Question

gravitational force between two bodies is 1 newton. if the distance between them is doubled, what will be the force between two bodies
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Answer 1

Explanation:

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Answer 2

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Answer :

If the distance between the bodies is reduced by 1/5 then gravitational force between these two bodies is 150N.

Step-by-step explanation :

Given :

The gravitation between two bodies is 6N

The distance between the bodies is reduced by 1/5

To Find :

Gravitational force between these two bodies when the distance between the bodies is reduced by 1/5?

Solution :

We know that gravitational force is given by,

$\hookrightarrow\:{\large{\underline{\boxed{\bf{F = \dfrac{Gm_{1}m_{2}}{r^2}}}}}}↪$

F= r 2 Gm 2 m 2

Where,F denotes gravitational force

G denotes gravitation constant

m₁ denotes mass of first body

m₂ denotes mass of second body

r denotes distance between bodies

We have,

Gravitational force (F) = 6N

Distance between bodies (r) = 1/5 of r as distance is reduced by 1/5)

Gravitational force after reducing distance (F') = ?

Putting all values,

$\begin{gathered}\\ :\implies\:\sf F' = \dfrac{Gm_{1}m_{2}}{\bigg(r\:\times\:\dfrac{1}{5}\bigg)^2}\end{gathered}$

:⟹F

′ = (r× 51 ) 2 Gm 1 m 2

$\begin{gathered}\\ :\implies\:\sf F' = \dfrac{Gm_{1}m_{2}}{\bigg(\dfrac{r\:\times\:1}{5}\bigg)^2}\end{gathered}$

:⟹F

′ =( 5r×1 ) 2 Gm 1 m 2

$\begin{gathered}\\ :\implies\:\sf F' = \dfrac{Gm_{1}m_{2}}{\bigg(\dfrac{r}{5}\bigg)^2}\end{gathered}$

:⟹F

′ = ( 5r ) 2Gm 1

$\begin{gathered}\\ :\implies\:\sf F' = \dfrac{Gm_{1}m_{2}}{\dfrac{r^2}{5^2}}\end{gathered}$

:⟹F

′ = 5 2 r2Gm 1 m 2

$\begin{gathered}\\ :\implies\:\sf F' = \dfrac{Gm_{1}m_{2}}{\dfrac{r^2}{25}}\end{gathered}$

:⟹F

′ = 25r 2

Gm 1 m 2

$\begin{gathered}\\ :\implies\:\sf F' = Gm_{1}m_{2}\:\div\:\dfrac{r^2}{25}\end{gathered}$

:⟹F

′ =Gm 1 m 2 ÷ 25r 2

$\begin{gathered}\\ :\implies\:\sf F' = Gm_{1}m_{2}\:\times\:\dfrac{25}{r^2}\end{gathered}$

:⟹F

′ =Gm 1 m 2 × r 225

We can write it as,

$\begin{gathered}\\ :\implies\:\sf F' = 25\:\times\:\dfrac{Gm_{1}m_{2}}{r^2}\end{gathered}$

:⟹F

′

=25× r 2

Gm 1 m 2

Now we know that,

$\begin{gathered}\\ \quad\qquad\bf \Bigg[F = \dfrac{Gm_{1}m_{2}}{r^2}\Bigg]\end{gathered}$

[F= r 2

Gm 1 m 2

Therefore,

$\begin{gathered}\\ :\implies\:\sf F' = 25\:\times\:F\end{gathered}$

:⟹F

′

=25×F

Putting value of gravitational force (F),

$\begin{gathered}\\ :\implies\:\sf F' = 25\:\times\:6\end{gathered}$

:⟹F

′

=25×6

$\begin{gathered}\\ :\implies\:{\large{\underline{\boxed{\bf{F' = 150N}}}}}\end{gathered}$

:⟹

F

′

=150N

Hence, if the distance between the bodies is reduced by 1/5 then gravitational force between these two bodies is 150N.

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