Problem in Newtonian gravity- 2nd order, integration problems

Romeo
Messages
13
Reaction score
0
The problem is this:

Given sun of mass M and a body of mass m (M>>m) a distance r from the sun, find the time for the body to 'fall' into the sun (initially ignoring the radius of the sun).


Our first equation is therefore \frac {d^2r}{dt^2} = \ddot{r} = \frac {GM}{r^2}.

I am able to integrate this, giving:
\dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}},

where R is the inital distance of the body from the sun. However, I am unable to integrate this again. I have shoved it into wolfram's integrator for an indicator of what to aim for, but cannot come close.

Any thoughts would be greatly appreciated.

Regards

Romeo
 
Physics news on Phys.org
This diff. eq. is separable.
We may write it as:
\sqrt{\frac{r}{R-r}}\frac{dr}{dt}=-\sqrt{\frac{2GM}{R}}
Integrating from t=0 to t=T, where T is the time when r=0, we have:
\int_{0}^{R}\sqrt{\frac{r}{R-r}}dr=\sqrt{\frac{2GM}{R}}T

Use the substitution u=\sqrt{\frac{r}{R-r}} to progress further:
We get: r=R\frac{u^{2}}{1+u^{2}}, \frac{dr}{du}=2R\frac{u}{(1+u^{2})^{2}}
And r=0\to{u}=0,r=R\to{u}=\infty

Therefore, we have:
\int_{0}^{R}\sqrt{\frac{r}{R-r}}dr=2R\int_{0}^{\infty}\frac{u^{2}du}{(1+u^{2})^{2}}=R(arctan(u)-\frac{u}{1+u^{2}})\mid_{u=\infty}-R(arctan(u)-\frac{u}{1+u^{2}})|_{u=0}=\frac{\pi{R}}{2}
That is:
T=\frac{\pi}{\sqrt{GM}}(\frac{R}{2})^{\frac{3}{2}}
 
Last edited:
Much appreciated Arildno and apologies if this is a little late coming. I hope my double posting was not too imposing- it would have been unnecessary had the original post in the College Homework forum taken a helpful direction.

Regards

Romeo
 
No problem.
I see from your post count that you are fairly new here, so the rule about not double posting have naturally escaped you (you're not alone in this..:wink:).

Welcome to PF, BTW.
 
Thanks Arildno. I was a little concerned about double (incidentely, it was a triple...) posting, but did so only because I thought patrons of the mathematics section may have a better insight- which seemed the case, since you very quickly responded :).

I'll keep it in mind for the future. Until then, if I have a problem that seems to be unresolved and decaying in one forum, is there any precedent for moving the thread to another forum, for fresh ideas?

Regards

Romeo
 
1. First "bump" your post (i.e, write a new reply like "Hello? Culd I have some help here, please?")

2. If that doesn't work, and it is really critical, you might consider PM'ing the moderator for the forum.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top