Question

Two point masses at a given distance exert a gravitational force on each other having a magnitude F. If one mass is doubled, the other mass is halved and the distance between them is tripled, what will be the resultant force?​

Answer 1

Solution :-

• Given the magnitude of the initial Gravitational force between two bodies is F

Masses are, m₁ and m₂

let initial distance be 'r'

Now,

m₁ is doubled and m₂ is halved

• We get new masses as,

2m₁ and m₂/2

and distance is tripled of initial one,

So, we get, r = 3r

As we know according to Newton's law of gravitation,

${\rightarrow\:\:\boxed{\bf{F=G\:\frac{m_1m_2}{r^2}}}$

• $\bf{As,\: initial\:force \:between \:then \:is\: F}$

Let the final force be f.

${\rightarrow\:\:\ \bf{f=G\:\frac{2m_1*\frac{m_2}{2} }{(3r)^2}}$

${\rightarrow\:\:\ \bf{f=G\:\frac{m_1*\frac{m_2}{1} }{9(r)^2}}$

${\rightarrow\:\:\ \bf{f=G\:\frac{m_1{m_2} }{9r^2}}$

${\rightarrow\:\:\ \bf{f=G\:\frac{m_1{m_2} }{r^2}*\frac{1}{9} }$

$As,\:\sf{Initial \:Force\:between\:them\:is\:F\:and$  ${\:\:\ {\bf{F=G\:\frac{m_1m_2}{r^2}}}$

So, f = F/9

i.e. new resultant force is F/9

Final Answer :-

The resultant force is F/9

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