Average Force (Singularity-Free)

Particle 1 1 has mass m m and is positioned in the region:

0 x 1 1 0 y 1 1 z 1 = 0 0 \leq x_1 \leq 1 \\ 0 \leq y_1 \leq 1 \\ z_1 = 0

Particle 2 2 has mass m m and is positioned in the region:

0 x 2 1 0 y 2 1 z 2 = 1 0 \leq x_2 \leq 1 \\ 0 \leq y_2 \leq 1 \\ z_2 = 1

Parameters ( x 1 , y 1 , x 2 , y 2 ) (x_1, y_1, x_2, y_2) are distributed randomly and uniformly within their ranges. What is the expected magnitude of the gravitational force between the particles?

Inspiration

Details and Assumptions:
1) Universal gravitational constant G = 1 G = 1
2) m = 1 m = 1
3) Take the average of the scalar force magnitude


The answer is 0.7793.

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2 solutions

Karan Chatrath
May 10, 2020

Applying Newton's law of gravitation, the magnitude of the force between two particles is:

F ( x 1 , y 1 , x 2 , y 2 ) = 1 ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 + 1 F(x_1,y_1,x_2,y_2) = \frac{1}{(x_2-x_1)^2 + (y_2-y_1)^2+1}

The coordinates of each particle are uniformly distributed and the expected value of force can be computed as such:

F a v = 0 1 0 1 0 1 0 1 F ( x 1 , y 1 , x 2 , y 2 ) d x 1 d y 1 d x 2 d y 2 0.779256 F_{av} = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} F(x_1,y_1,x_2,y_2) \ dx_1 \ dy_1 \ dx_2 \ dy_2 \approx 0.779256

Integral evaluated using Wolfram-Alpha. Using a script of code, a step size of 0.01 does not converge to a reasonably accurate result and decreasing the step size would significantly increase the time taken for computation ( O ( n 4 ) \mathrm{O}(n^4) ) .

@Karan Chatrath Nice. I just upvoted. Can you help me in this problem. In the solution part I have mentioned my difficulty in this problem https://brilliant.org/problems/circuit-for-practicing-numerical-solution/

A Former Brilliant Member - 1 year, 1 month ago

Pretty hard problem!

Krishna Karthik - 1 year, 1 month ago

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@Krishna Karthik Did you solve it with hand?

A Former Brilliant Member - 1 year, 1 month ago

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Nope, I made a guess based on a bad numerical method XD

Krishna Karthik - 1 year, 1 month ago

I don't expect anyone to solve it by hand. It's a quad integral.

Krishna Karthik - 1 year, 1 month ago

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@Krishna Karthik @Krishna Karthik Then i don't think the problem is hard.

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member The theory behind the problem is quite hard, I would easily rate it as a super-hard problem. Also the computation is another thing.

Krishna Karthik - 1 year, 1 month ago

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@Krishna Karthik @Krishna Karthik i agree with you

A Former Brilliant Member - 1 year, 1 month ago
Steven Chase
May 10, 2020

The proper way to solve this is with a quadruple integral, as shown in the solution by @Karan Chatrath . The downside, at least when solving numerically, is that a quadruple integral is very computationally expensive. To get a decent result, one would likely need between 1 0 8 10^8 and 1 0 12 10^{12} loop iterations. By contrast, the Monte Carlo simulation method is very cheap. With just 1 0 6 10^6 random trials, I can get a very good result. The Monte Carlo method requires the generation of random numbers with a uniform probability distribution. See code below:

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import math
import random

N = 10**6

Fav = 0.0

G = 1.0
m = 1.0

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

for j in range(0,N):

    x1 = random.uniform(0.0,1.0)
    y1 = random.uniform(0.0,1.0)
    z1 = 0.0

    x2 = random.uniform(0.0,1.0)
    y2 = random.uniform(0.0,1.0)
    z2 = 1.0

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

    Dsq = Dx**2.0 + Dy**2.0 + Dz**2.0

    F = G*m*m/Dsq

    Fav = Fav + F

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

Fav = Fav/N

print N
print Fav


#>>> 
#1000000
#0.779365934848
#>>> ================================ RESTART ================================
#>>> 
#1000000
#0.779299868788
#>>> ================================ RESTART ================================
#>>> 
#1000000
#0.779052645254
#>>> ================================ RESTART ================================
#>>> 
#1000000
#0.779407268843
#>>> 

@Steven Chase Sir post a problem electromechanics part 3 or RLC circuit. I want to practice daily problems.

A Former Brilliant Member - 1 year, 1 month ago

I really like your classical mechanics problems; I'd like to see more of that!

Krishna Karthik - 1 year ago

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Thanks. You can check out my profile. There are many there

Steven Chase - 1 year ago

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That's true. I'm practicing your problems on dynamics of more complicated systems.

Krishna Karthik - 1 year ago

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