Question

The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls every kilogram of water in our rivers.

Answer 1

Answer:

Approximately $5.19 \times 10^{-5}\; \rm N$.

Explanation:

Let $G$ denote the gravitational constant. ($G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}$.)

Let $M$ and $m$ denote the mass of two objects separated by $r$.

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

$\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}$.

In this question: $M = 7 \times 10^{22}\; \rm kg$ is the mass of the moon, while $m = 1\; \rm kg$ is the mass of the water. The two are $r = 3\times 10^{5}\; \rm km$ apart from one another.

Important: convert the unit of $r$ to standard units (meters, not kilometers) to reflect the unit of the gravitational constant $G$.

$\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m$.

$\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}$.