Question

It is known that Earth is 81 times the mass of the moon. Show that the point of weightlessness between the earth and the moon for a spacecraft occurs at 9/10 of the moon. HINT: Me/b²=Mm/( a-b)²

Answer 1

At the equilibrium point

$$G\frac{M_Mm}{b^2}=G\frac{M_Em}{(r_{EM}-b)^2}$$

$$r_{EM}/b-1=\sqrt{M_E/M_M}$$

$$r_{EM}/b-1=\sqrt{81}=9$$

Hence

$$b=\frac{1}{10}r_{EM}$$

Distance from the Earth

$$r=r_{EM}-b=\frac{9}{10}r_{EM}$$