Question

the gravitational force between two objects is 10N find the force when one of the masses is doubled and the distance between them is reduced to 1/4 of the value​

Answer 1

Explanation:

Gravitational force = G m1 m2 / r^2

Thus, on doubling any one mass and making distance r to 4th part makes F to 32F

Answer 2

Given :-

• The gravitational force between two objects is 10N find the force when one of the masses is doubled and the distance between them is reduced to 1/4 of the value​

Solution :-

• First of all, we know that, the gravitational force between two bodies ,

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

      Where :-

    → F = Gravitational force between the bodies

    → m₁ and m₂ are the masses of the two objects

    → r  = distance between the objects

    → G = universal gravitational constant

• Now, in the first case, it is given that the gravitational force between two objects.

   ⇒ $\sf\ \frac{Gm_{1}m_{2}}{r^{2} }$ = 10    ------------------- (1)

• Now consider the second case. Here, one of the masses are doubled. Let us assume that m₁ is doubled (You can assume that m₂ is doubled too). Distance between them is reducted to 1/4 of the value. It means here r is r/4 . Now, let's substitute the values

   ⇒ F = $\sf\ \frac{G(2m_{1})m_{2}}{(\frac{r}{4}) ^{2} }$

   ⇒  F = $\sf\ \frac{2Gm_{1}m_{2}}{\frac{r^{2} }{4} }$

   ⇒ F = $\sf\ \frac{8Gm_{1}m_{2}}{r^{2} }$

    → But, as we got in the previous case , $\sf\ \frac{Gm_{1}m_{2}}{r^{2} }$ = 10

   ⇒ F = 8(10) = 80 N

∴ The gravitational force between them is 80N

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→ Sathvik :)