Answer:
Approximately [tex]3.46 \times 10^{8}\; {\rm m}[/tex], assuming that the astronauts are travelling along a straight line between the Earth and the Moon.
Approximately [tex]3.34 \times 10^{-3}\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Let [tex]r[/tex] 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 [tex](3.84 \times 10^{8} - r)[/tex] meters.
Let [tex]G[/tex] denote the universal gravitational constant. Let [tex]m[/tex] denote the mass of the astronauts. Magnitude of the gravitational attraction from the Earth would be:
[tex]\begin{aligned}\frac{G\, m\, M_\text{earth}}{r^{2}}\end{aligned}[/tex].
Magnitude of the gravitational attraction from the Moon would be:
[tex]\begin{aligned}\frac{G\, m\, M_\text{moon}}{(3.84 \times 10^{8} - r)^{2}}\end{aligned}[/tex].
Equate the two expressions and solve for the distance [tex]r[/tex]:
[tex]\begin{aligned}\frac{G\, m\, M_\text{earth}}{r^{2}} = \frac{G\, m\, M_\text{moon}}{(3.84 \times 10^{8} - r)^{2}}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{(3.84 \times 10^{8} - r)^{2}}{r^{2}} = \frac{G\, m\, M_\text{moon}}{G\, m\, M_\text{earth}}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{3.84 \times 10^{8} - r}{r} &= \sqrt{\frac{M_\text{moon}}{M_\text{earth}}} \\ &= \sqrt{\frac{5.98 \times 10^{24}\; {\rm kg}}{7.36 \times 10^{22}\; {\rm kg}}}\end{aligned}[/tex].
[tex]\begin{aligned}r &= \frac{3.84 \times 10^{8}\; {\rm m}}{\displaystyle 1 + \sqrt{\frac{5.98 \times 10^{24}\; {\rm kg}}{7.36 \times 10^{22}\; {\rm kg}}}} \approx 3.45653\times 10^{8}\; {\rm m}\end{aligned}[/tex].
In other words, the distance between the astronaut and the center of the Earth should be approximately [tex]3.46 \times 10^{8}\; {\rm m}[/tex].
At [tex]r \approx 3.45653\times 10^{8}\; {\rm m}[/tex] from the center of the Earth, the gravitational field strength of the Earth would be:
[tex]\begin{aligned}g &= \frac{G\, M_{\text{earth}}}{r^{2}} \\ &\approx \frac{(6.672 \times 10^{-11}\; {\rm N \cdot m^{2}\cdot kg^{-2}})\, (5.98 \times 10^{24}\; {\rm kg})}{(3.45653 \times 10^{8}\; {\rm m})^{2}} \\ &\approx 3.34 \times 10^{-3}\; {\rm N\cdot kg^{-1}} = 3.34 \times 10^{-3}\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
