Solveeit Logo

Question

Question: Infinite number of bodies, each of mass \(2\,kg\) , are situated on x-axis at distances \(1\,m\) ,...

Infinite number of bodies, each of mass 2kg2\,kg , are situated on x-axis at distances
1m1\,m , 2m2\,m , 4m4\,m , 8m8\,m ,……..respectively, from the origin. The resulting gravitational potential due to this system at the origin will be
A. G - G
B. 83G - \dfrac{8}{3}G
C. 43G - \dfrac{4}{3}G
D. 4G - 4G

Explanation

Solution

Gravitational potential is defined as the amount of work done to move a given object from the reference point to a given point. Here, there are an infinite number of bodies, therefore, the gravitational potential will be calculated infinite times. Therefore, we will add the gravitational potentials of different potentials upto infinity.

FORMULA USED:
The formula of gravitational potential is given by
V=GMdV = \dfrac{{ - GM}}{d}
Here, VV is the gravitational potential, GG is the gravitational constant, MM is the mass of the object and dd is the distance of objects.

COMPLETE STEP BY STEP ANSWER:
Consider an infinite number of bodies that are placed on x-axis. Each body will have a mass of 2kg2\,kg and they are placed at distances 1m1\,m , 2m2\,m , 4m4\,m , 8m8\,m ……. from the origin.

Now, the gravitational potential can be calculated by using the following formula
V=GMdV = \dfrac{{ - GM}}{d}
The gravitational potential of object at a distance of 1m1\,m from the origin is, V1=GM1{V_1} = \dfrac{{ - GM}}{1}
Also, the gravitational potential of object at a distance of 2m2\,m from the origin is, V2=\-GM2{V_2} = \dfrac{{ \- GM}}{2}
The gravitational potential of object at a distance of 4m4\,m from the origin is, V3=GM4{V_3} = \dfrac{{ - GM}}{4}

Now, the gravitational potential due to infinite number of bodies can be calculated as
V=V1+V2+V3+.........V = {V_1} + {V_2} + {V_3} + .........\infty
V=[GM1+GM2+GM4+.......]\Rightarrow \,V = - \left[ {\dfrac{{GM}}{1} + \dfrac{{GM}}{2} + \dfrac{{GM}}{4} + .......\infty } \right]

Now, the mass of each body is, M=2kgM = 2\,kg
V=[2G1+2G2+2G4+.......]\therefore \,\,\,V = - \left[ {\dfrac{{2G}}{1} + \dfrac{{2G}}{2} + \dfrac{{2G}}{4} + .......\infty } \right]

V=2G[G1+G2+G4+........] \Rightarrow \,V = - 2G\left[ {\dfrac{G}{1} + \dfrac{G}{2} + \dfrac{G}{4} + ........\infty } \right]
V=2G[11+12+122+.........]\Rightarrow \,V = - 2G\left[ {\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + .........\infty } \right]
Now, using binomial expansion (1x)1=1+x+x2+......{\left( {1 - x} \right)^{ - 1}} = 1 + x + {x^2} + ...... , we get
V=2G[112]1V = - 2G{\left[ {1 - \dfrac{1}{2}} \right]^{ - 1}}
V=2G(112)\Rightarrow \,V = \dfrac{{ - 2G}}{{\left( {1 - \dfrac{1}{2}} \right)}}
V=2G12\Rightarrow \,V = \dfrac{{ - 2G}}{{\dfrac{1}{2}}}
V=4G\Rightarrow \,V = - 4G

Therefore, gravitational potential due to the infinite number of bodies at the origin is 4G - 4G .
Hence, option (D) is the correct option.

NOTE: Here, gravitational potential is always taken as negative because the work done to bring an object to infinity is done by gravitational pull and not the gravitational potential. Therefore, the answer is also negative. Here, we want to bring an infinite number of bodies to the origin, that is why we have added all the gravitational potentials.