# Location where gravitational force from Moon is 1/3 of that from Earth?

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Determine the location where an object with mass m0 when placed between the moon and earth experiences a gravitational force from the moon equal to a third of the gravitational force from earth. The distance from the earth to moon is 384,400 km. The ratio of mases between moon and earth is 0.0123. The radii of moon and earth are 1738.1 km and 6378.1 km, respectively.

edited Mar 6
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Suppose the object is placed at point P on the line EM joining the centres E, M of Earth and Moon. Let distance $PM=r$. Then distance $EP=R-r$ where $R=384,400km$ is the distance $EM$.

Suppose the mass of the Earth is $m$. Then the mass of the Moon is $\mu m$ where $\mu=0.0123$. The gravitational forces on an object of mass $m_0$ at point P from the Earth and Moon are respectively $$F_E=G\frac{mm_0}{(R-r)^2}$$ $$F_M=G\frac{\mu m m_0}{r^2}$$

We need to find a distance $r$ such that $$F_M=\frac13 F_E$$ $$G\frac{mm_0}{(R-r)^2}=\frac13 G\frac{\mu m m_0}{r^2}$$ Rearranging : $$\frac{3}{\mu}r^2= (R-r)^2$$ $$kr=R-r$$ $$r=\frac{R}{k+1}$$

Here $$k=\sqrt{\frac{3}{\mu}}=\sqrt{\frac{3}{0.0123}} \approx 15.617$$ hence $$r=\frac{384,400}{16.617} km\approx 23,100 km$$

2. This calculation assumes that P lies outside of the Moon and the Earth. Inside the Moon and Earth the $1/r^2$ law does not apply; instead the gravitational force is proportional to $r$. Another method would have to be used. So check that the calculated value of $r$ is greater than the radius of the Moon. It is. Phew.