Shadow Posted April 20, 2009 Share Posted April 20, 2009 Hey all, Let [math]M_A[/math] and [math]M_B[/math] denote the masses of objects [math]A[/math] and [math]B[/math], and let [math]R[/math] denote the distance between them at [math]t_0[/math]. Is there any way to calculate the instantaneous [math]F_G[/math] acting upon the two at time [math]t[/math] (from a Newtonian perspective)? What confuses me is that the distance between them changes with the distance traveled, which in turn changes according to the force acting upon the two, which in turn changes according to the distance between the two, and on we go in a vicious circle. What I'm a little afraid of is that I'm touching upon the n-body problem, but I'm not sure, that's why I ask. Cheers, Gabe Link to comment Share on other sites More sharing options...

swansont Posted April 20, 2009 Share Posted April 20, 2009 As long as n=2 you should be OK Link to comment Share on other sites More sharing options...

Shadow Posted April 20, 2009 Author Share Posted April 20, 2009 In that case, what would such a function look like? Link to comment Share on other sites More sharing options...

Megabrain Posted April 20, 2009 Share Posted April 20, 2009 you need to specify they are point masses otherwise it gets even more complex. As an example when you stand on earth the gravity you feel is a complex vector of all the mass of the earth from that at your feet to that on the opposite side of the earth, from horizon to horizon in all directions, from all points within the spehere that is earth. thus you are being pulled in all directions. specify a point mass and you lose all that extra complexity. Link to comment Share on other sites More sharing options...

Shadow Posted April 20, 2009 Author Share Posted April 20, 2009 Yup, it's a point mass ) I'm a simple person. Link to comment Share on other sites More sharing options...

Megabrain Posted April 20, 2009 Share Posted April 20, 2009 Gravity, thou art a heartless b*tch I remember saying that after falling off a wall ! Link to comment Share on other sites More sharing options...

Shadow Posted April 20, 2009 Author Share Posted April 20, 2009 I remember it from Big Bang ) Anyhow, any clue as to what that function would look like? Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted April 20, 2009 Share Posted April 20, 2009 You'd have to model it with calculus, I'd think. I'll try to work out the math when I have some time. Link to comment Share on other sites More sharing options...

Shadow Posted April 20, 2009 Author Share Posted April 20, 2009 That's what I tried, but I keep getting nowhere...the problem is, I can't find an abstract solution, as in I have no idea what I want to get to, so solving the problem is rather hard. But thanks a million in advance ) Link to comment Share on other sites More sharing options...

Norman Albers Posted April 20, 2009 Share Posted April 20, 2009 This sounds like a "retarded time" issue as we do in electrodynamics. As long as you are not moving at a huge relativistic velocity, say [math] \beta[/math] greater than one-tenth, you'll get a close answer. Link to comment Share on other sites More sharing options...

John Cuthber Posted April 21, 2009 Share Posted April 21, 2009 As far as I can see you haven't ruled out a system like the sun and the earth. You need to specify the original conditions but, in any event this is a bit of calculus that must have been done before. Link to comment Share on other sites More sharing options...

Shadow Posted April 21, 2009 Author Share Posted April 21, 2009 I don't have anything against calculus, I'm all for it. The original conditions: Two point masses, A and B, with masses [math]M_A[/math] and [math]M_B[/math]. Initial velocities = 0, distance between them is [math]R[/math] and, since all I need is 2D, the position of A is [math][x_a;y_a][/math] at [math]t_0[/math] and the position of B is [math][x_b;y_b][/math] at [math]t_0[/math]. What I'm looking for is a function [math]S(t)[/math] that will tell me where the given object will be at time [math]t[/math], and a way to convert that to Cartesian coordinates. Hope I'm not asking for too much :-D I tried looking up the two body problem, unfortunately all the explanations are fact based, as in "take this for granted", and due to this and also partly the level of math used, I'm having extreme difficulty in understanding even the most basic parts. Thanks in advance for any hints. Link to comment Share on other sites More sharing options...

Mr Skeptic Posted April 21, 2009 Share Posted April 21, 2009 That makes things much simpler. They should travel a straight line. Be sure to set your coordinate system starting at the center of mass, and an axis through both. You should have need of only 1D. Link to comment Share on other sites More sharing options...

swansont Posted April 21, 2009 Share Posted April 21, 2009 You'll need to set up a differential equation, with acceleration as the second differential of position. Link to comment Share on other sites More sharing options...

D H Posted April 22, 2009 Share Posted April 22, 2009 I don't have anything against calculus, I'm all for it. This problem is why calculus was invented. The original conditions:Two point masses, A and B, with masses [math]M_A[/math] and [math]M_B[/math]. Initial velocities = 0, distance between them is [math]R[/math] and, since all I need is 2D, the position of A is [math][x_a;y_a][/math] at [math]t_0[/math] and the position of B is [math][x_b;y_b][/math] at [math]t_0[/math]. What I'm looking for is a function [math]S(t)[/math] that will tell me where the given object will be at time [math]t[/math], and a way to convert that to Cartesian coordinates. Hope I'm not asking for too much :-D You picked a singular case. Much easier is to start with an initial relative velocity vector that has a non-zero component normal to the initial relative position vector. The first step is to write a differential equation. Do you need help doing this? Link to comment Share on other sites More sharing options...

Norman Albers Posted April 22, 2009 Share Posted April 22, 2009 (edited) What's the big deal starting at rest, D H ? Yes we work in a center of mass system so momenta are equal and opposite. Shadow, I rebuild acoustic pianos. I try to not need to move them. Edited April 22, 2009 by Norman Albers Link to comment Share on other sites More sharing options...

D H Posted April 22, 2009 Share Posted April 22, 2009 Start with a non-zero normal component to the velocity and you get Kepler's laws. Nice and simple. Start with a zero normal component and you get something quite different and quite a bit uglier. Start with zero velocity period (or zero normal component and a axial velocity smaller than escape velocity) and you get something real ugly. Think of it this way: At some point the particles will collide (or pass through one another). At that instant the gravitational acceleration is instantaneously infinite. That's an improper integral at best. The resulting 1D differential equation does have an analytic solution, but it is quite ugly. Link to comment Share on other sites More sharing options...

Shadow Posted April 22, 2009 Author Share Posted April 22, 2009 That makes things much simpler. They should travel a straight line. Be sure to set your coordinate system starting at the center of mass, and an axis through both. You should have need of only 1D. Eventually I will be needing a function for non-zero initial velocities, but I want to start simple. The first step is to write a differential equation. Do you need help doing this? This is all probably going to be used in a program one day, the program that I've already made a couple of posts about lately. At first I was going to deal with the problem the way I described here. Unfortunately, as Atheist pointed out, I assumed r being independent of time. At the time I didn't care much, because as I said, it was going to be used in a program where r was recalculated in a loop. But it took me a couple of days to realize that having a function where r was not independent of time would be much better, because it would open up lots of new possibilities for my program, and I'd probably learn a thing or two as well. Problem is, I just can't find a way to account for the change in r. I know abstractly it can be done, but I just can't find a way, and unfortunately me and calculus involves mostly lots of shooting in the dark, so I didn't really expect to find anything. However I also didn't expect me not being able to solve the problem abstractly; I keep getting stuck in that vicious triangle I talked about above. To cap it up, this may still be above my level of knowledge, but I still want to try, if only because of the fact that I could learn quite a few handy tricks in the process So, I guess the answer to your question is yes, I probably will. Maybe a nudge in the right direction will do, maybe not; as I said, I don't even know what I'm shooting for, which may be the only problem, or more likely just part of the problem. Think of it this way: At some point the particles will collide (or pass through one another). At that instant the gravitational acceleration is instantaneously infinite. What I'd like to do, at least for now, is to turn off gravity when they pass through each other. Eg. I'd have one function that would describe the motion right up to the moment they touch, then I would just let them pass through each other, and then make a new function with their current positions and velocities as their starting ones. Is that possible? Link to comment Share on other sites More sharing options...

Mr Skeptic Posted April 22, 2009 Share Posted April 22, 2009 If you don't actually care to have everything a function of time, it again gets much simpler. You can calculate the potential energy, and that will give you the velocity. Obviously, the force will also be easy to calculate from position. Link to comment Share on other sites More sharing options...

Shadow Posted April 23, 2009 Author Share Posted April 23, 2009 Even though I'm not sure how to calculate velocity from PE (I though that was escape velocity) that still wouldn't help me, because I would have to recalculate R each time, which is what I'm trying to avoid. The reason why I'm doing this is that I'd like the user of my program to be able to project the path of the object (star, in this case), from the point of origin up to t units of time in the future (like they do in star Trek ). That would be computationally very taxing, since I'd just have to pause the simulation and have the computer go through the steps in advance, and if I wanted to project the path the object took in the past, I'd have to make the computer remember every step the object took. But if I had a function, I could just plot the function and be done with it, or so I believe. And that's what I'm after. If there's a simpler way to do this, I'm all ears. Cheers, Gabe Merged post follows: Consecutive posts mergedI talked with my teacher today, and if I understand correctly, she told me to find F(t), getting a(t) from that and then a'(t). Unfortunately, it was all in a rush, and I'm not sure I understood everything correctly (I'm also not all that certain she understood me correctly). Does that make any sense? Link to comment Share on other sites More sharing options...

Mr Skeptic Posted April 23, 2009 Share Posted April 23, 2009 Well, what you have is F®, which gives you a®=r''®, PE® which gives you v®=r'®. r is the radius and is a function of t. Link to comment Share on other sites More sharing options...

Shadow Posted April 24, 2009 Author Share Posted April 24, 2009 That's what I got to, although in a slightly different way. So I guess my question is, how do I make a function r(t) that gives me radius dependent on time? Because this is the part where I start getting confused; radius is dependent not only on time, but also on distance traveled...I know in the end it's all dependent on time, but how do I put that relationship into math-speak? Link to comment Share on other sites More sharing options...

D H Posted April 24, 2009 Share Posted April 24, 2009 As has been said many times in this thread, you need to write a differential equation. Since this appears to be homework, I am not going to tell you what the answer is until you make some kind of attempt at doing so. Link to comment Share on other sites More sharing options...

Shadow Posted April 24, 2009 Author Share Posted April 24, 2009 It's not homework (if it were I'd be posting in the homework section), we haven't gotten even close to this level in class. We haven't even started calculus yet, nor are we going to for another 2 years (which is not to say I don't know some, it's just that my knowledge is very limited and incomplete, since it consists mostly of odds and ends picked up around the internet). I'm just doing this for the fun of it, and also to learn; if I can also find an application in my program, it'd be great. Unfortunately, since this is so far above my normal area of interest I really don't have the reasoning to make even an attempt; as I said, I don't even know exactly what I'm shooting for. I understand there's absolutely no reason for you to believe me, so it's really up to you whether or not to trust me. While nothing will happen if you don't help me, because I don't really need this for anything, I'm very curious to find out, and so are two or three friends I've discussed this with, one of them also being a member here at SF. If you tell me I don't have the necessary prerequisites, that's another story entirely; I know I have a sometimes annoying tendency to make huge leaps ahead without bothering about anything in between. But even though it might be out of my league, I still think there's a chance I might learn something in the process, which is always a good enough reason for me. Now, the way I see it, since I want to get a function that gives position at a certain time, I need to be able to calculate the force at any given time. After that, I know that [math] F=m\frac {d^2s}{dt}[/math], so it should just be a matter of dividing force by m, which means I get acceleration, and then integrating twice with respect to t. Now here is where I get unsure. When I look at F, I can see that F is dependent on the distance between the two objects, r. So what I have to do is make a function that gives me the distance between the objects at a certain time, right? If so, then this is the part I get lost in; how do I give the distance between the two objects without knowing how much distance was already traveled? If I were to think about this logically, then I can't, which means that there has to be a way to calculate the distance traveled. And that should be dependent on time if my reasoning is correct, so there must be some way to get the distance traveled from the time. And since the distance traveled is dependent on the force, which is also dependent on the masses of the two objects, I would guess that I'm going to be doing something with the force equation. Now this is the part were shooting in the dark is an understatement; would taking the derivative with respect to r and then integrating with respect to t get me anywhere? Because the derivative with respect to r should tell me how much the force changes when the radius changes, the same way the derivative of distance over time gives me how much the distance changes when time changes (velocity). When I take the derivative, it comes out negative, which makes sense; the larger r is, the smaller F will be. As for the integration part...well, to be quite honest the only reason I did that was to get t in the equation. So what I end up with is [math]\int \frac {dF}{dr} dt = \frac{-2 \cdot G \cdot m_1 \cdot m_2 \cdot t}{r^3}[/math]. Am I going in the right direction, or is this absolute rubbish? If so, what did that equation give me? Cheers and thanks for any tips, Gabe Link to comment Share on other sites More sharing options...

Mr Skeptic Posted April 24, 2009 Share Posted April 24, 2009 For the above, since r is a function of t, you can't threat it as a constant when integrating with respect to t. Link to comment Share on other sites More sharing options...

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