Question

Two objects, with masses m1 and m2 , are originally a distance r apart. The magnitude of the gravitational between is F. The masses are changed to 2m1 and 2m2 , and the distance is changed to 4r. What is the magnitude of the new gravitational force?

Answer 1

According to Newton's gravitational law, the magnitude of the gravitational force is:

$F=-G\frac{m_1m_2}{r^2}$

In this case, we have: $m_1'=2m_1,m_2'=2m_2, r'=4r$

Therefore, replacing this values and solving for F' in function of F:

$F'=-G\frac{m_1'm_2'}{r'^2}\\F'=-G\frac{2m_1(2m_2)}{(4r)^2}\\F'=-G\frac{4m_1m_2}{16r^2}\\F'=\frac{1}{4}(-G\frac{m_1m_2}{r^2})\\F'=\frac{1}{4}F$

Answer 2

Answer:

$\frac{1}{4}$ F

Explanation:

Newton's law of gravitation states that the force, F, that exists between the two objects of masses m₁ and m₂ is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance, r, between the bodies. i.e

F ∝ (m₁m₂ / r²)

F = Gm₁m₂/r²           ------------------(i)

Where;

G = proportionality constant called the gravitational constant.

Now, when the masses are changed to 2m₁ and 2m₂, and the distance is changed to 4r, then according to equation (i) the new force (say F₂) becomes;

F₂ = G (2m₁)(2m₂) / (4r)²

F₂ = 4Gm₁m₂ / 16r²

F₂ = Gm₁m₂ / 4r²

F₂ = $\frac{1}{4}$ Gm₁m₂/r²     -------------------------(ii)

Comparing equations (i) and (ii), equation (ii) can be re-written as;

F₂ = $\frac{1}{4}$ F

Therefore, the magnitude of the new gravitational force is a quarter of the old gravitational force, F. i.e   $\frac{1}{4}$ F