Question

Two neutral bodies of masses $m_1$ and $m_2$ are kept at a distance of r cm from one another in a vacuum medium. A gravitational force F acts between the two bodies. The entire set-up is then transferred, as it is, to a water medium. The force between the bodies is then $F_w$. The ratio of F to $F_w$ is:

• A. $1:3$
• B. $1:1.33$
• C. $1:1$
• D. $3:2$

Hint:

Gravity is a universal force! It does not care whether objects are separated by air, water, glass, or solid lead blocks—the gravitational force between two fixed masses always remains exactly identical, giving a ratio of $1:1$.

Answer 1

The Correct Option is C

Concept: Newton's law of universal gravitation states that the gravitational attraction force operating between two massive bodies is dictated exclusively by their masses, their separation distance, and the universal gravitational constant ($G$): \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] Unlike electrostatic or magnetic forces, which depend strongly on the permittivity or permeability of the surrounding medium, the universal gravitational constant $G$ is a fundamental constant of nature. It is completely independent of the intervening medium between the objects.

Step 1: Analyze the force behavior across both environments.

• In a vacuum environment, the mutual gravitational force is $F = \frac{G \cdot m_1 \cdot m_2}{r^2}$.
• When submerged in a water environment, because mass values, separation distance, and the constant $G$ remain completely unchanged, the force is $F_w = \frac{G \cdot m_1 \cdot m_2}{r^2}$. Therefore, $F = F_w$.

Step 2: Compute the final force ratio.
\[ \text{Ratio} = \frac{F}{F_w} = \frac{1}{1} \implies 1:1 \]