Question

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s.

(Part 1/2) Find the distance of this point from the center of the Earth.

The masses of the Earth and the Moon are 5.98 × 1024 kg and 7.36 × 1022 kg, respectively, and the distance from the Earth to the Moon is 3.84 × 108 m.

Answer in units of m.

(Part 2/2) What would the acceleration of the astronaut be due to the Earth’s gravity at this point if the moon was not there?

The value of the universal gravitational constant is 6.672 × 10−11 N · m^2/kg^2.

Answer in units of m/s^2.

Answer 1

Approximately
`3.46 * 10^(8)\; {\rm m}`, assuming that the astronauts are travelling along a straight line between the Earth and the Moon.

Approximately
`3.34 * 10^(-3)\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Let
`r` denote the distance (in meters) between the astronaut and the center of the Earth. Under the assumptions, the distance between the astronaut and the Moon would be
`(3.84 * 10^(8) - r)` meters.

Let
`G` denote the universal gravitational constant. Let
`m` denote the mass of the astronauts. Magnitude of the gravitational attraction from the Earth would be:

`\begin{aligned}\frac{G\, m\, M_\text{earth}}{r^(2)}\end{aligned}`.

Magnitude of the gravitational attraction from the Moon would be:

`\begin{aligned}\frac{G\, m\, M_\text{moon}}{(3.84 * 10^(8) - r)^(2)}\end{aligned}`.

Equate the two expressions and solve for the distance
`r`:

`\begin{aligned}\frac{G\, m\, M_\text{earth}}{r^(2)} = \frac{G\, m\, M_\text{moon}}{(3.84 * 10^(8) - r)^(2)}\end{aligned}`.

`\begin{aligned}((3.84 * 10^(8) - r)^(2))/(r^(2)) = \frac{G\, m\, M_\text{moon}}{G\, m\, M_\text{earth}}\end{aligned}`.

`\begin{aligned}(3.84 * 10^(8) - r)/(r) &= \sqrt{\frac{M_\text{moon}}{M_\text{earth}}} \\ &= \sqrt{\frac{5.98 * 10^(24)\; {\rm kg}}{7.36 * 10^(22)\; {\rm kg}}}\end{aligned}`.

`\begin{aligned}r &= \frac{3.84 * 10^(8)\; {\rm m}}{\displaystyle 1 + \sqrt{\frac{5.98 * 10^(24)\; {\rm kg}}{7.36 * 10^(22)\; {\rm kg}}}} \approx 3.45653* 10^(8)\; {\rm m}\end{aligned}`.

In other words, the distance between the astronaut and the center of the Earth should be approximately
`3.46 * 10^(8)\; {\rm m}`.

At
`r \approx 3.45653* 10^(8)\; {\rm m}` from the center of the Earth, the gravitational field strength of the Earth would be:

`\begin{aligned}g &= \frac{G\, M_{\text{earth}}}{r^(2)} \\ &\approx \frac{(6.672 * 10^(-11)\; {\rm N \cdot m^(2)\cdot kg^(-2)})\, (5.98 * 10^(24)\; {\rm kg})}{(3.45653 * 10^(8)\; {\rm m})^(2)} \\ &\approx 3.34 * 10^(-3)\; {\rm N\cdot kg^(-1)} = 3.34 * 10^(-3)\; {\rm m\cdot s^(-2)}\end{aligned}`.