User talk:Amitavakala

Physics- Gravity
First I will start with a straight strip with negligible thickness and depth. It has length L and mass per unit length M. The other object has mass "m" and has negligible dimension. The other object lies at a distance R from the strip on a line perpendicular to the strip and going through the mid of the strip. Now I want to study the nature of gravitational force by the strip on the object "m". Let us consider an infinitesimal length dx on this strip at a distance x from mid point. The gravitational pull of this infinitesimal portion on the mass “m" will have two components. One is straight towards the strip and another is perpendicular to it. The second force gets cancelled out if we consider both sides of the mid point. Only the first one will add up. Now writing the equation here for the force which is directed towards the strip.

(GmMdx/(R²+ x²))*(R/√(R²+ x²)). Now integrating it over the half length (L/2) we get

∫(GmMdx/(R²+ x²))*(R/√(R²+ x²)) = (GmMR)∫dx/(R²+ x²)3/2

Considering x/R = tanθ, we get dx = Rdθ/cos2θ,

substituting and rearranging

(GmMR)∫dx/(R²+ x²)3/2 = (GmMR)*(1/R²)∫cosθdθ = (GmMR)*(1/R²)*sinθ =(GmM)*(1/R)*sinΦ,

considering θ changes from 0 to Φ over L/2, Now multiply and divide by L/2, we get Gm(Ṁ/2)/(RṘ), Where Ṁ = M*L And (L/2)/Ṙ= sinΦ, considering both sides of the mid point it becomes GmṀ/(RṘ). Now comparing with normal equation GmṀ/R², we can see GmṀ/R² > GmṀ/(RṘ), as Ṙ > R.

2nd case - Now the second mass "m" lies on the line of the strip but at some distance from the strip. In this case the integration is from R1 (nearest to "m") to R2 (farthest from "m") and very straightforward.

∫(GmMdx/x2 = -∫(GmM*(1/x) = ∫(GmM/(1/R1 - 1/R2)= GmM(R2 - R1)/(R1R2)= GmṀ/(R1R2)

and GmṀ/(R1R2) > GmṀ/R2, where R = (R1 + R2)/2