Gravitational Interaction Between Infinite Rods

Classical Mechanics Level 5

Two very long (infinite) and uniform rods with identical linear mass density $\rho$ do not intersect. Their directions form an angle $\alpha$ and their shortest separation is $a$ . Find the force of attraction between them due to Newton's law of gravity.

Note:

This problem is not original.
Take $a = 1$ , $\alpha = \pi/4 \ rad$ .
For simplicity take the product $G\rho^2 = 1$ , where $G$ is the universal gravitational constant.

Bonus: Take $a = 10$ and recompute the result. What do you observe? Why is that so?

The answer is 8.88577.

1 solution

Steven Chase
Oct 26, 2019

I integrated over a length of $200$ units ( $100$ in each direction) for each rod. This yielded a result of $\approx 8.8$ for $a = 1$ . For $a = 10$ , the result is about $\approx 7.8$ . This is a fairly modest change, given that the separation has increased by a factor of ten. But this is not surprising, given that the rods are infinite in length. The smaller the value of $a$ , the higher the relative importance of the mass near the "center". As $a$ increases, the mass farther away from the center contributes more meaningfully, even as the central mass portions attract each other with less force. The result is that the overall attraction decreases only modestly with increasing separation.

import math

N = 10000

G = 1.0
rho = 1.0

C1x = 0.0
C1y = 0.0
C1z = 0.0

C2x = 0.0
C2y = 0.0
#C2z = 1.0
C2z = 10.0

u1x = 1.0
u1y = 0.0
u1z = 0.0

u2x = math.cos(math.pi/4.0)
u2y = math.sin(math.pi/4.0)
u2z = 0.0

############################################

dsigma1 = 200.0/N
dsigma2 = dsigma1

dm1 = dsigma1 * rho
dm2 = dsigma2 * rho

############################################

count = 0

Fx = 0.0
Fy = 0.0
Fz = 0.0

sigma1 = -100.0

while sigma1 <= 100.0:

    x1 = C1x + sigma1*u1x
    y1 = C1y + sigma1*u1y
    z1 = C1z + sigma1*u1z

    sigma2 = -100.0

    while sigma2 <= 100.0:

        x2 = C2x + sigma2*u2x
        y2 = C2y + sigma2*u2y
        z2 = C2z + sigma2*u2z

        Dx = x2 - x1
        Dy = y2 - y1
        Dz = z2 - z1

        D = math.sqrt(Dx**2.0 + Dy**2.0 + Dz**2.0)

        ux = Dx/D
        uy = Dy/D
        uz = Dz/D

        dF = G*dm1*dm2/(D**2.0)

        dFx = dF * ux
        dFy = dF * uy
        dFz = dF * uz

        Fx = Fx + dFx
        Fy = Fy + dFy
        Fz = Fz + dFz

        sigma2 = sigma2 + dsigma2

        count = count + 1

        if count%10**6 == 0:
            print count/10**6

    sigma1 = sigma1 + dsigma1

############################################

print ""
print ""

print Fx
print Fy
print Fz
print ""
Fmag = math.sqrt(Fx**2.0 + Fy**2.0 + Fz**2.0)

print Fmag
#approx 8.8 for a = 1
#approx 7.8 for a = 10

I did this in a different way. I evaluated a closed-form solution. First, I derived an expression for the gravitational field due to an infinite uniform rod at a point. Using that expression, I computed the force on an element length on the other wire, resolved it into components and integrated the resulting expression which came out to be:

$F_z = 2G\rho^2a\int_{-\infty}^{\infty}\frac{dx}{x^2\sin^2{\alpha} + a^2} = 2\sqrt{2}\pi \approx 8.88577$

So the result is actually independent of $a$ . I will post a detailed solution later if necessary. But the point of the bonus question was to illustrate that the result is independent of the separation.

Karan Chatrath - 1 year, 7 months ago

That's interesting. Yeah, some of the qualitative features of a problem can get lost when doing a brute numerical approximation. I guess part of the art of problem solving is knowing when certain methods will be strong and when they will be weak.

Steven Chase - 1 year, 7 months ago

Rightly said. Different problems warrant different approaches.

Karan Chatrath - 1 year, 7 months ago

@Karan Chatrath – @Karan Chatrath sir congratulations for 300 DS

A Former Brilliant Member - 1 year, 7 months ago

@A Former Brilliant Member – Thank you!

Karan Chatrath - 1 year, 7 months ago

Gravitational Interaction Between Infinite Rods

The answer is 8.88577.

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