Question

Infinite number of bodies, each of mass 6kg, are situated at a distance 1m, 2m, 4m, 8m ........, from the origin on the y-axis. The resultant gravitational field intensity at the origin is ?

Answer 1

As you know, gravitational field 8s given by
Gravitational Field , G.F= GM/r² , where M is mass of body and r is the location of point where gravitational field act by body .
∴ G.F₁ = G(6)/1² = 6G/1²
G.F₂ = G(6)/(2)² = 6G/2²
G.F₃ = G(6)/4² = 6G/4²
............
....................

We know, gravitational field is a vector quantity , so resultant of it follow vector rule . G.Fnet = G.F₁ + G.F₂ + G.F₃ + G.F₄ + ......... ∞
G.Fnet = 6G/1² + 6G/2² + 6G/4² + .............. ∞
= 6G[1/1² + 1/2² + 1/3² + 1/4² + ....... ∞ ]
∵ 1/1² , 1/2² , 1/4² , 1/8² ..... Are in GP where a = 1 and common ratio , r = 1/4
so, sum of infinite terms = a/(1 - r)

Hence, G.Fnet = 6G × 1/(1 - r) = 6G/(1 - 1/4) = 8G

Hence, net resultant = 8G = 8 × 6.6 × 10⁻¹¹ N/kg
= 5.28 × 10⁻¹⁰ N/kg

Answer 2

Resultant gravitational field GF=GM/r^2
GF1=6G/1^2 =6G/1
GF2=6G/2^2=6G/4=6G/4^1
GF3=6G/4^2=6G/4^2
GF4=6G/8^2=6G/4^3
and so on
GF=GF1+GF2+GF3+GF4...upto infinity
GF=6G{1+1/4^1+1/4^2+1/4^3...upto infinity }
The stuff inside the bracket is GP with a=1 and r=1/4
sum of infinite terms of GP are=a/(1-r)
=1/(1-1/4)
therefore
GF=6G{1/(1-1/4)
GF=6G{4/3}
GF=8G