# Problem no 41

3 gravitating bodies; find the work done by gravity.

What answer did you get for this question? And if you don't mind, can you pls share your method?

I'm getting $\displaystyle W = -\frac{3GM^2}{2l}$;

I integrated the force by the displacement; this results in this complicated integral:

$\large \displaystyle \int_{0}^{\frac{2}{\sqrt{3}} l} \frac{\sqrt{3}gm^2}{ \left(2l - \frac{2}{\sqrt{3}}r \right)^2 } dr$

That's 'cause I totally forgot about energy-work theorem lol

Note by Krishna Karthik
1 week, 4 days ago

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I just woke up.

- 1 week, 3 days ago

Very good! Btw I solved the problem using a really complicated integral lol. I integrated the force by the displacement along the path of the objects, and thereby got the final solution.

- 1 week, 3 days ago

@Krishna Karthik very nice.
The formula which I have used comes from your method only. . I just have used shortcut.

- 1 week, 3 days ago

Yeah; I really need to familiarise myself with work energy theorem lol.

- 1 week, 3 days ago

Actually, this can be solved using process of elimination as well; firstly, we know that the work done must be negative, and for that there are only two options.

Secondly, we know that the $M$ term of the gravitational force must be of order two ($M^2$), therefore the only suitable option for the work must be (A).

- 1 week, 3 days ago

I just ran a numerical time-domain simulation and looked at the total kinetic energy at the end, assuming that no external force is trying to slow it down. And then the answer is just the negative of that, which turns out to be the first option (A).

  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 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 import math # Constants M = 2.0 L = math.sqrt(3.0)/2.0 G = 1.0 dt = 10.0**(-5.0) ################################### # Initialize simulation t = 0.0 y = 0.0 x = 1.0 # start on x axis at 1 distance unit away xd = 0.0 xdd = 0.0 # Run simulation while x >= 0.5: x = x + xd*dt # numerical integration xd = xd + xdd*dt x2 = -0.5*x # location of other particle y2 = math.sqrt(3.0)*x/2.0 Dx = x2 - x Dy = y2 - y D = math.hypot(Dx,Dy) ux = Dx/D uy = Dy/D F = 2.0*G*M*M/(D**2.0) # double the force for two particles Fx = F*ux # net force only in horizontal direction xdd = Fx/M t = t + dt ################################### E = 3.0*0.5*M*(xd**2.0) # total kinetic energy of three particles opt1 = -3.0*G*(M**2.0)/(2.0*L) # compare against options opt2 = 3.0*G*(M**2.0)/(2.0*L) opt3 = G*(M**2.0)/L opt4 = -3.0*G*M/(2.0*L) print dt print E print "" print opt1 print opt2 print opt3 print opt4 #>>> #1e-05 #6.92824539557 #-6.92820323028 #6.92820323028 #4.61880215352 #-3.46410161514 #>>> 

- 1 week, 3 days ago

Thanks! Yes; turns out I did get the correct answer. Thanks for this beautiful solution :) Computers are always brilliant.

- 1 week, 3 days ago

@Krishna Karthik whenever he uses the variable of problem as constant.
Like in this problem he has used $M=2.0$ it made me laugh 😂
He is a very smart guy😎.

- 1 week, 3 days ago

yea lol; but literally any variable you plug into values of $M , G, L$ it will give you same thing. Programmers like to be efficient😎

- 1 week, 3 days ago

@Krishna Karthik yeah I used this smart tricks in maths in my exam.
The way he chooses random values make me laugh.

- 1 week, 3 days ago

@Krishna Karthik Hii sir. Try this problem

- 1 week, 3 days ago

Will do it soon! :)

- 1 week, 2 days ago

@Krishna Karthik would you like to play chess now?

- 1 week, 2 days ago

Oh lol I'm at school, but we're doing some dumb enrichment thing. I have no problem playing chess :)

- 1 week, 2 days ago

Ok, let's play 10 min game

- 1 week, 2 days ago

@Krishna Karthik it is not work done by gravity. It is workdone by external agent..

- 1 week, 3 days ago

Which is gravity, basically.

- 1 week, 3 days ago

@Krishna Karthik it can be other different forces also.

- 1 week, 3 days ago

Yeah, but why would there be a gravitational constant in the equation otherwise? Or, is it possible that a mechanical force would need the exact same amount of work to move the balls inward by half?

- 1 week, 3 days ago

@Krishna Karthik yes it is possible. If the balls come up with their own interacting force then they all have some velocity.

- 1 week, 3 days ago

Oh ok. Thanks for explaining. Bro you are expert in physics.

- 1 week, 3 days ago

@Neeraj Anand Badgujar @NJ STAR Ah, just for your interest, here is the integral which I calculated to solve the problem:

$\large \displaystyle \int_{0}^{\frac{2}{\sqrt{3}} l} \frac{\sqrt{3}gm^2}{ \left(2l - \frac{2}{\sqrt{3}}r \right)^2 } dr$

- 1 week, 3 days ago

Hey guys; I kinda need help with a problem (problem 41). I've posted it in a new note. If you can help me that would be awesome. Cheers!

- 1 week, 2 days ago