Question

The gravitational force between two bodies separated by a distance is F what will be the gravitational force between the same bodies if distance between them is 4hrs

Answer 1

Answer:

The gravitational force will remain: F

Explanation:

The gravitational force is independent of time.

Henc, in no matter how the time changes, the gravitational force remains the same.

(Untill and unless the mass or the distance is changed)

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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