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

A

`-G`

B

`-(8//3)G`

C

`-(4//3)G`

D

`-4G`

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To solve the problem of finding the gravitational potential at the origin due to an infinite number of bodies each of mass \(2 \, \text{kg}\) situated on the x-axis at distances \(1 \, \text{m}, 2 \, \text{m}, 4 \, \text{m}, 8 \, \text{m}, \ldots\), we can follow these steps: ### Step 1: Understand the formula for gravitational potential The 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. ### Step 2: Identify the distances and masses The masses are \(m = 2 \, \text{kg}\) and the distances from the origin are \(r_n = 1, 2, 4, 8, \ldots\). These distances can be expressed as \(r_n = 2^{n-1}\) for \(n = 1, 2, 3, \ldots\). ### Step 3: Calculate the total gravitational potential at the origin The total gravitational potential at the origin due to all the masses can be expressed as: \[ V_{\text{total}} = \sum_{n=1}^{\infty} -\frac{G \cdot 2}{r_n} \] Substituting \(r_n\): \[ V_{\text{total}} = \sum_{n=1}^{\infty} -\frac{G \cdot 2}{2^{n-1}} \] ### Step 4: Factor out constants We can factor out the constants from the summation: \[ V_{\text{total}} = -2G \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \] ### Step 5: Recognize the summation as a geometric series The series \(\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}\) is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{2}\). The sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] Substituting the values: \[ S = \frac{1}{1 - \frac{1}{2}} = 2 \] ### Step 6: Substitute back into the potential equation Now substituting back into the potential equation: \[ V_{\text{total}} = -2G \cdot 2 = -4G \] ### Conclusion Thus, the resulting gravitational potential at the origin due to this infinite system of bodies is: \[ \boxed{-4G} \]

To solve the problem of finding the gravitational potential at the origin due to an infinite number of bodies each of mass \(2 \, \text{kg}\) situated on the x-axis at distances \(1 \, \text{m}, 2 \, \text{m}, 4 \, \text{m}, 8 \, \text{m}, \ldots\), we can follow these steps: ### Step 1: Understand the formula for gravitational potential The 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. ...
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