
Infinite number of masses, each of 1kg, are placed along the x-axis at\[x = \pm 1m\],\[ \pm 2m\], \[ \pm 4m\],\[ \pm 8m\],\[ \pm 16m\]. Find the gravitational potential of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0).
A. \[\dfrac{G}{2}\]
B. G
C. 2G
D. 4G
E. 8G
Answer
232.8k+ views
Hint:Before we start addressing the problem let’s understand the gravitational potential. The gravitational potential at a location is equal to the work per unit mass that is needed to move an object to that location from a fixed reference location.
Formula Used:
To find the gravitational potential, we have,
\[V = - \dfrac{{Gm}}{r}\]
Where, G is gravitational constant, m is mass and r is the distance.
Complete step by step solution:
Consider an infinite number of masses, each of 1kg, that are placed along the x-axis at a distance of\[x = \pm 1m\],\[ \pm 2m\], \[ \pm 4m\],\[ \pm 8m\],\[ \pm 16m\]. We need to find the gravitational potential of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0).
When the x will be equal to zero at origin, then the resultant potential is given by,
\[V = 2\left[ { - \dfrac{{G \times 1}}{1} - \dfrac{{G \times 1}}{2} - \dfrac{{G \times 1}}{4}........} \right] \\ \]
\[\Rightarrow V = - 2G\left[ {1 + \dfrac{1}{2} - \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}}........} \right] \\ \]
\[\Rightarrow V = - 2G\left[ {\dfrac{1}{{1 - \dfrac{1}{2}}}} \right] \\ \]
\[\Rightarrow V = - 2G\left[ {\dfrac{1}{{\dfrac{1}{2}}}} \right] \\ \]
\[\Rightarrow V = - 2G\left[ 2 \right]\]
\[\therefore V = - 4G\]
If we take the magnitude of resultant gravitational potential then, the value is, \[V = 4G\]. Therefore, the gravitational potential of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0) is 4G.
Hence, option D is the correct answer.
Note:Gravitational potential energy is a type of potential energy and is equal to the product of the object's mass, the acceleration caused by gravity, and the object's height as the distance from the surface of the ground.
Formula Used:
To find the gravitational potential, we have,
\[V = - \dfrac{{Gm}}{r}\]
Where, G is gravitational constant, m is mass and r is the distance.
Complete step by step solution:
Consider an infinite number of masses, each of 1kg, that are placed along the x-axis at a distance of\[x = \pm 1m\],\[ \pm 2m\], \[ \pm 4m\],\[ \pm 8m\],\[ \pm 16m\]. We need to find the gravitational potential of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0).
When the x will be equal to zero at origin, then the resultant potential is given by,
\[V = 2\left[ { - \dfrac{{G \times 1}}{1} - \dfrac{{G \times 1}}{2} - \dfrac{{G \times 1}}{4}........} \right] \\ \]
\[\Rightarrow V = - 2G\left[ {1 + \dfrac{1}{2} - \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}}........} \right] \\ \]
\[\Rightarrow V = - 2G\left[ {\dfrac{1}{{1 - \dfrac{1}{2}}}} \right] \\ \]
\[\Rightarrow V = - 2G\left[ {\dfrac{1}{{\dfrac{1}{2}}}} \right] \\ \]
\[\Rightarrow V = - 2G\left[ 2 \right]\]
\[\therefore V = - 4G\]
If we take the magnitude of resultant gravitational potential then, the value is, \[V = 4G\]. Therefore, the gravitational potential of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0) is 4G.
Hence, option D is the correct answer.
Note:Gravitational potential energy is a type of potential energy and is equal to the product of the object's mass, the acceleration caused by gravity, and the object's height as the distance from the surface of the ground.
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