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

#Mechanics

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Comments

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I just woke up.

Lil Doug - 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.

Krishna Karthik - 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.

Lil Doug - 1 week, 3 days ago

@Lil Doug – Yeah; I really need to familiarise myself with work energy theorem lol.

Krishna Karthik - 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).

Krishna Karthik - 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).

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
#>>>

Steven Chase - 1 week, 3 days ago

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

Krishna Karthik - 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😎.

Lil Doug - 1 week, 3 days ago

@Lil Doug – 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😎

Krishna Karthik - 1 week, 3 days ago

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

Lil Doug - 1 week, 3 days ago

@Krishna Karthik Hii sir. Try this problem

Lil Doug - 1 week, 3 days ago

Will do it soon! :)

Krishna Karthik - 1 week, 2 days ago

@Krishna Karthik would you like to play chess now?

Lil Doug - 1 week, 2 days ago

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

Krishna Karthik - 1 week, 2 days ago

@Lil Doug – Ok, let's play 10 min game

Krishna Karthik - 1 week, 2 days ago

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

Lil Doug - 1 week, 3 days ago

Which is gravity, basically.

Krishna Karthik - 1 week, 3 days ago

@Krishna Karthik it can be other different forces also.

Lil Doug - 1 week, 3 days ago

@Lil Doug – 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?

Krishna Karthik - 1 week, 3 days ago

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

Lil Doug - 1 week, 3 days ago

@Lil Doug – Oh ok. Thanks for explaining. Bro you are expert in physics.

Krishna Karthik - 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:

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

Krishna Karthik - 1 week, 3 days ago

@NJ STAR @Steven Chase @Neeraj Anand Badgujar

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!

Krishna Karthik - 1 week, 2 days ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$