Question

If one of the mass is doubled and another mass is reduced to its one fourth value. Also the distance is increased by 4 times. What will be the new force between the object if initial force was F.​

Answer 1

Answer:

reduce by 32 times

Explanation:

F=Gm1m2/r*2

Answer 2

Answer:

The new force will be reduced by 32 times the initial force.

Step-by-step-explanation:

Let initially,

Mass of first object = m₁

Mass of second object = m₂

Distance between objects = d₁

We know that,

The gravitational force between two objects is given by

$\displaystyle{\boxed{\pink{\sf\:F\:=\:G\:\dfrac{m_1\:m_2}{d^2}\:}}}$

Initial force F will be

$\displaystyle{\sf\:F\:=\:G\:\dfrac{m_1\:m_2}{d_1^2}\:}$

Now,

New mass of first object ( M₁ ) = 2 * Initial mass

M₁ = 2m₁

New mass of second object ( M₂ ) = ( 1 / 4 ) * Initial mass

M₂ = m₂ / 4

New distance between two objects ( d₂ ) = 4 * d₁

d₂ = 4d₁

Now, the new gravitational force ( F' ) between the objects will be

$\displaystyle{\sf\:F'\:=\:G\:\dfrac{M1\:M2}{d_2^2}}$

$\displaystyle{\implies\sf\:F'\:=\:G\:\dfrac{\cancel{2}\:m_1\:\times\:\dfrac{m_2}{\cancel{4}}}{(\:4d_1\:)^2}}$

$\displaystyle{\implies\sf\:F'\:=\:G\:\dfrac{\dfrac{m_1\:m_2}{2}}{16\:d_1^2}}$

$\displaystyle{\implies\sf\:F'\:=\:G \: \dfrac{m_1\:m_2}{2\:\times\:16\:d_1^2}}$

$\displaystyle{\implies\sf\:F'\:=\:G\:\dfrac{m_1\:m_2}{32\:d_1^2}}$

$\displaystyle{\implies\sf\:F'\:=\:\dfrac{1}{32}\:\left(\:G\:\dfrac{m_1\:m_2}{d_1^2}\:\right)}$

$\displaystyle{\implies\sf\:F'\:=\:\dfrac{1}{32}\:\times\:F}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:F'\:=\:\dfrac{F}{32}}}}}$

∴ The new force will be reduced by 32 times the initial force.