1-dimensional problem in Newtonian gravity- HELP

[Romeo]

1-dimensional problem in Newtonian gravity- HELP!

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

 

[pervect]

You can think of this as a zero-width elliptical orbit, so you can use Kepler's law relating the orbital period to the length semi-major axis, and then take 1/4 of the orbital period of any orbit with the same semi-major axis.

If you insist on carrying out the integration, my textbook also suggests the substition

r = a(1-e*cos(psi))

in this case I think the eccentricity, e would be 1, so you'd have

r = a(1 - cos(psi))

I haven't tried this out, though.

 

[Romeo]

Thank you pervect, it's an approach I hadn't considered.

Regards

Romeo

 

[arildno]

Aah, the triple post.
This has already been solved in Diff-eq.

 

[CharlesP]

Here is my try at it. Can you check for errors?
A rock is dropped from r_0 and falls straight down to R.
How long does it take. Where R is Earth radius and r_0 >> R.

The gravity force equation F= \frac{GMm}{r^2} ,

integrates to a specific energy equation W= -\frac{u}{r}.

The velocity at some intermediate r is given by:

V=\sqrt{V_0^2+\frac{2u}{r} -\frac{2u}{r_0}} .

This is a hard differential equation to solve for r(t). I believe I have a form which resembles your integral.

Change of symbols:

A=2u,\ \ \ \ B = -\frac{2u}{r_0},\ \ \ \ V_0=0

V(r) = \sqrt{\frac{A}{r}+B}= \frac{dr}{dt}

Now the part which I have never trusted.

\frac{dt}{dr}=\frac{1}{\sqrt{\frac{A}{r}+B}}

Then \int_R^r dt= \int_\frac{A}{R}^\frac{A}{r} \frac{1}{\sqrt{\frac{A}{r}+B}}\ \ dr

So T = \int_\frac{A}{R}^\frac{A}{r} \frac{1}{\sqrt{\frac{A}{r}+B}}\ \ dr

A change of variables \frac{A}{r}=x,\ \ r=\frac{A}{x},\ \ dr=\frac{-A}{x^2}dx

<br /> T = \int_\frac{A}{R}^\frac{A}{r} \frac{\frac{-A}{x^2}}{\sqrt{x+B}}\ \ dx = -A \left[ \frac{-\sqrt{B+x}}{Bx} -\frac{1}{2B} \left( \frac{2}{\sqrt{-B}} \arctan{ \sqrt{\frac{B+x}{-B}}} \right) \right]_\frac{A}{R}^\frac{A}{r}

= \left[ \frac{-r_0 \sqrt{x-\frac{2u}{r_0}}}{x} -\frac{2r_0}{\sqrt{\frac{2u}{r_0}}} \arctan{\sqrt{\frac{x-\frac{2u}{r_0}}{\frac{2u}{r_0}}}} \right]_\frac{2u}{R}^\frac{2u}{r_0} =<br />

= \frac{-r_0 \sqrt{\frac{2u}{r_0}-\frac{2u}{r_0}}}{\frac{2u}{r_0}} -\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{\frac{2u}{r_0}-\frac{2u}{r_0}}{\frac{2u}{r_0}}}} \ <br /> +\ \frac{r_0 \sqrt{\frac{2u}{R}-\frac{2u}{r_0}}}{\frac{2u}{R}} +\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{\frac{2u}{R}-\frac{2u}{r_0}}{\frac{2u}{r_0}}}}<br />

= \frac{Rr_0}{\sqrt{2u}} \sqrt{\frac{1}{R}-\frac{1}{r_0}}+\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{r_0}{R}-1}}= \sqrt{\frac{Rr_0}{2u}} \sqrt{r_0-R} +\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{r_0}{R}-1}}
This of course means you have to work the problem backwards usually as an iteration, but I would use a binary search which I found converges rapidly for the Kepler problem. It should be ok here too.