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- Thread starter WraithM
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tiny-tim

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Hi WraithM![tex]\frac{d^2 r}{dt^2}=\frac{4}{3}\pi a G(r^2 - r)[/tex]. I don't really know much beyond solving simple differential equations. Is this solvable? If there is a solution, how do you get it?

Use the standard trick of writing dr/dt = v, and then

d

= dv/dt

= dv/dr dr/dt [chain rule]

= v dv/dr

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Does that actually work? Do you just separate variables from there? I see the logic of making [tex]\frac{d^2 r}{dt^2} = v \frac{dv}{dr}[/tex], but I don't want to believe that it's that easy.

For example, take [tex]\frac{d^2 r}{dt^2} = g[/tex] (constant acceleration). Separating variables using v dv/dr: v dv = g dr, that's: v^2/2 = gr

[tex]\frac{dr}{dt} = \sqrt{2gr}[/tex]

[tex]\frac{dr}{\sqrt{2gr}} = dt[/tex]

[tex]1/g \sqrt{2gr}=t[/tex]

[tex]r = \frac{gt^2}{2}[/tex]

Wow, so that actually works?

:/ I'm feeling stupid for not knowing this.

EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?

For example, take [tex]\frac{d^2 r}{dt^2} = g[/tex] (constant acceleration). Separating variables using v dv/dr: v dv = g dr, that's: v^2/2 = gr

[tex]\frac{dr}{dt} = \sqrt{2gr}[/tex]

[tex]\frac{dr}{\sqrt{2gr}} = dt[/tex]

[tex]1/g \sqrt{2gr}=t[/tex]

[tex]r = \frac{gt^2}{2}[/tex]

Wow, so that actually works?

:/ I'm feeling stupid for not knowing this.

EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?

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tiny-tim

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Wow, so that actually works?

Woohoo!

(btw, that's how 1/2 mv

hmm … let's see …EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?

you didn't get vdv = 4/3 πaG (r

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You integrate from ([tex]r_{0}[/tex] to [tex]r[/tex]) that to get [tex]v = \sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}[/tex]. That's all good, but then if you try to go another step further, it blows up. Now I'm solving for r(t). You get: [tex]\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}} = \int_{0}^{t} dt[/tex] I don't know how to do that integral. I don't think it's doable. Do you know how to do it? Am I perhaps doing it wrong?

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tiny-tim

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ah, I didn't realise you wanted to go as far as solving for r as a function of t… You get: [tex]\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}} = \int_{0}^{t} dt[/tex] I don't know how to do that integral. I don't think it's doable. Do you know how to do it? Am I perhaps doing it wrong?

Yes, that looks correct … unfortunately, not all integrals are "doable"

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nasu

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If the density is not uniform, the force acting on the mass when is at distance r from the center will be

G*m*M(r)/r^2

where M(r) is the mass contained in the sphere of radius r.

This is not 4/3pi*r^3*density . Density is not constant. You need to integrate from 0 to r in order to find the mass.

M(r) will be 4*pi(a/3*r^3-r^4/4) - you should do it again do double check.

But before you do this, check again your density. Of course, if you are interested in a math problem only, it does not matter.

But for a physics problem...

density=a-r ?

So density is zero at r=a and then it becomes negative for r<a.

Not mentioning the units. You will have density equal to distance. (a and r are disatnces, I suppose) Not very physical...

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[tex]\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{2}{3} \pi G(r^3 - 2ar^2-r_{0}^3+2ar_{0}^2)}}=\int_{0}^{t}dt[/tex]

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nasu

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The integral that you show in the last post can be solved numerically or maybe can be expressed in term of elliptical functions, I am not sure. Try Mathematica or Maple.

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