Question

The mass of a body is increased 4 fold and mass if another body is increased 16 fold. how should the distance betweenot them be changed to keep the same gravitational force between them?

Answer 1

Answer:

Let the masses be m1 and m2 and separation be "r"

So initial gravitational force

= (G × m1 × m2)/r²,

New masses are 4(m1) and 16(m2)

let the new separation distance be d

so gravitational force has to be same as initial gravitational force.

F1 = F2

=>(G×m1×m2)/r² = {G×4(m1)×16(m2)}/d²

=> d² = 64r²

=> d = 8r,

so the separation distance has to be made 8 times.

Answer 2

Question :

Given above ↑

Explanation :-

Le the give masses are "m" and "n" ,and distance between them is "r" ,

Apply the universal law of gravitation,for initial forces ,

$\rightarrow \sf{ f_{1} \: = g \: \frac{mn}{ {r}^{2} } } \: \: .....(1)$

According to the question, now masses

are 4m and 16n,let distance between these masses is "a",

Now force between them is ,

$\rightarrow \sf{ f_{2} \: = g \: \frac{(4m)(16n)}{ {a}^{2} } } \: \: ......(2)$

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•.• internal forces are same hence,

$\implies \sf{f_{1} \: = f_{2} \: } \\ \\ \sf{ from \: eq.(1) \: and \: eq.(2)} \\ \\ \rightarrow \sf{g \: \frac{mn}{ {r}^{2} } = g \: \frac{4m \times 16n}{ {a}^{2} } } \\ \\ \rightarrow \sf{\frac{1}{ {r}^{2} } = \frac{64}{ {a}^{2} } } \\ \\ \rightarrow \sf{ {a}^{2} = 64 {r}^{2} } \\ \\ \rightarrow \boxed{ \sf{ a \: = 8r}}$

hence distance is 8 times to second one.

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