Two Rings Attracting (Part 2)

Classical Mechanics Level pending

There are two circular rings, each with 1 1 unit of mass per unit length.

Ring 1 1 has a radius of 1 1 and center ( x , y , z ) = ( 0 , 0 , 0 ) (x,y,z) = (0,0,0) . The vector ( 1 , 2 , 3 ) (1,-2,3) is normal to the plane of ring 1.
Ring 2 2 has a radius of 2 2 and center ( x , y , z ) = ( 1 , 2 , 3 ) (x,y,z) = (1,2,3) . The vector ( 5 , 4 , 1 ) (-5,4,1) is normal to the plane of ring 2.

What is the magnitude of the gravitational force exerted by one ring on the other?

Details and Assumptions:
1) Universal gravitational constant G = 1 G = 1 , for simplicity


The answer is 7.26.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Karan Chatrath
Nov 8, 2019

Attached simulation code (MATLAB) below. I will include an explanation involving analytical expressions maybe later, as well.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
clear all
clc

% Unit vectors characterising the larger ring:
R1 = 2;
k1 = [-5;4;1];
k1_hat = k1/norm(k1);
i1 = [1;1;7] - [1;2;3];
i1_hat = i1/norm(i1);
j1_hat = cross(k1_hat,i1_hat);

% Unit vectors characterising the smaller ring:
R2 = 1;
k2 = [1;-2;3];
k2_hat = k2/norm(k2);
i2 = [1;1;1/3] - [0;0;0];
i2_hat = i2/norm(i2);
j2_hat = cross(k2_hat,i2_hat);

Fx = 0;
Fy = 0;
Fz = 0;
dx = pi/5000;
dy = pi/5000;

for x = 0:dx:2*pi
    for y = 0:dy:2*pi

        % Parameterisation of the larger ring:
        r1 = [1;2;3] + R1*cos(x)*i1_hat + R1*sin(x)*j1_hat;

        % Parameterisation of the smaller ring:
        r2 = [0;0;0] + R2*cos(y)*i2_hat + R2*sin(y)*j2_hat;

        % Newton's Law of Gravitation:
        rd = r2 - r1;
        rm = sqrt(rd(1)^2 + rd(2)^2 + rd(3)^2);
        dF = (((R1*R2)/rm^3)*rd);

        % Extracting x,y,and z components
        dFx = dF(1);
        dFy = dF(2);
        dFz = dF(3);

        % Solving Double Integrals
        Fx  = Fx + dFx*dx*dy;
        Fy  = Fy + dFy*dx*dy;
        Fz  = Fz + dFz*dx*dy;
    end
end

Fr = norm([Fx;Fy;Fz])

% Fr = 7.262608239158482 

@Karan chatrath Sir please expain these problems using analytical expressions. Please

A Former Brilliant Member - 1 year, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...