Question

It is known that the mass of the earth is 81 times the mass of the moon.Show that the point of weightless between the earth and the moon for a spacecraft occur at 9/10 of the distance to the moon Hint: Me/b²=Mm/(a-b)²?

Answer 1

At the equilibrium point

$$G\frac{M_em}{b^2}=G\frac{M_mm}{(a-b)^2}$$

Hence

$$\frac{a}{b}-1=\sqrt{\frac{M_m}{M_e}}=9$$

$$\frac{a}{b}=10$$

$$b=1/10a$$

Here $a$ is the distance from Earth to the Moon.

Finally, the distance from earth to the equilibrium point is

$$a-b=a-1/10a=9/10a$$