Question

The gravitational force between two particles is F, if the objects are stationary and separated by a distance of 1m. If the object starts moving in opposite directions, from rest with uniform acceleration of $a = 1m/{s^2}$, then the force between them after 3 seconds will be
A. $\dfrac{{4F}}{{81}}$
B. $\dfrac{{F}}{{100}}$
C. $\dfrac{{4F}}{{121}}$
D.$\dfrac{{F}}{{2}}$

Answer 1

In the problem, two particles are given that are separated by certain distance. The gravitational force between two particles is attractive and can be given by Newton’s Law of Universal Gravitation. When these two particles move apart in opposite directions, the gravitational force between them will decrease as it inversely depends on the distance between two particles.

Formula Used:
Newton’s Law of Universal Gravitation is given as:
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
where, $G$ is the gravitational constant, $F$ is the gravitational force , ${m_1}$ is the mass of particle 1, ${m_2}$ is the mass of particle 2 and $r$ is the distance between two particles.
Distance between two particles (if travelling in uniform acceleration):
$r' = \dfrac{1}{2}a{t^2}$
$r'$ is the distance between two particles (after uniform acceleration), $a$ is uniform acceleration and $t$ is the time period.

Complete step by step answer:
Two particles are given with a gravitational force $F$acting between them. Let the mass of these two particles be ${m_1}$and ${m_2}$respectively. They are stationary and the distance between them is $1m$. Let the distance be denoted as $r$. Therefore, $r = 1m$.
By Newton’s Law of Universal Gravitation, every object having a mass attracts every other object by a force $F$ that is directly proportional to its masses ${m_1}$ and ${m_2}$and inversely proportional to the square of the distance $r$between them. Thus, Force $F$is written as
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
where, $G$ is the gravitational constant.
Substituting the value $r = 1m$. Therefore,

\Rightarrow F = G{m_1}{m_2}$$ $$ \to (1)$$ Now, these particles start moving in opposite directions, from rest with uniform acceleration of $$a = 1m/{s^2}$$ . The distance between them after 3 seconds can be calculated by the following formula. Let $$t = 3s$$. Let $$r'$$ be the distance between these two particles. Then, $$r'$$is given as $$r' = \dfrac{1}{2}a{t^2}$$ where, $$a$$ is the acceleration and $$t$$ is the time. Thus, $$r' = \dfrac{1}{2}a{t^2} \\\ \Rightarrow r' = \dfrac{1}{2}\left( 1 \right){\left( 3 \right)^2} \\\ \Rightarrow r' = \dfrac{9}{2}$$ $$ \to (2)$$ Let the gravitational force after they move a distance apart in 3 seconds be $$F'$$. Then, $$F'$$ is given as $$F' = \dfrac{{G{m_1}{m_2}}}{{{{r'}^2}}}$$ Substituting the value of $$r'$$from equation (2) $$F' = \dfrac{{G{m_1}{m_2}}}{{{{\left( {\dfrac{9}{2}} \right)}^2}}} \\\ \Rightarrow F' = \dfrac{{G{m_1}{m_2}}}{{\left( {\dfrac{{81}}{4}} \right)}} \\\ \Rightarrow F' = \dfrac{{4G{m_1}{m_2}}}{{81}}$$ $$ \to (3)$$ Equation (2) can also be written as $$G{m_1}{m_2} = \dfrac{{81F'}}{4}$$ $$ \to (4)$$ Equating equations (1) and (3) $$F = \dfrac{{81F'}}{4} \\\ \therefore F' = \dfrac{{4F}}{{81}}$$ **Hence, option A is the correct answer.** **Note:** Consider equations (1) and (2). If the particles move apart, the force and the distance between them will change. Note that the values of $$G$$ (the gravitational constant), $${m_1}$$(mass of particle 1) and $${m_2}$$(mass of particle 2) will remain unchanged. Greater the distance between two particles, lesser is the gravitational force acting within them. The gravitational force is always attractive.