A physics problem, simple or subtle?

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Discussion Overview

The discussion revolves around the gravitational interaction between two point masses initially at rest and separated by a distance of 2r. Participants explore how to calculate the time it takes for the masses to collide under the influence of gravity, considering various approaches and underlying physics concepts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the gravitational force formula and poses the question of how long it takes for two point masses to collide when released from rest.
  • Another participant derives an expression for the force acting on one mass and suggests forming differential equations to solve the problem.
  • Some participants propose using Newton's second law (F=ma) to calculate acceleration and subsequently the time to collision.
  • Concerns are raised regarding the assumption that both masses will meet at the midpoint, especially if their masses are unequal.
  • Discussion includes the concept of viewing the problem from an inertial frame versus a non-inertial frame, with differing opinions on which approach is more effective for solving the problem.
  • One participant introduces the idea of space-time curvature as an explanation for gravitational attraction and questions the mechanism behind it.
  • Alternative theories of gravity, such as the exchange of gravitons and Le Sage's pushing gravity, are mentioned, highlighting different conceptual frameworks for understanding gravitational interactions.
  • Participants note the non-linear nature of the problem due to the changing force with separation and suggest simulation as a potential method for analysis.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions made regarding the masses and their motion, particularly concerning whether they will meet at the midpoint. There is no consensus on the best approach to solve the problem, with multiple competing perspectives on the underlying physics and mathematical methods.

Contextual Notes

The discussion highlights limitations in assumptions about mass equality and the constancy of acceleration, as well as the complexity introduced by non-linear dynamics in gravitational interactions.

Jeebus
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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?

A friend asked me this and I couldn't figure it out, can you?
 
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[tex]F_1 = \frac{GM_1M_2}{(2R - x_1 - x_2)^2}[/tex]

Where [itex]F_1[/itex] is the force on object 1, [itex]M_1[/itex] and [itex]M_2[/itex] are the masses of objects 1 and 2, respectively, [itex]2R[/itex] is their initial separation, and [itex]x_1[/itex] is the distance towards the center that object 1 has traveled from its initial position (and you can guess [itex]x_2[/itex]). Now, divide both sides by [itex]M_1[/itex]

[tex]a_1 = \frac{GM_2}{(2R - x_1 - x_2)^2}[/tex]

[tex]x_1'' = \frac{GM_2}{(2R - x_1 - x_2)^2}[/tex]

You can make a similar equation for [itex]x_2''[/itex], and maybe you can get two differential equations you can solve.
 
if you have the force which you have calculated and you have the mass then surely you could use F=MA

REGARDS jamie
 
yes, once you calculate force, then both objects will have that force on eachoter. The distance they will need to ravel is r, since they will both meet at midway. use the formula F = ma, calculate a for one body. The sub that a, and r into the formula:
d = volt + 1/2at^2
this becomes: r = 1/2at^2
therefore t = sqrt(2r/a)
or if you want to do it the long way
t = sqrt(2r/(G M*m/r^2)/m))
 
Nenad

Why would both objects meet in the middle? If one has much greater mass, then it will accelerate much less, and not make it to the middle. It was not specified that the masses are the same.
 
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.
 
That's the math, what about the physics? What is the mechanism that causes the masses to move toward each other is what I'd like to know. GR's answer is space-time curvature no? So how does that work? Let's see, the masses start off by distorting space-time. Then what? If they were initially at rest, what would cause them to accellerate? Say the balls are on the edges of the grand canyon (representing the space-time curvature), if nobody pushes them why would they fall in? Now, if space has some properties like it can attract matter it might be a different story. Say perhaps not only does matter attract space and vice vera but space can have different densities so that in the presence of two masses, the density of space in between them becomes greater (more effective mass), the attraction between matter and space bringing the masses together. On a second thought, this probably begs the question since we then have to propse a mechansim by which space attracts matter.

Another mechanism I've read is that these masses exchange force particles called gravitons that moves them toward each other. This seems pretty strange also- I can't picture how anything with forward momentum could hit something and cause it to move towards the collision. Still easier to picture than space-time curvature though.

Then there's the Le Sage "pushing gravity" theories that explain this acceleration as a result of a shortage of "pushing particles" in between the masses (shadowing effect) causing the two masses to be pushed toward each other. This is probably the easiest to picture since transfer of momentum is very familiar to us.

Any comments on these or any other bright ideas out there?
 
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eyesaw said:
GR's answer is space-time curvature no? So how does that work? Let's see, the masses start off by distorting space-time. Then what? If they were initially at rest, what would cause them to accellerate? Say the balls are on the edges of the grand canyon (representing the space-time curvature), if nobody pushes them why would they fall in?

The masses cause the curvature of spacetime to change, in that new geometry the new shortest paths give the masses a natural, zero-effort path to approach each other, and while they approach they continue to chage the curvature, changing the geodesics, changing their paths, it's dynamic. But if they are prevented from following the path, like a weight on a cliff then they won't follow it. Duh! General Relativity is not bone stupid.
 
I'd take a slightly different approach, to Doc Al and say it's best to view them from a non-inertial frame, as (without actually attempting solving the problem) an inertial frame is fibne for finding out where they collide, but you'd have to do a little bit of extra work to find the equation that is trivially found in a non-inertial refernce frame.

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.
 
  • #10
Nenad said:
yes, once you calculate force, then both objects will have that force on eachoter. The distance they will need to ravel is r, since they will both meet at midway. use the formula F = ma, calculate a for one body. The sub that a, and r into the formula:
d = volt + 1/2at^2
this becomes: r = 1/2at^2
therefore t = sqrt(2r/a)
or if you want to do it the long way
t = sqrt(2r/(G M*m/r^2)/m))

That's wrong, remember the accelration is not constant.
 
  • #11
hehe, the joys of the 2 body problem.. In this case, its easy b/c its not really a 2 body problem, but more like a 1.5 body problem (in difficulty) =) Its quite a bit tougher if you specify different initial conditions..
 
  • #12
two roads, one destination :-)

jcsd said:
I'd take a slightly different approach, to Doc Al and say it's best to view them from a non-inertial frame, as (without actually attempting solving the problem) an inertial frame is fibne for finding out where they collide, but you'd have to do a little bit of extra work to find the equation that is trivially found in a non-inertial refernce frame.

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.
Good point, but I only start from that inertial frame. First I find the acceleration of each mass. Then the relative acceleration, which is just the sum of the two individual accelerations. Which brings me to the the same equation that you found. :smile:
 
  • #13
add

This problem is NON-Linear as the force changes with their separation , it is easier to simulate using small time steps especially if they form an orbit rather than meeting head on.
WRT. cuvature , the rubber sheet example is a reasonable analogy ,( that is two masses placed on an infinite sheet depressing it due to their masses) it's clear that the overall curvature does not 'suddenly' start so that in the initial condition there is already a 'force' caused by the sheet curvature in general due to their combined mass.
In that example the observer is a third person viewing the system from outside it is convenient for that observer to be attached to the mass center of the system , in that view he sees the larger mass accellerated more slowly than the lighter one both heading towards the mass center.
The mass center is simply the view of someone with a weighing scale
M.d1 = m.d2 this point has no tendency to move because it is at the gravitational nul point.
Ray.
 
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  • #14
Since r changes, the force between the bodies changes, and so the acceleration changes. This makes the problem much tougher than expected.

This problem is solved numerically in the book Computational Physics by Rubin Landau.
 
  • #15
As has been pointed out, not such an easy problem. I'll sketch out one approach. The first step is to eliminate the Center of mass coord, and go to the relative coordinate which I'll call r = r1-r2 for masses 1 and 2 with coords r1 and r2. if they move along a line then there's no angular momentum, and the energy equation becomes(I'm being sloppy with constant factors)

(dr/dt)**2 = E + g/r where E is the energy(up to some scale factor) and g is the appropriate normalized gravitational coupling constant. Note that if the particles start at rest, then E=-g/r0, r0 is the initial separation. Now the rest is algebra

Take the square root to get dr/dt= SQRT(E + g/r),

and then rewrite dr/SQRT(E + g/r) = dt. On the lhs, will have

SQRT(r) dr/SQRT(Er + g), and the integral of this this can be found in Tables of integrals. Note that the time integral goes from 0 to T, while the spatial integral goes from s0, the initial separation, to 0. And, most likely, being careful with signs will be crucial. That's the gist of it.

Regards,
Reilly Atkinson
 
  • #16
btw the equation I posted (just in case anyone does want to solve it), though it is a non-linear second order differential equation, is one that is (relatively) easy to solve :

better stated as:

[tex]\renewcommand{\vec}[1]{\mbox{\boldmath $ #1 $}} <br /> \vec{\ddot{r}} = \frac{-G(M+m)}{r^2}\vec{\hat{r}}[/tex]
 
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  • #17
Another mechanism I've read is that these masses exchange force particles called gravitons that moves them toward each other. This seems pretty strange also- I can't picture how anything with forward momentum could hit something and cause it to move towards the collision. Still easier to picture than space-time curvature though.

The way we were taught to think of this was as the exchange particle being like a boomerang. If two people face in opposite directions, and one throws a boomerang, which loops back round and the other person catches it, they have exchanged this "particle" and will move towards each other.
 
  • #18
I hope all realize that two bodies meeting like this will accellerate to infinite velocity if they have zero radius and happen to travel 'in line' ------- i often wondered ( idly) why two photons ( which have momentun ( motional mass) ) don't do the same thing.
Ray
 
  • #19
my mistake. I thought the masses equaled each other.
 
  • #20
Guys, honestly, I got lost pretty quick. Who has the right answer? And can't you use the Keplerian equation of motion to solve the collision, even though it's trivial? Wouldn't it be at the barycenter?
 
  • #21
Jeebus said:
Guys, honestly, I got lost pretty quick. Who has the right answer? And can't you use the Keplerian equation of motion to solve the collision, even though it's trivial? Wouldn't it be at the barycenter?

I know 'my' equation is correct and it looks like (without checking tho') Reilly has an equation equivalent to the first-order DE that you get when solving 'my' equation. Also the equation i posted does occur in the the normal two-body problem where the bodies have angular momentum too.
 

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