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Question: Infinite number of masses, each 1 kg, are placed along the x -axis at x = ±1 m, ±2m, ±8 m, ±16 m... ...

Infinite number of masses, each 1 kg, are placed along the x -axis at x = ±1 m, ±2m, ±8 m, ±16 m... The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x = 0 ) is

A

-G/2

B

-G

C

-2 G

D

-4 G

Answer

-4 G

Explanation

Solution

The gravitational potential VV at a point due to a point mass mm at a distance rr is given by V=GmrV = -G \frac{m}{r}. Since gravitational potential is a scalar quantity, the resultant gravitational potential at a point is the algebraic sum of the potentials due to all individual masses.

The masses are placed at x=±1x = \pm 1 m, ±2\pm 2 m, ±8\pm 8 m, ±16\pm 16 m, ... . Each mass is m=1m = 1 kg. The distances of these masses from the origin (x=0x=0) are r=1,2,8,16,...r = 1, 2, 8, 16, ... meters. For each positive distance dd, there is a mass at +d+d and a mass at d-d.

The total gravitational potential at the origin is the sum of potentials due to all masses: Vtotal=all massesGmrV_{total} = \sum_{\text{all masses}} -G \frac{m}{r} Given m=1m=1 kg for all masses, this becomes: Vtotal=G(11+12+18+116+...)G(11+12+18+116+...)V_{total} = -G \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + ... \right) - G \left( \frac{1}{|-1|} + \frac{1}{|-2|} + \frac{1}{|-8|} + \frac{1}{|-16|} + ... \right) Vtotal=2G(11+12+18+116+...)V_{total} = -2G \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + ... \right)

The sequence of distances 1,2,8,16,...1, 2, 8, 16, ... suggests a pattern related to powers of 2. Assuming the intended sequence of distances is a geometric progression 20,21,22,23,24,...2^0, 2^1, 2^2, 2^3, 2^4, ..., i.e., 1,2,4,8,16,...1, 2, 4, 8, 16, .... The sum of the reciprocals of these distances is an infinite geometric series: S=11+12+14+18+116+...S = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ... This is a geometric series with the first term a=1a=1 and the common ratio r=1/2r=1/2. The sum of an infinite geometric series is given by S=a1rS = \frac{a}{1-r}. S=111/2=11/2=2S = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2.

Substituting this sum back into the expression for VtotalV_{total}: Vtotal=2G×S=2G×2=4GV_{total} = -2G \times S = -2G \times 2 = -4G.

The question asks for the "magnitude of the resultant gravitational potential". The calculated potential is 4G-4G. If "magnitude" implies the absolute value, it would be 4G=4G|-4G| = 4G. However, 4G4G is not an option, and all options are negative. This indicates that the question is likely asking for the value of the resultant gravitational potential itself, and the term "magnitude" might be used loosely or in the context of the negative sign being inherent to gravitational potential. Therefore, we choose the option that matches the calculated potential value.