Question

The force of attraction between two bodies at a
certain separation is 10 N. What will be the force
of attraction between them if the separation is
reduced to half ?​

Answer 1

To find :

The change in force of attraction.

solution :

Let the mass of two objects be $\rm{M_{1}}$ and $\rm{M_{2}}$ .

Let the distance between the two bodies be R.

We know about the Newton's law of gravitation i.e,

• The Gravitational between two bodies is directly proportional to the product of their masses.

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{F \propto m_{1}m_{2}}$

• The Gravitational force between two bodies is inversely proportional to the square of distance between them.

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{F \propto \dfrac{1}{r^{2}}}$

Now , by combining them we get :-

$\boxed{\bf{F_{1} \propto \dfrac{2M_{1}M_{2}}{r^{2}}}}$

Or/

$\boxed{:\implies \bf{F = G\dfrac{m_{1}m_{2}}{r^{2}}}}$

Where :-

• G = Universal Gravitational constant.
• m = Mass of the bodies.
• r = Distance between the two bodies.
• F = Force of Gravitation

Now to find the force of seperation between the two bodies we have to first the orginal Gravitational force between the bodies and the Gravitational force between the two bodies when distance is halved after that by dividing them we can obtain the required answer.

Gravitational force (Orginal) :-

Using the Equation for force of Gravitation and substituting the values (in terms of variables) , we get :-

$:\implies \bf{F = G\dfrac{m_{1}m_{2}}{r^{2}}}$

$:\implies \bf{F_{1} = G\dfrac{M_{1}M_{2}}{R^{2}}}$

$\boxed{\therefore \bf{F_{1} = G\dfrac{m_{1}m_{2}}{r^{2}}}}$

Hence, the Gravitational force between the two bodies is :-

$\bf{G\dfrac{M_{1}M_{2}}{R^{2}}}$

Gravitational force (When the distance is halved) :-

Using the Equation for force of Gravitation and substituting the values (in terms of variables) , we get :-

$:\implies \bf{F = G\dfrac{m_{1}m_{2}}{r^{2}}}$

$:\implies \bf{F_{2} = G\dfrac{M_{1}M_{2}}{\dfrac{1}{2}R^{2}}}$

$:\implies \bf{F_{2} = G\dfrac{M_{1}M_{2}}{R^{2}}}$

$\boxed{\therefore \bf{F_{1} = G\dfrac{2M_{1}M_{2}}{R^{2}}}}$

Hence, the Gravitational force between the two bodies is $\bf{G\dfrac{M_{1}M_{2}}{R^{2}}}$

Now by dividing $\bf{F_{1}}$ by $\bf{F_{2}}$, we get :-

$:\implies \bf{\dfrac{F_{1}}{F_{2}} = \dfrac{G\dfrac{M_{1}M_{2}}{\dfrac{1}{2}R^{2}}}{G\dfrac{2M_{1}M_{2}}{R^{2}}}}$

$:\implies \bf{\dfrac{F_{1}}{F_{2}} = G\dfrac{M_{1}M_{2}}{R^{2}} \times \dfrac{R_{2}}{2GM_{1}M_{2}}}$

$:\implies \bf{\dfrac{F_{1}}{F_{2}} = \not{G}\dfrac{\not{M_{1}}\not{M_{2}}}{\not{R^{2}}} \times \dfrac{\not{R_{2}}}{2\not{G}\not{M_{1}}\not{M_{2}}}}$

$:\implies \bf{\dfrac{F_{1}}{F_{2}} = 1 \times \dfrac{1}{2}}$

$:\implies \bf{\dfrac{F_{1}}{F_{2}} = \dfrac{1}{2}}$

$:\implies \bf{F_{1} = \dfrac{F_{2}}{2}}$

$:\implies \bf{2F_{1} = F_{2}}$

$\underline{\boxed{\bf{2F_{1} = F_{2}}}}$

Hence, we get that if the distance is halved then the force of Gravitation becomes 2 times of the orginal force.

Now , putting the value of $\bf{F_{1}}$ , we get :-

$:\implies \bf{2F_{1} = F_{2}}$

$:\implies \bf{2 \times 10 = F_{2}}$

$:\implies \bf{20 = F_{2}}$

$\underline{\bf{Force\:(F_{2}) = 20\:}}$

Hence, the Force of Gravitation when distance is reduced to half is 20 N