Answer:
Approximately [tex]5.19 \times 10^{-5}\; \rm N[/tex].
Explanation:
Let [tex]G[/tex] denote the gravitational constant. ([tex]G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}[/tex].)
Let [tex]M[/tex] and [tex]m[/tex] denote the mass of two objects separated by [tex]r[/tex].
By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:
[tex]\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}[/tex].
In this question: [tex]M = 7 \times 10^{22}\; \rm kg[/tex] is the mass of the moon, while [tex]m = 1\; \rm kg[/tex] is the mass of the water. The two are [tex]r = 3\times 10^{5}\; \rm km[/tex] apart from one another.
Important: convert the unit of [tex]r[/tex] to standard units (meters, not kilometers) to reflect the unit of the gravitational constant [tex]G[/tex].
[tex]\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m[/tex].
[tex]\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}[/tex].
