Question

Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1m, 2m, 4m, 8m, ...., respectively, from the origin. The resulting gravitational potential due to this system at the origin will be

• A. $ - 4 G $
• B. $ - G$
• C. $- \frac{8}{3} G$
• D. $ - \frac{4}{3} G$

Answer 1

The Correct Option is A

To find the gravitational potential due to an infinite number of masses along the x-axis, we can use the following approach:

1. Gravitational potential \( V \) at a point due to a mass \( m \) at a distance \( r \) is given by V = -\frac{Gm}{r}, where \( G \) is the gravitational constant.
2. In this problem, we have an infinite series of masses \( m = 2 \text{ kg} \) located at distances \( 1 \text{ m}, 2 \text{ m}, 4 \text{ m}, 8 \text{ m}, \ldots \) (following a geometric sequence) from the origin along the x-axis.
3. The potential \( V \) at the origin due to all these masses can be calculated by summing up the potentials due to each mass: V_{\text{total}} = -\frac{G \times 2}{1} - \frac{G \times 2}{2} - \frac{G \times 2}{4} - \frac{G \times 2}{8} - \ldots
4. This series can be simplified into: V_{\text{total}} = -2G \left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right)
5. This is a geometric series with the first term \( a = 1 \) and common ratio \( r = \frac{1}{2} \). The sum \( S \) of an infinite geometric series is given by \( S = \frac{a}{1-r} \), so: 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = \frac{1}{1-\frac{1}{2}} = 2
6. Therefore, plugging this result back into our expression for \( V_{\text{total}} \): V_{\text{total}} = -2G \times 2 = -4G

Thus, the resulting gravitational potential at the origin due to the infinite series of masses is -4G.

The correct answer is $ - 4 G $.