At a particular instant the magnitude of the gravitational force exerted by a planet on one of its moons is $3\times {10}^{23}N$. If the mass of the moon were three times as large, what would be the magnitude of the force? If instead the distance between the moon and the planet were three times as large (no change in mass), what would be the magnitude of the force?

Gravitational Force on the moon$F=3\times {10}^{23}N$

Definition

The force attraction between every object in the universe is called gravitation force.

For example: throwing the ball in the air and the ball comes to the ground.

Gravitation force can be written as

$F=\frac{G{M}_1{M}_2}{R^2}.................................................\left(1\right)$

Where G is the universal gravitational constant, $M_1$is the mass of the moon, $M_2$is the mass of the planet, and $R$is the distance between moon and planet.

Now, $M"=3M,$

Where$M"$ is the new mass of the moon.

From equation no (1), we get,

$$\begin{gathered} F=\frac{G3M_1{M}_2}{R^2} \\ =3F \end{gathered}$$

Substituting the$3\times {10}^{23}N$ for F in the above equation,

$$\begin{gathered} F\text{'}=3\times 3\times {10}^{23}N \\ F\text{'}=9\times {10}^{23}N \end{gathered}$$

Thus, if the mass of the moon were three times as large, then the magnitude of the force is$9\times {10}^{23}N$.

Now for the distance between the moon and planet three times large,

$R"=3R$

From the equation no (1), we get,

$$\begin{gathered} F\text{'}=\frac{G{M}_1{M}_2}{{\left(3R\right)}^2} \\ =\frac{F}{9} \end{gathered}$$

Substituting the$3\times {10}^{23}N$ for F in the above equation

$$\begin{gathered} F\text{'}=\frac{3\times {10}^{23}N}{9} \\ =3.33\times {10}^{22}N \end{gathered}$$

Thus, if the distance between the moon and the planet were three times as large (no change in mass), then the magnitude of the force is$3.33\times {10}^{22}N.$