Question

Let $F$ be the gravitational force between two objects. If the masses of the two objects are halved without changing the distance between them, find the gravitational force between them.
A) $\dfrac{F}{4}$
B) $\dfrac{F}{2}$
C) $F$
D) $\dfrac{2}{F}$

Answer 1

The gravitational force refers to the force of attraction between two objects that have mass. It is proportional to the product of the masses of the two objects and inversely proportional to the square of the distance of separation between them.

Formula used:
-The gravitational force between two objects of masses $m$ and $M$ is given by, $F = \dfrac{{GMm}}{{{r^2}}}$ where $G$ is the gravitational constant and $r$ is the distance of separation between them.

Complete step by step answer.
Step 1: List the given data.
We have two cases. Case 1 is when two objects of masses $m$ and $M$ have a gravitational force $F$ and case 2 is when the masses of the objects get halved.
Let $r$ be the distance of separation between them.
Step 2: Express the gravitational force between the two objects in the first case.
In the first case, the masses of the two objects are denoted by $m$ and $M$ , $r$ is the distance of separation between them.
Now the gravitational force between them is given by, $F = \dfrac{{GMm}}{{{r^2}}}$ -----------(1)
Here, $G$ is the gravitational constant.
Step 3: Express the gravitational force between the two objects in the second case.
In the second case, the masses of the two objects get halved. So their mass will be $\dfrac{m}{2}$ and $\dfrac{M}{2}$ , but the distance of separation between them hasn’t changed i.e., $r$ remains the same.
Let $F'$ be the new gravitational force between the two objects.
Then, $F' = \dfrac{{G\left( {\dfrac{M}{2}} \right)\left( {\dfrac{m}{2}} \right)}}{{{r^2}}}$ --------- (2)
Here, $G$ is the gravitational constant.
Step 4: Express the new gravitational force $(F')$ between the two objects in terms of the gravitational force in the first case $(F)$ .
Equation (1) gives $F = \dfrac{{GMm}}{{{r^2}}}$ and equation (2) gives $F' = \dfrac{{G\left( {\dfrac{M}{2}} \right)\left( {\dfrac{m}{2}} \right)}}{{{r^2}}}$
Now let’s simplify equation (2) to get, $F' = \dfrac{{GMm}}{{4{r^2}}}$ -------- (3).
Using equation (1) we can replace $\dfrac{{GMm}}{{{r^2}}}$ in (3) by $F$ to express $F'$ in terms of $F$ .
Then we have, $F' = \dfrac{{GMm}}{{4{r^2}}} = \dfrac{F}{4}$

Therefore, when the masses of the two objects get halved the gravitational force between them will be $\dfrac{F}{4}$ .

Note: The gravitational force is a force of attraction. It constantly tries to pull the masses together. It never repels or pushes them apart. So we can say the force is directed along $- r$ . Its strength is high if the objects are massive. Here, the masses of the objects were reduced and thus, the gravitational force between them also got reduced.