Question

What happens to the force of gravitation between two objects, if
(a). The mass of one object is tripled
(b). Distance between the object is halved
(c). Masses of both the objects got tripled
(d). The mass of one object is doubled and the mass of the other is tripled and the distance between them is halved.

Answer 1

Answer:

a)If the mass of one of the objects is tripled, then the force of gravity between them is tripled.

b)then the force of gravitation become 4 times

c)force become 9 times

d)the force become 24 times

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Answer 2

$\sf \bold {Formula\:used\::}$

$\sf \bullet F = \dfrac{G\times m_{1}\times m_{2}}{d^{2} }$

Here

$\bullet$ F = Gravitational force between two body

$\bullet$ G = gravitational constant

$\bullet$ $\sf m_{1}$ = mass of 1st body

$\bullet$ $\sf m_{2}$ = mass of 2nd body

$\bullet$ d = distance between them

$\sf \bold {Solution\::}$

(a). The mass of one object is tripled

Given, $\sf m_{1}'$ = 3 × $\sf m_{1}$

As other parameters are not changed

F is directly proportional to $\sf m_{1}$

$\sf \dfrac{F'}{F} =\: \dfrac{ m_{1}'}{m_{1}} \\\\\\\sf \implies \dfrac{F'}{F} =\: \dfrac{ 3m_{1}}{m_{1}}\\\\\\\sf \implies \dfrac{F'}{F} = 3\\\\\sf \implies \bold {F'\: =\:3F}$

(b). Distance between the object is halved

Given, d' = (1/2)d

As other parameters are not changed

F is indirectly proportional to d

$\sf \implies \dfrac{F'}{F} = \dfrac{d^{2}}{d'^{2}} \\\\\\\sf\implies \dfrac{F'}{F} = \dfrac{d^{2}}{(\dfrac{d}{2})^{2}}\\\\\\\sf \implies \dfrac{F'}{F} = \dfrac{d^{2}\times 2^{2}}{(d) ^{2}}\\\\\\\sf \implies \dfrac{F'}{F} = 2^{2}\\\\\sf \implies \bold {F' = \:4F}$

(c). Masses of both the objects got tripled

Given, $\sf m_{1}'$ = 3 × $\sf m_{1}$

$\sf m_{2}'$ = 3 × $\sf m_{2}$

As other parameters are not changed

F is directly proportional to $\sf m_{1} \times m_{2}$

$\sf \implies \dfrac{F'}{F} = \dfrac{m_{1}'\times m_{2}'}{m_{1}\times m_{2}}\\\\\\\sf \implies \dfrac{F'}{F} = \dfrac{3m_{1}\times 3m_{2}}{m_{1}\times m_{2}}\\\\\\\sf \implies \dfrac{F'}{F} = 3\times 3\\\\\sf \implies \bold {F'\:=\:9F}$

(d). The mass of one object is doubled and the mass of the other is tripled and the distance between them is halved.

Given,

$\sf m_{1}'$ = 2 × $\sf m_{1}$

$\sf m_{2}'$ = 3 × $\sf m_{2}$

d' = (1/2)d

F is directly proportional to $\sf \dfrac{m_{1}\times m_{2}}{d^{2}}$

$\sf \implies \dfrac{F'}{F} = \dfrac{\dfrac{m_{1}' \times m_{2}' }{d'^{2}} }{\dfrac{m_{1} \times m_{2} }{d^{2}} } \\\\\\\sf \implies \dfrac{F'}{F} = \dfrac{m_{1}' \times m_{2}' \times d^{2}}{m_{1} \times m_{2} \times d'^{2} }\\\\\\\sf \implies \dfrac{F'}{F} = \dfrac{2m_{1} \times 3m_{2} \times d^{2}}{m_{1} \times m_{2} \times (\dfrac{d}{2})^{2} }\\\\\sf \implies \dfrac{F'}{F} = 2 \times 3\times 2^{2}\\\\\sf \implies \bold {F'\:=\: 24F}$