Question

How does the gravitational force between two bodies changes if the distance between them is tripled .explain ?

Answer 1

Given:

• The distance between two bodies is tripled.

To find:

• How the gravitational force changes?

Calculation:

The general expression of Newton's Law of Gravitational Force :

$F = \dfrac{Gm_{1}m_{2}}{ {r}^{2} }$

Now, the distance between the masses is tripled (i.e. 3r )

So, new force will be :

$F_{2}= \dfrac{Gm_{1}m_{2}}{ {(3r)}^{2} }$

$\implies F_{2}= \dfrac{Gm_{1}m_{2}}{9 {r}^{2} }$

$\implies F_{2}= \dfrac{1}{9} \times \dfrac{Gm_{1}m_{2}}{ {r}^{2} }$

$\implies F_{2}= \dfrac{1}{9} \times F$

So, the new gravitational force becomes ⅑ times the initial value.

Answer 2

Explanation:

Given:

The distance between two bodies is tripled.

To find:

How the gravitational force changes?

Calculation:

The general expression of Newton's Law of Gravitational Force :

F = \dfrac{Gm_{1}m_{2}}{ {r}^{2} }F=

r

2

Gm

1

m

2

Now, the distance between the masses is tripled (i.e. 3r )

So, new force will be :

F_{2}= \dfrac{Gm_{1}m_{2}}{ {(3r)}^{2} }F

2

=

(3r)

2

Gm

1

m

2

\implies F_{2}= \dfrac{Gm_{1}m_{2}}{9 {r}^{2} }⟹F

2

=

9r

2

Gm

1

m

2

\implies F_{2}= \dfrac{1}{9} \times \dfrac{Gm_{1}m_{2}}{ {r}^{2} }⟹F

2

=

9

1

×

r

2

Gm

1

m

2

\implies F_{2}= \dfrac{1}{9} \times F⟹F

2

=

9

1

×F

So, the new gravitational force becomes ⅑ times the initial value.