Question

Two objects, one 4 times as massive the other are approaching each other under their mutual gravitational attraction. When the separation between the objects is 100 km, the acceleration of the lighter object is $1{\text{ }}m/{s^2}$. When the separation between them is 25km, the acceleration of the heavier object is
A. $16m/{s^2}$
B. $2m/{s^2}$
C. $8m/{s^2}$
D. $4m/{s^2}$

Answer 1

Hint: Gravitational attraction is defined as the force of attraction between all the masses in the universe. To find the solution of the given question write down all the given physical quantities and apply the formula of Newton’s law of gravitation.

Formula used: $F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$

Complete step by step solution:
Newton’s law of gravitation states that every particle in the universe attracts every other particle with a force. The magnitude of the force is given as,
It is directly proportional to the product of their masses, i.e. $F \propto Mm$
And is inversely proportional to the square of the distance between their centre, i.e. $F \propto \dfrac{1}{{{r^2}}}$
On combining the above relations, we get,
$ \Rightarrow F \propto \dfrac{{{m_1}{m_2}}}{{{r^2}}}$
$ \Rightarrow F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
 Now, according to Newton’s law of gravitation
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where ‘F’ is expressed as the force, ‘G’ is the gravitational constant, ‘m’ and ‘M’ are the masses and ‘r’ is the distance between the masses.
Let us assume that the mass of the other object be ‘m’. i.e. ${m_1} = m$. It is given that the mass of one object is 4 times the mass of the other object. i.e. ${m_1} = 4{m_2}$.
r = 100km= 100000m
${a_1} = 1m/{s^2}$
Now, we know that force is given by the formula,
F = ma Where ‘F’ is the force, ‘m’ is the mass of the object or the particle and ‘a’ is the acceleration.
Equating the equation for the lighter object, we get,
$F = {m_1}a = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Now, substituting the given values we get,
$\eqalign{
  & \Rightarrow ma = \dfrac{{G4{m^2}}}{{{{\left( {100} \right)}^2} \times {{10}^6}}} \cr
  & \Rightarrow 4Gm = {10^{10}} \cr} $
Equating the equation for the heavier object, we get,
$\eqalign{
  & {F_2} = m{a_2} = \dfrac{{G4{m^2}}}{{{{\left( {25} \right)}^2} \times {{10}^6}}} \cr
  & \therefore {a_2} = \dfrac{{{{10}^{10}}}}{{{{\left( {25} \right)}^2} \times {{10}^6}}} = 16m/{s^2} \cr} $
Thus, when the separation between them is 25km, then the acceleration of the heavier object is $16m/{s^2}$.

Hence, option (A) is the correct answer.

Note:
The acceleration due to gravity ‘g' depends only upon the mass of the gravitating object and the distance from it. It does not depend upon the mass of the object which is being pulled.