Answer to Question #306697 in Mechanics | Relativity for ED Angelo

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

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