First order differential equation involving a square root

[Lilian Sa]

Summary:: solution of first order derivatives

we had in the class a first order derivative equation:
$\frac{dR(t)}{dt}=-\sqrt{\frac{2GM(R)}{R}}$
in which R dependent of time.
and I don't understand why the solution to this equation is:
$R(t)={\frac{3}{2}}^{\frac{3}{2}}{2GM}{\frac{1}{3}}{(t_*-t)}^{\frac{2}{3}}$
thanks

[Moderator's note: moved from a technical forum.]

 

[Office_Shredder]

Are you saying you don't know how to plug that in and check it's a solution, or you don't know how to derive it without already knowing the solution?

Also, why is there an R in the numerator and denominator of the right hand side? Is there a typo?

 

[Orodruin]

What is the function $M(R)$? Note that this differential equation in general is separable.

 

[Lilian Sa]

Are you saying you don't know how to plug that in and check it's a solution, or you don't know how to derive it without already knowing the solution?

Also, why is there an R in the numerator and denominator of the right hand side? Is there a typo?

let me be honest, both.
I just did not understand it.
In the numerator there is M as a function of R, and in the denominator there is the function R.

 

[Office_Shredder]

M is an arbitrary function of R? How does it relate to the M in the solution that you are given? Is that a constant?

I assumed it's supposed to be mass and this has something to do with orbital rotation given the general shape of it, which makes me surprised that M is a function of R to begin with.

 

[Mark44]

I assumed it's supposed to be mass and this has something to do with orbital rotation given the general shape of it, which makes me surprised that M is a function of R to begin with.

Same here. The only thing I can come up with is that the scenario represents a rocket whose mass steadily decreases as the fuel is used up. However, that's just a guess, as there was no additional information provided in the problem statement.

 

[pasmith]

$M$ being an aribtrary function of $R$ is not consistent with the proposed solution.

Rearranging the ODE gives $$\frac12 \left(\frac{dR}{dt}\right)^2 = \frac{GM(R)}{R}$$ which is a gravitational potential (assuming that the tangential velocity vanishes and $M$ is constant).

 

[Lilian Sa]

Mentor note: Fixed the LaTeX
It is a collapse of a non relativistic star under its own gravity.
$M(R)=4\pi \int\rho(t)r^2dr$

 

[Mark44]

Mentor note: Fixed the LaTeX
It is a collapse of a non relativistic star under its own gravity.
$M(R)=4\pi \int\rho(t)r^2dr$

This doesn't make sense to me. M is a function of R, but R doesn't appear in the integral.

 

[Lilian Sa]

This doesn't make sense to me. M is a function of R, but R doesn't appear in the integral.

replace r in R.

 

[Mark44]

replace r in R.

Is this what you mean?
$M(R)=4\pi \int\rho(t)R^2dR$

 

[Lilian Sa]

Ok I am on it now, but why we have in the answer (t_*-t)^(2\3)

 

[Office_Shredder]

Surely it's supposed to be $M(R)= \int_0^R 4\pi \rho(r) r^2 dr$.

I'm still not sure what M in the proposed solution is supposed to be, since I guess it's not supposed to depend on R.

Edit to add: wait, $\rho(r)$ maybe is supposed to be $\rho(t)$ because as the star collapses its density increases. But if that's the case the mass is probably constant the whole time?

 

[Lilian Sa]

Surely it's supposed to be $M(R)= \int_0^R 4\pi \rho(r) r^2 dr$.

I'm still not sure what M in the proposed solution is supposed to be, since I guess it's not supposed to depend on R.

Edit to add: wait, $\rho(r)$ maybe is supposed to be $\rho(t)$ because as the star collapses its density increases. But if that's the case the mass is probably constant the whole time?

yes rho depends on time

 

[pasmith]

Surely it's supposed to be $M(R)= \int_0^R 4\pi \rho(r) r^2 dr$.

I'm still not sure what M in the proposed solution is supposed to be, since I guess it's not supposed to depend on R.

Edit to add: wait, $\rho(r)$ maybe is supposed to be $\rho(t)$ because as the star collapses its density increases. But if that's the case the mass is probably constant the whole time?

That would seem to require $$\frac{d}{dt} \left( \frac43 \pi \rho(t) R^3(t) \right) = 0$$ and hence $$\frac 1\rho \frac{d\rho}{dt} = - \frac{3}{R} \frac{dR}{dt}$$ which can be wrangled into the ODE in the OP by taking $$\frac{d\rho}{dt} = 3 \rho\sqrt{\frac{GM}{R^3}} = 3 \rho \sqrt{\frac{4\pi G \rho}{3}}.$$ I've no idea if that is physically justified.

 

[William Crawford]

Mentor note: Fixed the LaTeX
It is a collapse of a non relativistic star under its own gravity.
$M(R)=4\pi \int\rho(t)r^2dr$

I would expect the equation to be
$$ \bigg(\frac{dR}{dt}\bigg)^2 = \frac{GM}{R} $$
if it where to describe the gravitational collapse of a non-relativistic star, as a fluid parcel located at distance $R$ is gravitational bound (the Viral theorem). But I might be wrong, I haven't really studied these type of systems since I went to gymnasium (college prep school, common in europe).

Nonetheless, I believe that the OP has made a typo and intended to write
$$ R(t) = \Big(\frac{3}{2}\Big)^\frac{2}{3}(2GM)^\frac{1}{3}(t_\ast - t)^\frac{2}{3}$$
as the solution. Which suggest that $M(R)$ is taken to be constant. If so, the original ODE can be rewritten as
$$ \frac{d}{dt}R^\frac{3}{2} = - \frac{3}{2}\sqrt{2GM} $$
which can be integrated to give
$$ R(t) = \bigg(\frac{3}{2}\sqrt{2GM}(t_0 - t) + R_0^\frac{3}{2}\bigg)^\frac{2}{3} $$
and simplifies to
$$ R(t) = \Big(\frac{3}{2}\Big)^\frac{2}{3}(2GM)^\frac{1}{3}(t_0 - t)^\frac{2}{3},$$
assuming that we are given the "initial" condition $R(t_0) = R_0 = 0$.