Question

Two particles, one heavy and the other light, placed at $50\text{ cm}$ from each other, are under the influence of gravitational force of one another. If mass of the heavier particle is $4\text{ kg}$ and its acceleration under the influence of gravitational force is $5\times10^{-10}\text{ m s}^{-2}$, find the mass of the lighter particle.

• A. $7.843\text{ kg}$
• B. $1.8728\text{ kg}$
• C. $3.675\text{ kg}$
• D. $0.364\text{ kg}$

Hint:

The acceleration of any body due to gravitational attraction depends strictly on the mass of the other body pulling it, not its own mass! This is why the $4\text{ kg}$ value can be completely ignored, allowing you to solve directly for $m_2$ using $a = \frac{G m_2}{r^2}$.

Answer 1

The Correct Option is B

Understanding the Concept: According to Newton's Law of Universal Gravitation, the mutual attractive force ($F$) acting between two point masses $m_1$ and $m_2$ separated by a distance $r$ is: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] where $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$. The acceleration of the heavier body ($a_1$) is produced by this mutual force acting on its mass: $a_1 = \frac{F}{m_1}$.
Step 1: Set up the force equation for the acceleration of the heavier particle.
Let $m_1 = 4\text{ kg}$ (heavier mass) and $m_2$ be the unknown lighter mass. The acceleration experienced by $m_1$ is: \[ a_1 = \frac{F}{m_1} = \frac{\left(\frac{G \cdot m_1 \cdot m_2}{r^2}\right)}{m_1} = \frac{G \cdot m_2}{r^2} \] Notice that the mass of the heavier body ($m_1$) cancels out completely!
Step 2: Isolate and solve for the unknown lighter mass ($m_2$).
We are given:
Separation distance, $r = 50\text{ cm} = 0.5\text{ m}$
Acceleration, $a_1 = 5\times10^{-10}\text{ m s}^{-2}$
Rearranging the equation to isolate $m_2$: \[ m_2 = \frac{a_1 \cdot r^2}{G} \] Substitute the values into the equation: \[ m_2 = \frac{(5\times10^{-10}) \times (0.5)^2}{6.67 \times 10^{-11}} = \frac{5\times10^{-10} \times 0.25}{6.67 \times 10^{-11}} \] \[ m_2 = \frac{1.25 \times 10^{-10}}{6.67 \times 10^{-11}} = \frac{1.25}{6.67} \times 10^1 = 0.1874 \times 10 = 1.874\text{ kg} \] This matches option B within standard rounding limits: \[ m_2 \approx 1.8728\text{ kg} \]