Answer
2 gravitational forces balance each other when the distance between the spacecraft and the moon equals $3.84\times10^7m$ or that between the spacecraft and Earth equals $3.46\times10^8m$.
Work Step by Step
Let's take:
- the mass of the spacecraft $m$
- the mass of Earth $M_E$
- the mass of the moon $M_M$
- the distance between the spacecraft and Earth $r_{se}$
- the distance between the spacecraft and the moon $r_{sm}$
From the given information, $M_E=81.4M_M$ and $r_{se}+r_{sm}=3.85\times10^8m$ (the distance between Earth and the moon)
The gravitational force between the spacecraft and Earth: $$F_{se}=G\frac{mM_E}{r_{se}^2}$$
The gravitational force between the spacecraft and Earth: $$F_{sm}=G\frac{mM_M}{r_{sm}^2}$$
Now we need to find the point where $F_{se}=F_{sm}$. In other words, $$G\frac{mM_E}{r_{se}^2}=G\frac{mM_M}{r_{sm}^2}$$ $$\frac{M_E}{r_{se}^2}=\frac{M_M}{r_{sm}^2}$$ $$\frac{r_{se}^2}{r_{sm}^2}=\frac{M_E}{M_M}=81.4$$ $$\frac{r_{se}}{r_{sm}}=\sqrt{81.4}=9.02$$
As we already know $r_{se}+r_{sm}=3.85\times10^8m$, we can find either $r_{sm}=3.84\times10^7m$ or $r_{se}=3.46\times10^8m$
So $F_{se}=F_{sm}$ when the distance between the spacecraft and the moon equals $3.84\times10^7m$ or that between the spacecraft and Earth equals $3.46\times10^8m$
