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Author Topic:   To all amateur physicists - a simple physics problem
Eta_Carinae
Member (Idle past 4374 days)
Posts: 547
From: US
Joined: 11-15-2003


Message 1 of 19 (123056)
07-08-2004 5:09 PM


Let's see if anyone likes to solve simple physics problems.
We all know (I hope) that the standard Newtonian formula for the gravitational attraction of two (point like) masses is:
F = G M*m/r^2 where G is the Gravitational Constant - M is the mass of
one object, m is the mass of the other and r is their separation.
So far so good.
Now imagine two point masses as above, separated by an initial distance 2r (it helps the algebra to make it 2r) and that they are initially at rest.
They are then 'let go' and allowed to move together by gravity alone. I could couch this in some fancy language about assuming the space is maximally symmetric blah blah blah - but the initial conditions are as stated above - it doesn't need GR or such.
Now the question is:
How long does it take them to collide?
Yep - that simple a question. Seems easy doesn't it? After all it's so basic it must be in any basic mechanics text - except it isn't in those books - or at least any I remember. In fact it is curiously absent - I wonder why?
It really isn't that difficult but it is subtle enough to make you think a tad.
Hint:
There are two ways of solving this I can think of. One uses a very fundamental theorem that dates back 300 years or so - the other is a clever way of recasting the problem that actually dates back 400 years or so.
(I don't know really what forum this belongs in - I thought Big Bang and Cosmology might be best just because most of the physics types post there but use your judgement.)

Replies to this message:
 Message 2 by Coragyps, posted 07-08-2004 5:18 PM Eta_Carinae has not replied
 Message 3 by Loudmouth, posted 07-08-2004 5:47 PM Eta_Carinae has replied

  
Coragyps
Member (Idle past 734 days)
Posts: 5553
From: Snyder, Texas, USA
Joined: 11-12-2002


Message 2 of 19 (123057)
07-08-2004 5:18 PM
Reply to: Message 1 by Eta_Carinae
07-08-2004 5:09 PM


I'll think about that while you go have a peek at this
non-standard treatment. Take your hip waders.

This message is a reply to:
 Message 1 by Eta_Carinae, posted 07-08-2004 5:09 PM Eta_Carinae has not replied

  
Loudmouth
Inactive Member


Message 3 of 19 (123066)
07-08-2004 5:47 PM
Reply to: Message 1 by Eta_Carinae
07-08-2004 5:09 PM


It's been a long time since calculus and physics, but let me see if I can figure out the basic concepts.
As the masses get closer, the gravitational force increases. Therefore, their acceleration will increase as they come closer together. If they had the same mass, then they maximal acceleration would be at 1r from their starting positions. In this instance I would set up the differential equation to go from 0 to 1r (don't ask me what the equation would look like, but pretty sure it would end with dt). However, if the masses are not the same I am not sure how it would be done. I am guessing that you would first have to figure out the point at which they would collide from their starting positions, and then calculate the time.
Second, they are point masses. So can they actually collide, or would they oscillate in an asymptotic pattern? Normally masses have an outer, physical limit. For instance, if you said that the masses have a radius of r and their centers of mass were 4r apart it might be a simpler question.

This message is a reply to:
 Message 1 by Eta_Carinae, posted 07-08-2004 5:09 PM Eta_Carinae has replied

Replies to this message:
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Eta_Carinae
Member (Idle past 4374 days)
Posts: 547
From: US
Joined: 11-15-2003


Message 4 of 19 (123069)
07-08-2004 6:19 PM
Reply to: Message 3 by Loudmouth
07-08-2004 5:47 PM


No just assume the point masses collide.
I specifically made them point masses so you can neglect anything related to finite source size effects.
Actually where they collide is trivial. It's at the barycentre. But the time to do so is not quite as trivial - hence the question.
The reason I asked for the collision time and not some other time to fall a given distance is that it simplifies the result a lot. The integral is pretty nasty then but for the time to collide at the barycentre the integral is trivial.
Hint #2:
Think Kepler

This message is a reply to:
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Maxwell's Demon
Member (Idle past 6229 days)
Posts: 59
From: Stockholm, Sweden
Joined: 05-09-2004


Message 5 of 19 (123320)
07-09-2004 10:38 AM


Well... I'm not sure I got it right, since I've never attempted to solve this problem before, but here goes:
If we are allowed to assume that the collision will occur at the centre of gravity for the two point masses, then we can define two vectors, r1 and r2, such that m1*r1 + m2*r2 = 0.
The vectors are therefore the vectors leading from respective point to the centre of gravity.
Also defining r = r1 - r2 we can write the equation of motion for one of the particles as:
m1*r1'' = Gm1m2/r^2 (where ' denotes the derivative with respect to time).
Using that r2 = -(m1/m2)r1, and that r = r1 - r2 we can eliminate r1 from the equation of motion.
We get:
r'' = G(m1 + m2)/r^2, where I'll let M = m1 + m2 from now on.
This yields:
r^2*dr^2 = G*M*dt^2
Integrating twice from 0 to 2r on the left hand side and from 0 to t on the right hand side we gain:
(1/12)*(2r)^4 = (1/2)G*M*t^2
Finally, we get:
t = sqrt(8/3GM)*r^2
I'm assuming I've made some form of mistake since using a distance of 2r didn't seem to aide me in the calculations whatsoever.
(Edited out horrendous error in last calculation)
This message has been edited by Kent, 07-09-2004 10:10 AM

Replies to this message:
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Eta_Carinae
Member (Idle past 4374 days)
Posts: 547
From: US
Joined: 11-15-2003


Message 6 of 19 (123325)
07-09-2004 11:12 AM
Reply to: Message 5 by Maxwell's Demon
07-09-2004 10:38 AM


No...
You cannot integrate twice like that anyway. The collision does occur at the barycentre but that is obvious since the centre of mass of this closed system cannot move because there are no external forces acting.
You were somewhat on the right track initially though. That is what I called the first method of solving this. The second method is the 'cooler' one so to speak.
Hint#3
the final answer has Pi in it.

This message is a reply to:
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Primordial Egg
Inactive Member


Message 7 of 19 (124211)
07-13-2004 9:34 AM


Has anyone got anywhere with this? I get stuck where I have to solve:
y'' = k/y^2 (k = const).
I reckon that the cunning method involves some sort of restating the problem into an orbital motion problem (Kepler + pi were my clues here), but haven't really got anywhere with this train of thought.
PE

Replies to this message:
 Message 8 by Eta_Carinae, posted 07-13-2004 12:22 PM Primordial Egg has replied

  
Eta_Carinae
Member (Idle past 4374 days)
Posts: 547
From: US
Joined: 11-15-2003


Message 8 of 19 (124224)
07-13-2004 12:22 PM
Reply to: Message 7 by Primordial Egg
07-13-2004 9:34 AM


Correct!
Restating it as an orbital problem is the 2nd method I referred to.

This message is a reply to:
 Message 7 by Primordial Egg, posted 07-13-2004 9:34 AM Primordial Egg has replied

Replies to this message:
 Message 10 by Primordial Egg, posted 07-13-2004 5:41 PM Eta_Carinae has replied

  
JIM
Inactive Member


Message 9 of 19 (124243)
07-13-2004 2:27 PM


Mhm.
F1= GM1M2 / (2R-x1-x2)^2
Where F1 is the force on object 1, M1 and M2 are the masses of objects 1 and 2, respectively, 2R is their initial separation, and x1 is the distance towards the center that object 1 has travelled from its initial position (and you can guess x2). Now, divide both sides by M1
a1 = GM2 / (2R - x1 - x2)^2
x"1 = GM2 / (2R - x1 - x2)^2
You can make a similar equation for x"2 , and maybe you can get two differential equations you can solve.

  
Primordial Egg
Inactive Member


Message 10 of 19 (124274)
07-13-2004 5:41 PM
Reply to: Message 8 by Eta_Carinae
07-13-2004 12:22 PM


Re: Correct!
Thanks Eta - but to be honest, I don't think that's helped me very much.
Here's what I remember about Kepler:
1. Orbital motion - equal areas swept out in equal times.
2. Orbital period T^2 is proportional to R^3
3. Orbital period for a planet is independent of its mass.
4. Planetary orbits are generally elliptical, with the sun at one focus.
Not sure how any of that helps me - the problem is two objects travelling towards one another in a straight line.
I've had two ideas, but since my physics is so rusty, I haven't been able to follow them through. The first is imagining that you were moving perpendicular to the plane of the two objects so that as they came together they would appear to you to trace out a semi-circle. The second is to picture the objects as the foci of an ellipse with ever decreasing eccentricity, and to use the parametric equation for an ellipse to determine the time taken for the two masses to collide, but it hurts my head just thinking about it.
PE

This message is a reply to:
 Message 8 by Eta_Carinae, posted 07-13-2004 12:22 PM Eta_Carinae has replied

Replies to this message:
 Message 11 by Eta_Carinae, posted 07-13-2004 7:20 PM Primordial Egg has not replied

  
Eta_Carinae
Member (Idle past 4374 days)
Posts: 547
From: US
Joined: 11-15-2003


Message 11 of 19 (124286)
07-13-2004 7:20 PM
Reply to: Message 10 by Primordial Egg
07-13-2004 5:41 PM


Big time hint:
Imagine the two bodies are in orbit around each other. From this you can set up the equations of motion.
Then just imagine the bodies are stopped in their orbits suddenly. Then they will fall inwards a la the described problem.
Think about it: The equations of motion are describing the orbits in polar coordinates. By stopping the bodies and letting them fall you are fixing the angular coordinate thus theta(dot) is set to zero.
Thus you can use the Keplerian equation of motion to solve the given problem.
I'll post the solution in a day or so if no one else does.

This message is a reply to:
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Replies to this message:
 Message 13 by joz, posted 07-14-2004 6:37 PM Eta_Carinae has replied

  
JIM
Inactive Member


Message 12 of 19 (124458)
07-14-2004 11:46 AM


The equation you want is a = -G(M + m)/r^2 (which the accelartion of the object which don't choose to be stationery in your non-inertial refernce frame). This is a seconnd order differential equation, which you must solve in order to find t.
Best to view them from the inertial frame in which the masses were initially at rest. They will of course meet at their center of mass.

  
joz
Inactive Member


Message 13 of 19 (124550)
07-14-2004 6:37 PM
Reply to: Message 11 by Eta_Carinae
07-13-2004 7:20 PM


Re: Big time hint:
Can you integrate the force with respect to time for both bodies to get momentum ( p = G.M.m.t/4r2 + c ) bodies are initially at res thus c = 0, due to conservation momentum is the same for each body....
Then p = mv so
VM = G.m.t/4r2
and
Vm = -G.M.t/4r2
Integrate to get equation for position constants will be r and -r,
xM = G.m.t2/8r2 - r
and
xm = r - G.M.t2/8r2
at the time they collide the positions will obviously be equal...
G.m.t2/8r2 - r = r - G.M.t2/8r2
2r = G.m.t2/8r2 + G.M.t2/8r2
16r3 = G.m.t2 + G.M.t2
t2 = 16r3G(M+m)
t = (16r3G(M+m))1/2
hows that?

This message is a reply to:
 Message 11 by Eta_Carinae, posted 07-13-2004 7:20 PM Eta_Carinae has replied

Replies to this message:
 Message 14 by Eta_Carinae, posted 07-14-2004 9:04 PM joz has not replied

  
Eta_Carinae
Member (Idle past 4374 days)
Posts: 547
From: US
Joined: 11-15-2003


Message 14 of 19 (124568)
07-14-2004 9:04 PM
Reply to: Message 13 by joz
07-14-2004 6:37 PM


Nope!
Sorry.

This message is a reply to:
 Message 13 by joz, posted 07-14-2004 6:37 PM joz has not replied

  
joz
Inactive Member


Message 15 of 19 (124590)
07-14-2004 11:03 PM


whats wrong with it?
Nvm just figured that out the distance between the two points doesn't stay 2r....
bugger...
This message has been edited by joz, 07-14-2004 10:06 PM

  
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