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Last updated date: 02nd Dec 2023
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# The gravitational force between two objects is $F$. How will this force change when (i) distance between them is reduced to half? (ii) the mass of each object is quadrupled ?

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Hint:We are surrounded by gravitational force. It determines how much we weigh and how far a basketball travels before returning to the ground when tossed. The force exerted by the Earth on you is equal to the force exerted by the Earth on you. The gravity force equals your weight when you're at rest on or near the Earth's surface. The acceleration of gravity on a particular planetary body, such as Venus or the Moon, is different than on Earth, because if you stood on a scale, it would show you weighing a different volume than on Earth.

Any particle in the universe attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their poles, according to Newton's law of universal gravitation. The theory's publication was dubbed the "first great convergence" because it brought together previously established gravity phenomena on Earth with known astronomical activities.We know that $F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$.

(i) The gravitational force $F$ between two objects separated by $r$, according to Newton's law of gravitation, is
$F \propto \dfrac{1}{{{r^2}}}$
When $r$ is halved
${r^\prime } = \dfrac{r}{2}$
Force becomes
$\dfrac{{{F^\prime }}}{F} = \dfrac{{{r^2}}}{{{r^{'2}}}} \Rightarrow \dfrac{{{F^\prime }}}{F}= \dfrac{{{r^2}}}{{{{(r/2)}^2}}} \Rightarrow \dfrac{{{F^\prime }}}{F}= 4$
$\therefore {F^\prime } = 4F$

(ii) The gravitational force F between two particles of mass and, according to Newton's law of gravitation, is
$F \propto {m_1}{m_2}$
${m_1}' = 4{m_1}{\rm{ and }}{m_2}' = 4{m_2}$
${F^\prime } \propto m_1^\prime m_2^\prime$
$\Rightarrow \dfrac{{{F^\prime }}}{F} = \dfrac{{m_1^\prime {m^\prime }2}}{{{m_1}{m_2}}} \\ \Rightarrow \dfrac{{{F^\prime }}}{F}= \dfrac{{\left( {4{m_1}} \right)\left( {4{m_2}} \right)}}{{{m_1}{m_2}}} \\ \Rightarrow \dfrac{{{F^\prime }}}{F}= 16$
$\therefore {F^\prime } = 16F$