*Notes to a video lecture on http://www.unizor.com*

__Gravity Integration 1__

*Determine the potential of the gravitational field of an infinitely thin solid rod at any point outside of it.*

Let's establish a system of coordinates with a rod and a point mass lying in the XY-plane with the rod on the X-axis with one end at point

*and another at point*

**A(a,0)***.*

**B(b,0)**Assume that the rod's length is

*and the mass is*

**L=b−a***, so the density of mass per unit of length is*

**M***.*

**ρ=M/L**Assume further that the coordinates of a point

*, where we want to calculate the gravitational potential, are*

**P***.*

**(p,q)**If, instead of a rod, we had a point mass

*concentrated in the midpoint of a rod at point*

**M***, its gravitational potential at a point*

**((a+b)/2,0)***would be*

**(p,q)**

**V**_{0}=G·M/rwhere

*is the distance between the midpoint of a rod and a point of measurement of gravitational potential*

**r***:*

**P***{*

**r =***[*

*]*

**(p−(a+b)/2***}*

^{2}+ q^{2}

^{1/2}Since the mass in our case is distributed along the rod, the gravitational potential will be different.

As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.

Therefore, to calculate a gravitational potential of a rod, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest

*and integrate all these potentials.*

**P**Consider a picture below (we recommend to save it locally to see in the bigger format).

As a variable, we will use an X-coordinate of a point on a rod

*and calculate the gravitational potential at point of interest*

**Q***from an infinitesimal segment of a rod of the length*

**P(p,q)***d*around point

**x***.*

**Q(x,0)**Knowing that, we will integrate the result by

*on a segment [*

**x***] to get the gravitational potential of the rod.*

**a;b**The infinitesimal segment of a rod

*d*, positioned around a point

**x***, has an infinitesimal mass*

**Q(x,0)***d*that can be calculated based on the total mass of a rod

**m***and its length*

**M***as*

**L=b−a***d*

**m = M·**d**x /L**The gravitational potential of this segment depends on its mass

*d*and its distance

**m***to a point of interest*

**r(x)***.*

**P(p,q)***d*

**V = G·**d**m /r(x)**Obviously,

*[*

**r(x) =***]*

**(p−x)**^{2}+q^{2}

^{1/2}Combining all this, the full gravitational potential of a rod [

*] of mass*

**a;b***at point*

**M***will then be*

**P(p,q)***{*

**V(p,q) = ∫**d_{a}^{b}G·**m /r(x) = ∫**d_{a}^{b}G·M·**x/***[*

**L·***]*

**(p−x)**^{2}+q^{2}*}*

^{1/2}We can use the known indefinite integral

**∫**d**t /√(t²+c²) = ln|t+√(t²+c²)|**Let's substitute in the integral for gravitational potential

*.*

**t=x−p**Then

*[*

**V(p,q) = ∫G·M·**d**t /L·***]*

**t**^{2}+q^{2}

^{1/2}where integration is from

*to*

**t=a−p***.*

**t=b−p***[*

**V(p,q) = (G·M/L)·***]*

**ln|b−p+√(b−p)²+q²| − ln|a−p+√(a−p)²+q²|**where

**L = b−a**Since the difference of logarithms is a logarithm of the result of division,

**V(p,q) = G·M·ln(R) /L**where

*and*

**L = b−a**

**R = |b−p+√(b−p)²+q²| / |a−p+√(a−p)²+q²|**