What happens to force between two masses if the distance between them is halved and masses are doubled?

Answer

Hint: Relation between gravitational force, mass and distance is,
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
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.
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Where G is the proportionality constant and it is universal constant.

(i) If the mass of an object is doubled:
$m{{'}_{1}}$ = ${{m}_{1}}$
$m'_{2}$ = $2{{m}_{2}}$
$F'=G\dfrac{{{m}_{1}}'{{m}_{2}}'}{{{\left( r{{'}^{{}}} \right)}^{2}}}$
$F'=G\dfrac{{{m}_{1}}\left( 2{{m}_{2}} \right)}{{{r}^{2}}}$
$F'=2\times G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$F'=2\times F$
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 $r'=2r$
Then $F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{r{{'}^{2}}}$
$F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{\left( 2r \right)}^{2}}}$
$F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{4{{r}^{2}}}$
$F'=\dfrac{G}{4}\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$F'=\dfrac{F}{4}$
When the distance between the objects is doubled then force between them is one fourth.
When $r'=3r$
Then $F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{\left( r' \right)}^{2}}}$
 $F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{\left( 3r \right)}^{2}}}$
 $F'=G\dfrac{{{m}_{1}}{{m}_{2}}}{9{{r}^{2}}}$
 $F'=\dfrac{F}{9}$
When the distance between the objects is tripled then force between them is one ninth.

(iii) The masses of both objects are doubled:
When $\begin{align}
& m{{'}_{1}}=2{{m}_{1}} \\
& m{{'}_{2}}=2{{m}_{2}} \\
\end{align}$
Then $F'=G\dfrac{m{{'}_{1}}m{{'}_{2}}}{{{r}^{2}}}$
$F'=G\dfrac{2{{m}_{1}}\times 2{{m}_{2}}}{{{r}^{2}}}$
$F'=4G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$F'=4F$
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.

What happens to the gravitational force between two objects if the mass of one object is doubled and the distance between them is also doubled?

This question was previously asked in

CDS 01/2022: GK Previous Paper (Held On 10 April 2022)

View all CDS Papers >

1. The force would remain the same
2. The force would be doubled
3. The force would be halved
4. The force would increase by a factor of 4

Answer (Detailed Solution Below)

Option 3 : The force would be halved

Free

Electric charges and coulomb's law (Basic)

10 Questions 10 Marks 10 Mins

CONCEPT:

• Newton's law of gravitation states that any two bodies having masses (m1 and m2) keeping at a distance (r) from each other exerts a Force of attraction on each other.
  • This force is directly proportional to the masses of bodies and inversely proportional to the square of the distance between them. 

\(\Rightarrow F \propto \frac{M_1M_2}{R^2} \Rightarrow F = \frac {GM_1M_2}{R^2} \)

Where G = 6.674 × 10-11 m3Kg-1s-2 is a universal constant

• The dimension of the force is MLT-2 and the SI unit is Newton (N).

CALCULATION:

Given: Mass of the one body (M1) = M, Mass of the another body (M2) = 2M, initial distance = R, Final distance = 2R,F = Initial Force, F' = Final force

M1 = M,  M2 = 2M, R' = 2R

\(\Rightarrow F \propto \frac{M_1M_2}{R^2}\)

\(F'=\frac{GM_1M_2}{R'^2} =\frac{G M_1(2M_2)}{(2R)^2}=\frac {2GM_1M_2}{4R^2}=\frac {1}{2}F\)

Hence, the Force would be halved is the correct answer.

Last updated on Sep 29, 2022

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