (see the diagram below).

The distance between the Earth and the Moon is $3.810\times 10^{8}$ m. When the satellite is at a distance $x$ from the Earth then it is at a distance $([3.810\times 10^{8}]-x)$ from the Moon. We are looking for a distance $x$ at which $F_{1}+F_{2}=0$, which happens when the magnitudes of these forces are equal. The gravitational force $F_{1}$ depends on the mass $m_{E}$ of the Earth, the mass $m_{s}$ of the satellite, and the distance $x$ between them; the force $F_{2}$ depends on the mass $m_{M}$ of the Moon , the mass $m_{s}$ of the satellite, and the distance $([3.810\times 10^{8}]-x)$. We will now write the equation $F_{1}=F_{2}$ and replace both $F_{1}$ and $F_{2}$ by appropriate formulas in the Newton Law of gravitation.

${Gm_{E}m_{s} \over x^{2}}={Gm_{M}m_{s} \over ([3.810\times 10^{8}]-x)^{2}}$

Given that the Earth is 81 times more massive than the Moon, we can write: $m_{E}=81m_{M}$. Upon substitution into the above, we get:

${G(81m_{M})m_{s} \over x^{2}}={Gm_{M}m_{s} \over ([3.810\times 10^{8}]-x)^{2}}$

The constants $G$, $m_{M}$ and $m_{s}$ cancel out to leave us with:

${81 \over x^{2}}={1 \over ([3.810\times 10^{8}]-x)^{2}}$

By taking the positive square root on either side of this equation, we get

${9 \over x}={1 \over (3.810\times 10^{8})-x}$

and after cross multiplication and isolation of $x$, we see that $x=3.4\times 10^{8}$ m.