Answer to Question #305001 in Physics for Ogar Precious

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

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