Question

The gravitation between two bodies is 6N. If the distance between the bodies is reduced by 1/5 what will the gravitational force between these bodies

Answer 1

Given that:

• The gravitation between two bodies is 6 N.
• The distance between the bodies is reduced to 1/5.

To Find:

• What will be the gravitational force between these bodies?

We know that:

• F = (Gm₁m₂)/r²

Where,

• F = Gravitational Force = 6 N
• G = Gravitation constant
• m₁ = Mass of first object
• m₂ = Mass of second object
• r = Distance between the object

When distance between the object is reduced to 1/5.

↠ F' = (Gm₁m₂)/(r/5)²

↠ F' = (Gm₁m₂)/(r²/25)

↠ F' = 25(Gm₁m₂)/r²

↠ F' = 25F [ ∴ F = (Gm₁m₂)/r²]

↠ F' = 25 × 6

↠ F' = 150

Hence,

• The gravitational force between these bodies will be 150 N.

Answer 2

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,

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

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,

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

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

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

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

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

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

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

We can write it as,

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

Now we know that,

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

Therefore,

$\\ :\implies\:\sf F' = 25\:\times\:F$

Putting value of gravitational force (F),

$\\ :\implies\:\sf F' = 25\:\times\:6$

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

• 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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