## changing acceleration

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thepopasmurf
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### changing acceleration

If an object is falling to the earth and its acceleration is $GM/r^2$ at any given point, how do find out it's velocity and position at any given time? I've been trying to figure this one out for ages but I don't know how to handle the change of [imath]r[/imath] as it gets closer. (To simplify I've been using [imath]a=1/r^2[/imath])

Charlie!
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### Re: changing acceleration

Do you know calculus? If not, this is much more difficult.
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### Re: changing acceleration

You need to integrate.

If [imath]f(x)[/imath] is the position then the velocity is given by
$\frac{d}{dx} \,f(x) = f\, \prime(x)$

and the acceleration is

$\frac{d}{dx}\, f\,\prime(x) = f\,\prime\prime(x)$

In other words, the velocity is the rate of change of position, and the acceleration is the rate of change of velocity.

Hope this helps.

thepopasmurf
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### Re: changing acceleration

I know calculus up to general integration. I am also aware of the relationship between acceleration, velocity and position. The problem is that I'm fairly sure that you can't just simply integrate [imath]GM/r^2[/imath] with respect to time. What you get (leaving out [imath]v_0[/imath]) is $GMt/r^2$
I don't think is right though because not only is time changing, but [imath]r[/imath] is changing as well. I was trying to get [imath]r[/imath] in terms of [imath]t[/imath] but I got confused.

Really Repeatedly Redundantly Redundant
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### Re: changing acceleration

How you do that is measure gravity as closely as you can get it, and then assume it stays constant.

In our everyday lives gravity can vary from 9.79 to 9.81 m/s2 depending on where you are, but 9.8's easily within our needed accuracy.

Since this is clearly not possible with space travel, I'm sure one or four of our resident rocket scientists will be along shortly with a complete answer. I'm not all that great with math, and it's been-- years-- since I've taken any calc classes, but I think you'll just need to integrate acceleration as a function of r then carry through the normal routine for position. Corrections, please, if I'm missing this; I'm just intuiting it.
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ThinkerEmeritus
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### Re: changing acceleration

thepopasmurf wrote:I know calculus up to general integration. I am also aware of the relationship between acceleration, velocity and position. The problem is that I'm fairly sure that you can't just simply integrate [imath]GM/r^2[/imath] with respect to time.

You are right that you can't just integrate. Since r is a function of t, you don't automatically know what the integral of [imath]GM/r^2[/imath] with respect to time is. What you have is a differential equation, and you can't even always find an analytic solution to a differential equation. In this case you can find the solution by trickery: Multiply the equation F = ma by v; you can then do [imath]\int GM/r^2 (dr/dt) dt = \int GM/r^2 dr[/imath] and the right-hand-side [imath]\int m v (dv/dt) dt = \int m v dv[/imath] . The result is energy conservation for the gravitational field. Solve this equation for v(t) and you can integrate directly to get r(t) .

For simplicity I have ignored the subscripts that indicate that everything is in terms of radial components. This procedure is dangerous but works OK here. The trick of multiplying by v works in one form or another for any conservative force (one that has a corresponding potential energy). For nonconservative forces the procedure is more difficult because you may have to worry about what path the particle takes. If the system is so complicated that all else fails, you do numerical integrations.
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thepopasmurf
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### Re: changing acceleration

Differential equations?.................Crap, I haven't done them yet. I guess I have to put this problem into the "to be completed pile"

Edit: On a side note, while trying to figure it out, I tried integrating with respect to [imath]r[/imath]. I came up with $-GM/r + c$ What does this represent, if anything?

Starside
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### Re: changing acceleration

ThinkerEmeritus wrote:Multiply the equation F = ma by v; you can then do [imath]\int GM/r^2 (dr/dt) dt = \int GM/r^2 dr[/imath] and the right-hand-side [imath]\int m v (dv/dt) dt = \int m v dv[/imath] . The result is energy conservation for the gravitational field. Solve this equation for v(t) and you can integrate directly to get r(t) .

Do you think you could finish this calculation? I am working it on paper and I am not getting r(t). I am pretty sure I am doing the integrals correctly, but I am not sure where I am going wrong.

sindustries
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### Re: changing acceleration

Your integration gives us the potential energy, well the potential, V, but U = mV. This is why we also use energy to solve problems.

Conservation of energy.

E = KE + PE = 1/2 m v^2 - GMm/r

lets say starts at rest and at some r = a.

Ei = -GMm/a

Ef = 1/2 m v^2 - GMm/r

Ei = Ef. Solve for v! Simple math, no annoying integrals.

thepopasmurf
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### Re: changing acceleration

It's potential energy(ish)?! I was trying to find a way to use conservation of energy to find the velocity at a point. Yay!

Klotz
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### Re: changing acceleration

Yup. In second year physics you learn that the force in a potential field is just the derivative (a special 3d kind of derivative called the gradient) of the potential.

The differential equation for r''=-GM/r2 is actually really hard, if not impossible, to solve. It's a funny example of a simple problem with a crazy solution.

Solt
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### Re: changing acceleration

I think you can account for the fact that F changes with r by accounting for the change in the gravitational constant g.

F = G*M*m/r^2. I'm guessing G*M/r^2 simplifies to acceleration, and the value of 9.81 m/s/s at r = sea level. That leaves F = g*m, which is the familiar third law formation. Thus, we know that having g as a function of r accounts for the law of gravitation.

Thus, if you write your equations with g = g(r), you should get the correct DE. I got a + r*g(r)+r = 0.
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ConMan
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### Re: changing acceleration

Klotz wrote:Yup. In second year physics you learn that the force in a potential field is just the derivative (a special 3d kind of derivative called the gradient) of the potential.

The differential equation for r''=-GM/r2 is actually really hard, if not impossible, to solve. It's a funny example of a simple problem with a crazy solution.

It's solvable, but not particularly pretty to do (at least it wasn't the last time I tried it). You can try using the substitution dv/dt = v dv/dr.
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nilkemorya
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### Re: changing acceleration

Couldn't you just substitute in r(t) = V(t)*t. Then use separation of variables to solve it, or did I go and oversimplify it? Man, I need to practice my calculus.
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ThinkerEmeritus
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### Re: changing acceleration

Starside wrote:
ThinkerEmeritus wrote:Multiply the equation F = ma by v; you can then do [imath]\int GM/r^2 (dr/dt) dt = \int GM/r^2 dr[/imath] and the right-hand-side [imath]\int m v (dv/dt) dt = \int m v dv[/imath] . The result is energy conservation for the gravitational field. Solve this equation for v(t) and you can integrate directly to get r(t) .

Do you think you could finish this calculation? I am working it on paper and I am not getting r(t). I am pretty sure I am doing the integrals correctly, but I am not sure where I am going wrong.

I'll try to do the whole thing with more detail. F=ma applied in the radial direction when there is no angular motion is

[imath]-GMm/r^2 = m (dv/dt)[/imath]

Then multiplying by v = (dr/dt) gives

[imath][-\int GMm/r^2] (dr/dt) dt = \int mv (dv/dt) dt[/imath]

The left-hand integral is just an integral of [...] over dr, and similarly the right-hand integral is of [mv] over dv, so we get

[imath]GMm/r + constant = (1/2) mv^2[/imath]

where the constants on the left and right sides have been combined*. This equation is really energy conservation:

[imath](1/2)mv^2 - GMm/r = constant[/imath]

This derivation of energy conservation is standard, but since it uses a procedure that seems mysterious to a student who has never solved a differential equation, energy conservation is seldom derived this generally in introductory courses. Notice that you don't know r(t) so you have a chance of getting the integral of f(r) over r, but you have no way to integrate r(t) over t.

thepopasmurf has already commented that the next step is to use this equation to get r(t). That is done by solving for
v(t) = [dr(t)/dt]. By algebra,

[imath](dr/dt) = sqrt[(2GM/r) + constant/m][/imath]

Naming C'= constant/m and getting all the r dependence on the left,

[imath]\int { dr / sqrt[(2GM/r) + C']} = \int dt = t - t0[/imath]

Now the integral on the left is not entirely trivial [meanng no one is likely to know it from memory], but if you start everything off a great distance from the earth and at v=0, you have C' = 0 and the integration is elementary.

This is a bit lengthy, but I hope it helps.

* As I was writing out the draft I wrote that the two constants had been confined. Is this a Freudian slip or what?
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### Re: changing acceleration

nilkemorya wrote:Couldn't you just substitute in r(t) = V(t)*t. Then use separation of variables to solve it, or did I go and oversimplify it? Man, I need to practice my calculus.

That would work if the velocity was constant, but in reality, you have to express [imath]r(t)[/imath] as $r(t)=\int_{t_0}^{t'} v(t)\,dt$

Which evaluates to

$r(t)=v(t)t$

if v(t) is a constant with respect to time, as it just gets pulled out of the integral and you're evaluating the integral of '1' with respect to time:

$r(t)=\int_{r_0}^{r'} v(t)\,dt=v(t)\int_{t_0}^{t'} \,dt=v(t)[t'-t_0]$

But velocity obviously isn't constant with respect to time, as the particle/body is accelerating.
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qinwamascot
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### Re: changing acceleration

I think someone has given this already, but the solution to the differential equation

r''=-GM/r^2

is extremely complicated. In fact, I could only find a power series solution. There might be a closed-form solution, but it certainly isn't easy. I will try mathematica to see if it can convert it to a somewhat nice answer once I get onto a computer that has it.

(incase you don't know, a power series solution is when you have functions that can not be represented using standard functions. It is basically just the taylor expansion of the solution, so we can find as many derivatives as we want, but they don't form any kind of a nice function)
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Starside
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### Re: changing acceleration

Ah cool, thanks for clarifying the solution. Now I see it. I do not think I would have. Diffeq is next semester.

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### Re: changing acceleration

ThinkerEmeritus wrote:<snip>
[imath]\int { dr / sqrt[(2GM/r) + C']} = \int dt = t - t0[/imath]

Now the integral on the left is not entirely trivial [meanng no one is likely to know it from memory], but if you start everything off a great distance from the earth and at v=0, you have C' = 0 and the integration is elementary.

Yeah, that looks pretty much like what I remember from when I last did this calculation. The next step is to rearrange things a little bit like so:

$\mbox{LHS } = \int{ \frac{\sqrt{r} dr}{\sqrt{2GM + C'r}} } = \frac{1}{\sqrt{C'}} \int{ \frac{\sqrt{r} dr}{\sqrt{2GM / C' + r}} } = \frac{1}{\sqrt{C'}} \int{ \frac{\sqrt{r} dr}{\sqrt{A + r}} }$

where A = 2GM/C'. Then, you fiddle around with that for a while, realise you're getting nowhere, and plug it into Mathematica - or, if you're like me and don't want to shell out huge amounts of money, the Wolfram Integrator. That then gives you:

$\int{\frac{\sqrt{r} dr}{\sqrt{A + r} }} = \sqrt{r(A + r)} - A\log{(\sqrt{x} + \sqrt{A + x})}$

which you can then substitute all the original constants and boundary conditions and such into.
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### Re: changing acceleration

qinwamascot wrote:There might be a closed-form solution, but it certainly isn't easy. I will try mathematica to see if it can convert it to a somewhat nice answer once I get onto a computer that has it.

I found a nice and simple solution for [imath]r(t)[/imath], but where [imath]r_0=\infty[/imath] and [imath]v_0=0[/imath], but for arbitrary initial conditions, it gets *extremely* complicated and appears to NOT have a closed-form solution.
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taby
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### Re: changing acceleration

Bizarre, I was just looking into this kind of thing the other week... radial infall, or dr/dt.

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### Re: changing acceleration

So what was the consensus on setting a = g(r), so the position (assuming it starts from rest) can be found with

$x = \frac{g(r)t^2}{2}$

And the velocity is

$v = g(t)t$

Isn't the D.E. that results from these easier to solve?

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### Re: changing acceleration

Just a bit off-topic but you guys are aware of the subscript and superscript tags right? Just to make things a little prettier

Gm1m2 / r2

edit: nvm, I feel like a douche.
Last edited by Aradae on Wed Oct 29, 2008 5:38 pm UTC, edited 1 time in total.
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### Re: changing acceleration

Aradae wrote:Just a bit off-topic but you guys are aware of the subscript and superscript tags right? Just to make things a little prettier

Gm1m2 / r2

Or the [ math ] environment?

To make things even prettier,

$\mathrm{G}\frac{m_1m_2}{r^2}$

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### Re: changing acceleration

Aradae wrote:Just a bit off-topic but you guys are aware of the subscript and superscript tags right? Just to make things a little prettier

Gm1m2 / r2

Or the [ math ] environment?

To make things even prettier,

$\mathrm{G}\frac{m_1m_2}{r^2}$

all I'm getting is text. Is there something I have to do to get it to look pretty on my browser?

edit: nvm.
Last edited by Aradae on Wed Oct 29, 2008 5:39 pm UTC, edited 1 time in total.
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ThinkerEmeritus
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### Re: changing acceleration

seladore wrote:So what was the consensus on setting a = g(r), so the position (assuming it starts from rest) can be found with

$x = \frac{g(r)t^2}{2}$

And the velocity is

$v = g(t)t$

Isn't the D.E. that results from these easier to solve?

Unfortunately that doesn't get the right answer. Strictly speaking, a = g( r(t) ), with time dependence implied by the time dependence of r . Therefore a is not a constant, and if a is not a constant, dv/dt = a does not imply that v = a t.

The closest you can get to a simpler differential equation is to substitute v = dx/dt into the energy conservation relation

(1/2) m v2 - GMm/r2 = constant.

The constant is the kinetic energy at large r.

My earlier post gives a direct derivation of this equation from F( r(t) ) = m (dv/dt) .
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Solt
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### Re: changing acceleration

seladore wrote:So what was the consensus on setting a = g(r), so the position (assuming it starts from rest) can be found with

$x = \frac{g(r)t^2}{2}$

And the velocity is

$v = g(t)t$

Isn't the D.E. that results from these easier to solve?

Sure, but what's g(t)? You only have g(r).
"Welding was faster, cheaper and, in theory,

produced a more reliable product. But sailors do

not float on theory, and the welded tankers had a

most annoying habit of splitting in two."

-J.W. Morris

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### Re: changing acceleration

Solt wrote:
seladore wrote:So what was the consensus on setting a = g(r), so the position (assuming it starts from rest) can be found with

$x = \frac{g(r)t^2}{2}$

And the velocity is

$v = g(t)t$

Isn't the D.E. that results from these easier to solve?

Sure, but what's g(t)? You only have g(r).

I meant g(r), sorry.

But ThinkerEmeritus has already set me straight.