Variable acceleration
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Variable acceleration
Two point masses are 1 meter apart in free space. How long will it take for them to collide?
 gmalivuk
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Re: Variable acceleration
What work have you done on this problem so far? Where are you getting stuck?
Anyway, the problem as stated doesn't have enough information.
Anyway, the problem as stated doesn't have enough information.
 meat.paste
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Re: Variable acceleration
Would the mass of the particles matter? The gravitational force scales with mass, but the acceleration should not. I'm assuming that only gravity is involved.
Huh? What?
 gmalivuk
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Re: Variable acceleration
The point is, are they identical pointmasses? Because if one's a kilogram and the other's the mass of the Earth, they'll collide a hell of a lot faster than if they're both 1kg.
Re: Variable acceleration
the mass would...... MATTER!!! get it?
anyway, yeah... we need to know the mass. you accelerate more quickly on earth than the moon.
anyway, yeah... we need to know the mass. you accelerate more quickly on earth than the moon.
 skeptical scientist
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Re: Variable acceleration
meat.paste wrote:Would the mass of the particles matter? The gravitational force scales with mass, but the acceleration should not. I'm assuming that only gravity is involved.
The force scales with the product of the two masses, and the acceleration on each particle scales with the mass of the other particle. So yes, the masses matter. (In fact only the total mass matters, if all you care about is the time until collision.)
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Re: Variable acceleration
aussie7us wrote:Two point masses are 1 meter apart in free space. How long will it take for them to collide?
Depends. How big are they (do we need to consider quantum effects or relativistic effects)? Are they charged? Is there any electric or magnetic fields present? Do they have a velocity relative to each other? Is there other gravitational fields present?
I edit my posts a lot and sometimes the words wrong order words appear in sentences get messed up.
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Re: Variable acceleration
You, sir, name? wrote:aussie7us wrote:Two point masses are 1 meter apart in free space. How long will it take for them to collide?
Depends. How big are they (do we need to consider quantum effects or relativistic effects)? Are they charged? Is there any electric or magnetic fields present? Do they have a velocity relative to each other? Is there other gravitational fields present?
I believe the answer to all your questions would be no. There isnt any reason to start complicating the question. If a test asks you how long it takes a mass to fall 10 meters, do you start asking about air resistance and length contraction?
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Re: Variable acceleration
BlackSails wrote:I believe the answer to all your questions would be no. There isnt any reason to start complicating the question. If a test asks you how long it takes a mass to fall 10 meters, do you start asking about air resistance and length contraction?
It largely depends on the test. If it's a test in general relativity or fluid dynamics, I start asking. And this question is completely out of context, so you really can't take anything for granted.
I edit my posts a lot and sometimes the words wrong order words appear in sentences get messed up.
 gmalivuk
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Re: Variable acceleration
Sure, but the general tendency on these boards when someone asks a homeworklooking question with insufficient information is to complicate things for the hell of it. And I can't say as I have a huge problem with that tendency, since we aren't here to do people's work for them, anyway.
 skeptical scientist
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Re: Variable acceleration
aussie7us wrote:Two point masses are 1 meter apart in free space. How long will it take for them to collide?
Since nobody's bothered to ask yet:
Is this homework? If so, what class is it for? How much do you know about differential equations?
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Re: Variable acceleration
Are the point masses the mass of a feather?
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Re: Variable acceleration
wow sorry for the 'late' response.
haha not for a class.
keep it simple, just gravity.
the question said two identical point masses. same mass of each object, and since gravityforce and acceleration are both proportional and inversely proportional to mass respectively, i think it doesnt matter..
just took AP Physics, high school... I am pretty good at calculus, for a high schooler, taken three years...
haha not for a class.
keep it simple, just gravity.
the question said two identical point masses. same mass of each object, and since gravityforce and acceleration are both proportional and inversely proportional to mass respectively, i think it doesnt matter..
just took AP Physics, high school... I am pretty good at calculus, for a high schooler, taken three years...
 gmalivuk
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Re: Variable acceleration
Yeah, if the masses are identical, it doesn't matter how big they are.
Re: Variable acceleration
mkay good.
so what ive gotten so far, is to take the derivative of the a=gm/r^2 with respect to R...and now im pulling a brain fart of sorts. should be simple....
so what ive gotten so far, is to take the derivative of the a=gm/r^2 with respect to R...and now im pulling a brain fart of sorts. should be simple....
Re: Variable acceleration
How is knowing the rate of change of acceleration going to help? Maybe you should be integrating. Trig substitution may be useful here.
Another fruitful approach is to find an expression for position wrt velocity.
Another fruitful approach is to find an expression for position wrt velocity.
 skeptical scientist
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Re: Variable acceleration
Assuming all the motion is taking place in one dimension, let x(t) be the distance between the two masses at time t. Then the force on each mass is Gm_{1}m_{2}/x(t)^{2}, so the acceleration of the two masses is Gm_{2}/x(t)^{2} and Gm_{1}/x(t)^{2}, respectively. So the acceleration DOES depend on the mass. This is just like the fact that if I'm in the gravitational field of the Earth near Earth's surface, the acceleration I experience is 9.8 m/s^{2} regardless of my mass, but it does depend on the mass of the Earth  if the Earth had half as much mass, I would only experience an acceleration of 4.9 m/s^{2}.
Anyways, we know the acceleration on each particle, so this tells us that x''(t)=Gm_{2}/x(t)^{2}Gm_{2}/x(t)^{2}, which we will rewrite as x''(t)=Gm/x(t)^{2}. This is a second order nonlinear differential equation, which in general can be very tricky to solve, and you may not have seen methods for solving them. The trick for this particular type of equation, since it does not depend on t, is to set y=x'. Then x''=y'=dy/dt=dy/dx*dx/dt=dy/dx*y by the chain rule. We can use this to transform our second order differential equation in x and t into a first order differential equation in x and y: y*dy/dx=Gm/x^{2}.
This equation is separable: we rewrite it as
and integrate:
Now we have y in terms of x, but we want x in terms of t, so we recall that y=x':
Again this is a separable equation, so we separate variables and integrate:
[math]x'=\sqrt{\frac{2Gm}{x}+C}[/math](Note that if we had preferred, we could have derived this same equation from energy considerations. This allows us to interpret C, the arbitrary constant from our first integration, as being proportional to the total energy of the system, under the usual convention that gravitational potential energy is zero when the distance separating our particles is infinite. In fact, the total energy of the system is [imath]\frac{1}{2}\mu C[/imath], where [imath]\mu=\frac{m_1m_2}{m_1+m_2}[/imath] is the reduced mass of the system.)[math]dt=\frac{dx}{\sqrt{\frac{2Gm}{x}+C}}[/math][math]dt=\frac{\sqrt{x}}{\sqrt{2Gm+Cx}}\,dx[/math]
[math]\int dt=\int \frac{\sqrt{x}}{\sqrt{2Gm+Cx}}\,dx[/math]
At this point I decide that integrating the right hand side by hand is going to be a complete pain, so I cheat.
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C}  \frac{2Gm}{C^{3/2}}\log\left( C\sqrt{x} + \sqrt{C}\sqrt{2Gm + Cx} \right)+D[/math]
There may be a way of getting this into the form x=x(t), but I doubt it (except in the special case C=0, when the total energy of the system is zero, in which case the solution is actually completely different and easy to derive analytically). In any case, I wouldn't know how to begin. Of course, it's quite possible that I made a mistake somewhere and the correct answer would be a good deal simpler.
Despite being such a simple question to state, this turns out to be a rather complicated question to solve, which is why I suspected, unlike gmalivuk and probably some other people in this thread, that it wasn't homework after all.
Anyways, we know the acceleration on each particle, so this tells us that x''(t)=Gm_{2}/x(t)^{2}Gm_{2}/x(t)^{2}, which we will rewrite as x''(t)=Gm/x(t)^{2}. This is a second order nonlinear differential equation, which in general can be very tricky to solve, and you may not have seen methods for solving them. The trick for this particular type of equation, since it does not depend on t, is to set y=x'. Then x''=y'=dy/dt=dy/dx*dx/dt=dy/dx*y by the chain rule. We can use this to transform our second order differential equation in x and t into a first order differential equation in x and y: y*dy/dx=Gm/x^{2}.
This equation is separable: we rewrite it as
ydy=Gm/x^{2} dx,
and integrate:
y^{2}/2=Gm/x + C.
Now we have y in terms of x, but we want x in terms of t, so we recall that y=x':
(x')^{2}/2=Gm/x + C.
Again this is a separable equation, so we separate variables and integrate:
[math]x'=\sqrt{\frac{2Gm}{x}+C}[/math](Note that if we had preferred, we could have derived this same equation from energy considerations. This allows us to interpret C, the arbitrary constant from our first integration, as being proportional to the total energy of the system, under the usual convention that gravitational potential energy is zero when the distance separating our particles is infinite. In fact, the total energy of the system is [imath]\frac{1}{2}\mu C[/imath], where [imath]\mu=\frac{m_1m_2}{m_1+m_2}[/imath] is the reduced mass of the system.)[math]dt=\frac{dx}{\sqrt{\frac{2Gm}{x}+C}}[/math][math]dt=\frac{\sqrt{x}}{\sqrt{2Gm+Cx}}\,dx[/math]
[math]\int dt=\int \frac{\sqrt{x}}{\sqrt{2Gm+Cx}}\,dx[/math]
At this point I decide that integrating the right hand side by hand is going to be a complete pain, so I cheat.
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C}  \frac{2Gm}{C^{3/2}}\log\left( C\sqrt{x} + \sqrt{C}\sqrt{2Gm + Cx} \right)+D[/math]
There may be a way of getting this into the form x=x(t), but I doubt it (except in the special case C=0, when the total energy of the system is zero, in which case the solution is actually completely different and easy to derive analytically). In any case, I wouldn't know how to begin. Of course, it's quite possible that I made a mistake somewhere and the correct answer would be a good deal simpler.
Despite being such a simple question to state, this turns out to be a rather complicated question to solve, which is why I suspected, unlike gmalivuk and probably some other people in this thread, that it wasn't homework after all.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
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Re: Variable acceleration
Despite being such a simple question to state, this turns out to be a rather complicated question to solve,
Certainly. IIRC, Newton took about 20 pages to solve this in Principia Mathematica, although he was using a geometrical approach to calculus, rather than our modern algebraic approach.
If you want to approximate a solution to this, I suggest using small time increments, perhaps making them even smaller as the separation distance decreases & velocity increases.
(I wrote a program to do this a few years back, but I can't locate it, ATM. I guess it'd help if I remembered what language I wrote it in. )
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Re: Variable acceleration
Oh, I forgot to mention. While you may not be able to solve the equation
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C}  \frac{2Gm}{C^{3/2}}\log\left( C\sqrt{x} + \sqrt{C}\sqrt{2Gm + Cx} \right)+D[/math]
for x=x(t), you can certainly use it to answer your initial question. Assuming the initial separation is x_{0}, and the initial velocities are 0, we have C=2Gm/x_{0}, and D is whatever will make t=0 when x=x_{0} (i.e. the negation of everything else, with each x replaced by x_{0}), and we want to find t when x=0. At this point I realize that when C is negative, the solution above is actually incorrect, and should instead be
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C} + \frac{2Gm}{C^{3/2}}\arctan\left( \frac{\sqrt{C}\sqrt{x}}{\sqrt{2Gm + Cx}} \right)+D.[/math]
Therefore, [imath]t_{\text{collision}}=D[/imath], and
[math]D=\frac{2Gm}{(2Gm/x_0)^{3/2}}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x_0}}{\sqrt{2Gm  (2Gm/x_0)x_0}} \right).[/math]
Of course, we now have a minor problem of dividing by 0, but we can find D instead by a limit, and in fact this should give us the right answer because while our differential equation does bad things when x=x_{0}, the physical situation is still perfectly fine. So we have
[math]D=\lim_{x \to x_0^}\frac{2Gm}{(2Gm/x_0)^{3/2}}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x}}{\sqrt{2Gm  (2Gm/x_0)x}} \right)=\frac{2Gm}{(2Gm/x_0)^{3/2}}\lim_{x \to x_0^}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x}}{\sqrt{2Gm  (2Gm/x_0)x}} \right)=\frac{x_0^{3/2}}{\sqrt{2Gm}}\frac{\pi}{2}.[/math]
But in fact our differential equation is symmetric in time, so we have
[math]t_{\text{collision}}=\frac{\pi x_0^{3/2}}{2\sqrt{2Gm}}.[/math]
Edit: Silly me, lim_{x > ∞} arctan x = π/2, not 1.
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C}  \frac{2Gm}{C^{3/2}}\log\left( C\sqrt{x} + \sqrt{C}\sqrt{2Gm + Cx} \right)+D[/math]
for x=x(t), you can certainly use it to answer your initial question. Assuming the initial separation is x_{0}, and the initial velocities are 0, we have C=2Gm/x_{0}, and D is whatever will make t=0 when x=x_{0} (i.e. the negation of everything else, with each x replaced by x_{0}), and we want to find t when x=0. At this point I realize that when C is negative, the solution above is actually incorrect, and should instead be
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C} + \frac{2Gm}{C^{3/2}}\arctan\left( \frac{\sqrt{C}\sqrt{x}}{\sqrt{2Gm + Cx}} \right)+D.[/math]
Therefore, [imath]t_{\text{collision}}=D[/imath], and
[math]D=\frac{2Gm}{(2Gm/x_0)^{3/2}}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x_0}}{\sqrt{2Gm  (2Gm/x_0)x_0}} \right).[/math]
Of course, we now have a minor problem of dividing by 0, but we can find D instead by a limit, and in fact this should give us the right answer because while our differential equation does bad things when x=x_{0}, the physical situation is still perfectly fine. So we have
[math]D=\lim_{x \to x_0^}\frac{2Gm}{(2Gm/x_0)^{3/2}}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x}}{\sqrt{2Gm  (2Gm/x_0)x}} \right)=\frac{2Gm}{(2Gm/x_0)^{3/2}}\lim_{x \to x_0^}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x}}{\sqrt{2Gm  (2Gm/x_0)x}} \right)=\frac{x_0^{3/2}}{\sqrt{2Gm}}\frac{\pi}{2}.[/math]
But in fact our differential equation is symmetric in time, so we have
[math]t_{\text{collision}}=\frac{\pi x_0^{3/2}}{2\sqrt{2Gm}}.[/math]
Edit: Silly me, lim_{x > ∞} arctan x = π/2, not 1.
Last edited by skeptical scientist on Sat May 23, 2009 3:40 pm UTC, edited 1 time in total.
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Re: Variable acceleration
If the point masses both have mass m, you can get quite a bit of the way there with some dimensional analysis.
The dimensional quantities in the problem are the time until impact t (T), x_{0} (L), m (M), and G (L^{3}M^{1}T^{2}). Solving for t, we must have [imath]t = Cx_0^{3/2} / \sqrt{Gm}[/imath] for some dimensionless C. You can establish a bound for C by considering that the acceleration is monotonic in x and by symmetry the particles collide at the midpoint; thus t is strictly less than the time taken for a body of mass m to travel a distance x_{0}/2 under a constant acceleration Gm/x_{0}^{2}, which is
[math]x_0/2 = (1/2)(Gm/x_0^2)t^2 \Longrightarrow t = x_0^{3/2}/\sqrt{Gm}[/math]
which implies that C < 1. (In fact, C = pi/4 from skeptical scientist's calculation [his m is double mine]).
The dimensional quantities in the problem are the time until impact t (T), x_{0} (L), m (M), and G (L^{3}M^{1}T^{2}). Solving for t, we must have [imath]t = Cx_0^{3/2} / \sqrt{Gm}[/imath] for some dimensionless C. You can establish a bound for C by considering that the acceleration is monotonic in x and by symmetry the particles collide at the midpoint; thus t is strictly less than the time taken for a body of mass m to travel a distance x_{0}/2 under a constant acceleration Gm/x_{0}^{2}, which is
[math]x_0/2 = (1/2)(Gm/x_0^2)t^2 \Longrightarrow t = x_0^{3/2}/\sqrt{Gm}[/math]
which implies that C < 1. (In fact, C = pi/4 from skeptical scientist's calculation [his m is double mine]).
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Re: Variable acceleration
I have a similar question, and thought it might be better to post it in an existing topic over creating a new one. How would I create an equation for the position of a rocket with respect to time? Gravitational force would be a function of mass and position. Rocket force would be a function of (the existance of?) fuel. Mass would be a function of fuel. Acceleration would be a function of the forces. Position would be a function of the acceleration. That would make it a function of the forces, which would make position a function of position... I can't figure this out, and I suspect that I need to know more calculus than I do now...
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Re: Variable acceleration
VDOgamez wrote:I have a similar question, and thought it might be better to post it in an existing topic over creating a new one. How would I create an equation for the position of a rocket with respect to time? Gravitational force would be a function of mass and position. Rocket force would be a function of (the existance of?) fuel. Mass would be a function of fuel. Acceleration would be a function of the forces. Position would be a function of the acceleration. That would make it a function of the forces, which would make position a function of position... I can't figure this out, and I suspect that I need to know more calculus than I do now...
Write out your forces first. F = Gmm/r^{2} + T (Thrust).
A = Gm_{e}/r^{2} + T/m
x'' = Gm_{e}/x^{2} + T/m
Substitute in your function for mass with respect to time. I don't know if the resulting differential equation is actually solvable, though.
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Re: Variable acceleration
skeptical scientist wrote:Oh, I forgot to mention. While you may not be able to solve the equation
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C}  \frac{2Gm}{C^{3/2}}\log\left( C\sqrt{x} + \sqrt{C}\sqrt{2Gm + Cx} \right)+D[/math]
for x=x(t), you can certainly use it to answer your initial question. Assuming the initial separation is x_{0}, and the initial velocities are 0, we have C=2Gm/x_{0}, and D is whatever will make t=0 when x=x_{0} (i.e. the negation of everything else, with each x replaced by x_{0}), and we want to find t when x=0. At this point I realize that when C is negative, the solution above is actually incorrect, and should instead be
[math]t=\frac{\sqrt{x}\sqrt{2Gm + Cx}}{C} + \frac{2Gm}{C^{3/2}}\arctan\left( \frac{\sqrt{C}\sqrt{x}}{\sqrt{2Gm + Cx}} \right)+D.[/math]
Therefore, [imath]t_{\text{collision}}=D[/imath], and
[math]D=\frac{2Gm}{(2Gm/x_0)^{3/2}}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x_0}}{\sqrt{2Gm  (2Gm/x_0)x_0}} \right).[/math]
Of course, we now have a minor problem of dividing by 0, but we can find D instead by a limit, and in fact this should give us the right answer because while our differential equation does bad things when x=x_{0}, the physical situation is still perfectly fine. So we have
[math]D=\lim_{x \to x_0^}\frac{2Gm}{(2Gm/x_0)^{3/2}}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x}}{\sqrt{2Gm  (2Gm/x_0)x}} \right)=\frac{2Gm}{(2Gm/x_0)^{3/2}}\lim_{x \to x_0^}\arctan\left( \frac{\sqrt{(2Gm/x_0)}\sqrt{x}}{\sqrt{2Gm  (2Gm/x_0)x}} \right)=\frac{x_0^{3/2}}{\sqrt{2Gm}}\frac{\pi}{2}.[/math]
But in fact our differential equation is symmetric in time, so we have
[math]t_{\text{collision}}=\frac{\pi x_0^{3/2}}{2\sqrt{2Gm}}.[/math]
Edit: Silly me, lim_{x > ∞} arctan x = π/2, not 1.
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Re: Variable acceleration
Yes,but that's just a drop in the bucket... I think I need to wait until next year when I learn a higher level of calculus to actually solve it, if even then... I tried to solve it during my English final, to no avail... I got CRAZY stuff that became unsolvable to my limited knowledge of calc. (BC level)
Re: Variable acceleration
Plenty of differential equations have no closed form solution. But they often can be solved using numerical approximation methods. Anyway, with a rocket launched from Earth, you probably want to use a multistage rocket, which is going to get very messy if you try to find a single closed form solution.
Re: Variable acceleration
PM 2Ring wrote:Plenty of differential equations have no closed form solution. But they often can be solved using numerical approximation methods. Anyway, with a rocket launched from Earth, you probably want to use a multistage rocket, which is going to get very messy if you try to find a single closed form solution.
Noooooo!
Re: Variable acceleration
Yesssssss. There's not even a closed form for the general 3 body problem, ie 3 bodies orbiting each other, with no friction, thrust, etc. But there are special 3 body cases that can be solved exactly.
http://en.wikipedia.org/wiki/Nbody_problem
http://math.ucr.edu/home/baez/week234.html
http://en.wikipedia.org/wiki/Nbody_problem
http://math.ucr.edu/home/baez/week234.html
John Baez wrote:1) Cris Moore, The 3body (and nbody) problem, http://www.santafe.edu/~moore/gallery.html
In 1993 Cris Moore discovered solutions of the gravitational nbody problem where the particles' paths lie in a plane and trace out braids in spacetime! I spoke about these in "week181".
More recently, Moore and Michael Nauenberg have found solutions with cubic symmetry and vanishing angular momentum, and made movies of these:Spoiler:
Re: Variable acceleration
Well, the collision depends on the surface types. If it's a flat surface, so collision may took to long and I think it depends also to the movement of the two objects.
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