Two bodies of mass M each are placed R distance apart. In another syst

Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed $\frac{R}{2}$ distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be

16 F
1 F
4 F
None of the above
This question was previously asked in
UPSC NDA-2 – 2019
The correct answer is A) 16 F.
The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton’s Law of Gravitation: $F = G \frac{m_1 m_2}{R^2}$, where G is the gravitational constant.
In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G \frac{M \times M}{R^2} = G \frac{M^2}{R^2}$. This force is given as F. So, $F = G \frac{M^2}{R^2}$.
In the second system: $m_1′ = 2M$, $m_2′ = 2M$, $R_2 = R/2$. The force is $F_2 = G \frac{(2M) \times (2M)}{(R/2)^2}$.
Calculate $F_2$: $F_2 = G \frac{4M^2}{R^2/4} = G \frac{4M^2}{R^2} \times 4 = 16 G \frac{M^2}{R^2}$.
Substitute the expression for F: $F_2 = 16 F$.
The gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. Doubling the masses quadruples the product of masses ($2M \times 2M = 4M^2$). Halving the distance quarters the squared distance ($(R/2)^2 = R^2/4$), meaning the force is multiplied by 4 due to the inverse square law ($1 / (1/4) = 4$). The combined effect is a multiplication by $4 \times 4 = 16$.
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