Particle 1 has mass m and is positioned in the region:
0 ≤ x 1 ≤ 1 0 ≤ y 1 ≤ 1 z 1 = 0
Particle 2 has mass m and is positioned in the region:
0 ≤ x 2 ≤ 1 0 ≤ y 2 ≤ 1 z 2 = 1
Parameters ( 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?
Details and Assumptions:
1)
Universal gravitational constant
G
=
1
2)
m
=
1
3)
Take the average of the scalar force magnitude
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@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/
Pretty hard problem!
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@Krishna Karthik Did you solve it with hand?
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Nope, I made a guess based on a bad numerical method XD
I don't expect anyone to solve it by hand. It's a quad integral.
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@Krishna Karthik – @Krishna Karthik Then i don't think the problem is hard.
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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.
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@Krishna Karthik – @Krishna Karthik i agree with you
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 and 1 0 1 2 loop iterations. By contrast, the Monte Carlo simulation method is very cheap. With just 1 0 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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@Steven Chase Sir post a problem electromechanics part 3 or RLC circuit. I want to practice daily problems.
I really like your classical mechanics problems; I'd like to see more of that!
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Thanks. You can check out my profile. There are many there
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That's true. I'm practicing your problems on dynamics of more complicated systems.
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Applying Newton's law of gravitation, the magnitude of the force between two particles is:
F ( x 1 , y 1 , x 2 , y 2 ) = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + 1 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 . 7 7 9 2 5 6
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 ) ) .