Question

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?

Short answer

If the mass of the moon were three times as large, then the magnitude of the force is$9\times {10}^{23}N.$ If instead, the distance between the moon and the planet were three times as large (no change in mass), then the magnitude of the force is $3.33\times {10}^{22}N.$

Step-by-step solution

Identification of given data

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

Gravitational force

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.

Step : Calculating gravitational force when the mass of the moon is three times large

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$9\times {10}^{23}N$.

Calculating gravitational force when the distance between the moon and planet is three times large

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$3.33\times {10}^{22}N.$