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

Approximately
`5.19 * 10^(-5)\; \rm N`.

Step-by-step explanation:

Let
`G` denote the gravitational constant. (
`G \approx 6.67 * 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 = (G \cdot M \cdot m)/(r^(2))`.

In this question:
`M = 7 * 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* 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 * 10^(5)\; \rm km * (10^(3)\; \rm m)/(1\; \rm km) = 3 * 10^(8)\; \rm m`.

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