Question

when the mass of one of the two objects is doubled. Ffind the force of attraction and the relation between them

Answer 1

If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

The gravitational force between two objects is inversely proportional to the square of distance between them.

So, if the distance between two objects is doubled, then the gravitational force becomes

1/2×2

That is, 1/4.Therefore, if the distance between two masses be doubled, then the force between will becomes 1/4.

Answer 2

Given:

Two object having certain masses

To Find:

Relation between initial and final force of attraction between the two objects

Solution:

We know that,

• The force of attraction F between two bodies is given by

$\pink{\boxed{\bf{F=\dfrac{Gm_{1}m_{2}}{r^{2}}}}}$

where,

m₁ is the mass of first body

m₂ is the mass of second body

r is the distance between two bodies

G is gravitational constant

$\rule{190}{1}$

Let the mass of first object be m₁, mass of second object be m₂ and the distance between them be 'r'  

Also, let the initial force of attraction between two objects be F

So,

$\longrightarrow\rm{F=\dfrac{Gm_{1}m_{2}}{r^{2}}}------(1)$

$\rule{190}{1}$

On doubling the mass of the first object, new mass of first object will become 2m₁

Let the Final force of attraction between two objects be F'

So,

$\longrightarrow\rm{F'=\dfrac{G(2m_{1})(m_{2})}{r^{2}}}$

$\longrightarrow\rm{F'=\dfrac{2Gm_{1}m_{2}}{r^{2}}}$

$\longrightarrow\rm{F'=2\bigg(\dfrac{Gm_{1}m_{2}}{r^{2}}\bigg)}$

From (1), we get

$\longrightarrow\rm\green{F'=2F}$

Hence, new or final force of attraction is twice the old or initial force of attraction.