Question

- What happen to the gravitational force between two object, if:
(i)The mass of one object is doubled?
(ii)The distance between the objects is doubled and tripled?
(iii)The masses of both objects are doubled?
Give reason in each case.

Answer 1

Hint: Relation between gravitational force, mass and distance is,
$F=G{\displaystyle \frac{m_1{m}_2}{r^2}}$
Where G is Newton’s gravitational constant
$m_1$ and $m_2$ are the masses
r is the distance.

Complete step by step solution:
Newton stated that in the universe each particle of matter attracts every other particle. This universal attractive force is called “Gravitational”.
Newton’s law:- Force of attraction between any two material particles is directly proportional to the product of masses of the particles and inversely proportional to the square of the distance between them. It acts along the line joining the particles.
$F\propto {\displaystyle \frac{m_1{m}_2}{r^2}}$
$F=G{\displaystyle \frac{m_1{m}_2}{r^2}}$
Where G is the proportionality constant and it is universal constant.

(i) If the mass of an object is doubled:
$m{{}^{\prime }}_1$ = $m_1$
$m_2^{\prime }$ = $2m_2$
$F^{\prime }=G{\displaystyle \frac{{m_1}^{\prime }{m_2}^{\prime }}{{\left(r{{}^{\prime }}^{}\right)}^2}}$
$F^{\prime }=G{\displaystyle \frac{m_1\left(2m_2\right)}{r^2}}$
$F^{\prime }=2\times G{\displaystyle \frac{m_1{m}_2}{r^2}}$
$F^{\prime }=2\times F$
When the mass of an object is doubled then the force between them is doubled.

(ii) The distance between object is doubled and tripled:
When $r^{\prime }=2r$
Then $F^{\prime }=G{\displaystyle \frac{m_1{m}_2}{r{{}^{\prime }}^2}}$
$F^{\prime }=G{\displaystyle \frac{m_1{m}_2}{{\left(2r\right)}^2}}$
$F^{\prime }=G{\displaystyle \frac{m_1{m}_2}{4r^2}}$
$F^{\prime }={\displaystyle \frac{G}{4}}{\displaystyle \frac{m_1{m}_2}{r^2}}$
$F^{\prime }={\displaystyle \frac{F}{4}}$
When the distance between the objects is doubled then force between them is one fourth.
When $r^{\prime }=3r$
Then $F^{\prime }=G{\displaystyle \frac{m_1{m}_2}{{\left(r^{\prime }\right)}^2}}$
 $F^{\prime }=G{\displaystyle \frac{m_1{m}_2}{{\left(3r\right)}^2}}$
 $F^{\prime }=G{\displaystyle \frac{m_1{m}_2}{9r^2}}$
 $F^{\prime }={\displaystyle \frac{F}{9}}$
When the distance between the objects is tripled then force between them is one ninth.

(iii) The masses of both objects are doubled:
When $\begin{array}{rl}& m{{}^{\prime }}_1=2m_1\\ & m{{}^{\prime }}_2=2m_2\end{array}$
Then $F^{\prime }=G{\displaystyle \frac{m{{}^{\prime }}_{1}m{{}^{\prime }}_2}{r^2}}$
$F^{\prime }=G{\displaystyle \frac{2m_1\times 2m_2}{r^2}}$
$F^{\prime }=4G{\displaystyle \frac{m_1{m}_2}{r^2}}$
$F^{\prime }=4F$
When the masses of both objects are doubled then the force between them is four times.

Note: This law is true for each particle of matter, each particle of matter attracts every other particle. Students should use the gravitational force formula carefully and write its term properly.