A physics problem, simple or subtle?

  • Thread starter Jeebus
  • Start date
  • #1
249
0
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?
 

Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
2,565
4
[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 travelled 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.
 
  • #3
59
0
if you have the force which you have calculated and you have the mass then surely you could use F=MA

REGARDS jamie
 
  • #4
698
0
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 = vot + 1/2at^2
this becomes: r = 1/2at^2
therefore t = sqrt(2r/a)
or if you wanna do it the long way
t = sqrt(2r/(G M*m/r^2)/m))
 
  • #5
AKG
Science Advisor
Homework Helper
2,565
4
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.
 
  • #6
Doc Al
Mentor
45,093
1,397
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.
 
  • #7
129
0
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?
 
Last edited:
  • #8
selfAdjoint
Staff Emeritus
Gold Member
Dearly Missed
6,786
9
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.
 
  • #9
jcsd
Science Advisor
Gold Member
2,090
12
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
jcsd
Science Advisor
Gold Member
2,090
12
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 = vot + 1/2at^2
this becomes: r = 1/2at^2
therefore t = sqrt(2r/a)
or if you wanna do it the long way
t = sqrt(2r/(G M*m/r^2)/m))

That's wrong, remember the accelration is not constant.
 
  • #11
Haelfix
Science Advisor
1,955
222
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
Doc Al
Mentor
45,093
1,397
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
283
0
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 tendancy to move because it is at the gravitational nul point.
Ray.
 
Last edited:
  • #14
467
1
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
reilly
Science Advisor
1,075
1
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
jcsd
Science Advisor
Gold Member
2,090
12
btw the equation I posted (just in case anyone does want to solve it), though it is a non-linear second order differntial equation, is one that is (relatively) easy to solve :

better stated as:

[tex]\renewcommand{\vec}[1]{\mbox{\boldmath $ #1 $}}
\vec{\ddot{r}} = \frac{-G(M+m)}{r^2}\vec{\hat{r}}[/tex]
 
Last edited:
  • #17
18
0
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
283
0
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
698
0
my mistake. I thought the masses equaled eachother.
 
  • #20
249
0
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
jcsd
Science Advisor
Gold Member
2,090
12
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.
 

Related Threads on A physics problem, simple or subtle?

  • Last Post
Replies
15
Views
3K
Replies
1
Views
997
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
0
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
Top