Crazy physics problem (big debate in my high school)

[littlebhawk]

I did model this using excel accurate to the second, i came out with the correct veloctiy at moment of impact, with gravity equal to 9.81 (like it should be) at moment of impact, with a time of 27916 seconds or about 7.75 hours. Ill post the excel spreadsheet in a little bit I am trying to crunch it down so the file size isn't insanely large ( the spreadsheet comes out to almost 30000 rows!)

 

[awvvu]

I did model this using excel accurate to the second, i came out with the correct veloctiy at moment of impact, with gravity equal to 9.81 (like it should be) at moment of impact, with a time of 27916 seconds or about 7.75 hours. Ill post the excel spreadsheet in a little bit I am trying to crunch it down so the file size isn't insanely large ( the spreadsheet comes out to almost 30000 rows!)

I think our answers agree, if I were to make it end at one earth-radius.

 

[littlebhawk]

Yeah I am sure they do, i finally uploaded the excel spreadsheet as a zip file, its VERY LARGE! and may take your computer a large amoutn of time to open, depending on comp.

http://www.youshare.com/view.php?file=VariableGravityField.zip

You can enter the initial values in the first row and itll work out everything else automatically.

 

[rohanprabhu]

How do you input something like this into Mathematica? I tried:

DSolve[x''[t] == -GM/x[t]^2, x[t], t]

I don't use DSolve generally.. it doesn't solve many problems. See.. we got this:

$$v = \sqrt{2} \sqrt{\frac{GM + rY}{r}}$$

$$\frac{dr}{dt} = \sqrt{2} \sqrt{\frac{r}{GM + rY}}$$

$$\sqrt{\frac{r}{GM + rY}}dr = \sqrt{2} dt$$

This is what I integrated in Matematica using:

Integrate[Sqrt[(r)/(G M + r Y)], r]

 

[rohanprabhu]

After a lot of battling with equations and a few mountain dews later.. I've arrived at a solution.. which i believe is quite correct. So.. i request you to please take a look and point out flaws [if any]:

Assumptions for this problem: Take for any arbitrary distance 'r' from the center, let the center be the origin.
Also, instead of taking the starting position as $10R_E$, I'm assuming an arbitrary position 's'.

The acceleration of the object is given by:

$$a = -\frac{GM}{r^2}$$

$$\frac{dv}{dt} = -\frac{GM}{r^2}$$

$$\frac{dv}{dr} \frac{dr}{dt} = -\frac{GM}{r^2}$$

$$vdv = -\frac{GM}{r^2}dr$$

$$\frac{v^2}{2} = \frac{GM}{r} + C$$

also, at $r = s$, $v = 0$ Hence,

$$GM + sC = 0$$

$$C = -\frac{GM}{s}$$

So, we have:

$$v = \sqrt{2GM} \sqrt{\frac{1}{r} - \frac{1}{s}}$$

Let, us just take out the $\sqrt{2GM} = \Phi$. We'll call it the 'planetary mass factor' [yup.. it's my turn now to use fancy terms :P], So, we have the 'planetary mass factor' as:

$$\Phi = \sqrt{2GM}$$

so..

$$\frac{dr}{dt} = \Phi \sqrt{\frac{1}{r} - \frac{1}{s}}$$

let the above equation be eqn. 1.

$$\sqrt{\frac{sr}{s - r}}dr = \Phi dt$$

Let,
$$\frac{sr}{s - r} = x$$

$$\frac{s^2}{(s - r)^2}dr = dx$$

Putting the values in eqn. 1, we get:

$$\frac{(s - r)^\frac{3}{2}}{s^2}dx = \Phi dt$$

Also, from the first assumption,

$$r = \frac{sx}{s + x}$$

So,

$$\frac{dx}{s + x} = \Phi dt$$

$$log_e|s + x| = \Phi t + J$$

J is again, a constant of integration. Now, substituting the value of 'x' in terms of r, we get:

$$\frac{s^2}{s - r} = e^{\Phi t + J}$$

$$\frac{s^2}{s - r} = e^J e^{\Phi t}$$

$$\frac{s^2}{s - r} = ke^{\Phi t}$$

On simplifying it a bit,

$$r(t) = s\left(1 - \frac{s}{ke^{\Phi t}}\right)$$

Now, r(0) = s . Solving for 'k',

$$k = 2s$$

Putting it in the equation, we get the answer as:

$$r(t) = s\left(1 - \frac{1}{2e^{\Phi t}}\right) = s\left(1 - \frac{1}{2e^{\sqrt{2GM} t}}\right)$$

This is the equation of motion I got. :D

 

[Andy Resnick]

After a lot of battling with equations and a few mountain dews later.. I've arrived at a solution.. which i believe is quite correct. So.. i request you to please take a look and point out flaws [if any]:
<snip>

Putting it in the equation, we get the answer as:

$$r(t) = s\left(1 - \frac{1}{2e^{\Phi t}}\right) = s\left(1 - \frac{1}{2e^{\sqrt{2GM} t}}\right)$$


This is the equation of motion I got. :D

I don't think this is correct- GM has units of L^3/T^2, so $$\sqrt{2GM}t$$ is not dimensionless. Not sure where the initial error occurs...

 

[K.J.Healey]

The integration looks wrong, in mathematica:

F[r_] = Expand[Integrate[(1/r - 1/s)^(-1/2), r]]

$$-\frac{1}{2} \tan ^{-1}\left(\frac{\sqrt{\frac{1}{r}-\frac{1}{s}} (2 r-s) \sqrt{s}}{2 (r-s)}\right) s^{3/2}-r \sqrt{\frac{1}{r}-\frac{1}{s}} s$$

 

[rahuldandekar]

Ok, people have tried excel, mathematica, maple...

I tried Python, pure and simple. No Runge-Kutta methods or anything, just pure dv = a*dt and dx = v*dt ...

I got the following results:

Final Velocity = 10607.0 m/s
Total Time Taken = 27943.16 seconds
Actual Final Velocity (from formula) = 10607.1 m/s

I think these confirm the earlier answers... I admit that trying to find the analytical solution is a much more interesting job ;)

 

[qspeechc]

What about my effort on page 2? Anyone bother to go over that?

 

[rahuldandekar]

I should've noticed it, gspeechc... I have been thinking along the exact same lines during the last hour. I'm too tired to develop it now, but I think that expression [ dx/ (lot of stuff) ] does have an anti derivative...

You made a mistake there, "x-10" should be replaced by x*10 (when cross multiplying R/x - 1/10) . I have been trying to get an expression substituting 10*R - x = z^2 (or -z^2 ... I haven't worked out the signs yet). Anyway, with such a subtitution, I find it becomes a solvable integral. Either that or I have made mistakes (more probable, that).

 

[D H]

In general,

$$\Delta t = \int_{r_i}^{r_f}\frac{dr}{v(r)}$$

where $r_i$ and $r_f$ are the initial and final distances and $v(r)$ is the velocity (dr/dt) expressed as a function of r.

Combining conservation of energy and gravitational potential energy expressions yields an expression for velocity in terms of radial distance:

$$v^2-v_0^2 = 2GM\left(\frac 1 r - \frac 1 {r_i}\right)$$

Here the velocity is negative (r is decreasing) until the object hits the Earth and the initial velocity is zero. Thus

$$v = -\sqrt{2GM}\sqrt{\frac{r_i-r}{r \, r_i}}$$

Applying this result to the initial integral (note the change in integration order to accommodate the negative velocity):

$$\Delta t = \frac 1 {\sqrt{2GM}} \int_{r_f}^{r_i}\sqrt{\frac{r \, r_i}{r_i-r}} \; dr$$

Using the the substitution $r=r_i\sin^2\phi$ ($\phi = \sin^{-1}\surd{\frac r {r_i}}$) and defining $\kappa\equiv r_f/r_i$,

$$\Delta t = \sqrt{\frac {r_i^3} {2GM}}<br /> \int_{\sin^{-1}\surd{\kappa}}^{\pi/2}2\,\sin^2\phi \; d\phi$$

from which

$$\Delta t = \sqrt{\frac {r_i^3} {2GM}}<br /> \left(\sqrt{\kappa(1-\kappa)} + \cos^{-1}\surd{\kappa}\right)$$

With $r_i=63781.37\,\text{km}$, $GM_e = 398600.4418\,\text{km}$, and $\kappa = 1/10$, the above evaluates to 7.7628 hours = http://www.google.com/search?hl=en&...^2))*(sqrt(0.09)+acos(sqrt(0.1)))+in+seconds". (Link to Google calculator result highlighted)

 

[rohanprabhu]

Combining conservation of energy and gravitational potential energy expressions yields an expression for velocity in terms of radial distance:

$$v^2-v_0^2 = 2GM\left(\frac 1 r - \frac 1 {r_i}\right)$$

Here the velocity is negative (r is decreasing) until the object hits the Earth and the initial velocity is zero. Thus

$$v = -\sqrt{2GM}\sqrt{\frac{r_i-r}{r \, r_i}}$$

absolute gem.. priceless. Great work!

and yes.. my method is wrong. The integration I did is wrong.. as pointed by a member in a previous post than this. Also, it is wrong for obvious reasons:

i] $\sqrt{2GM}t$ is used in exponentiation and is not dimensionless [pointed by Andy Resnick]
ii] The equation suggests that the object will gradually move slower.. whereas it should gradually move faster.

I did a really stupid mistake in my integration. While writing down.. i did not copy the 3/2 power and hence it came out as a simple equation derivate of Log[f[x]]...

 

[awvvu]

In general,

$$\Delta t = \int_{r_i}^{r_f}\frac{dr}{v(r)}$$

This equation seems pretty useful. Where's it derived from?

 

[rohanprabhu]

This equation seems pretty useful. Where's it derived from?

That is the 'definition' of velocity:

$$v = \frac{dr}{dt}$$

 

[awvvu]

That is the 'definition' of velocity:

$$v = \frac{dr}{dt}$$

Oh, doh.

 

[D H]

$$\Delta t = \sqrt{\frac {r_i^3} {2GM}}<br /> \int_{\sin^{-1}\surd{\kappa}}^{\pi/2}2\,\sin^2\phi \; d\phi$$

from which

$$\Delta t = \sqrt{\frac {r_i^3} {2GM}}<br /> \left(\sqrt{\kappa(1-\kappa)} + \cos^{-1}\surd{\kappa}\right)$$

Just to stave off questions:

First, the parameter $\phi$ is obviously in the range$[0,\pi/2]$. Phi starts at pi/2 at the release point and would drop to zero at the center of the Earth if the Earth's surface didn't stop the object first.

I integrated $2\,\sin^2\phi$ is done by the means of the identity $2\,\sin^2\phi=1-\cos(2\,\phi)$, yielding $\phi-1/2\sin(2\,\phi)$. Evaluating wrt the limits leads to (phi part first) $\pi/2-\sin^{-1}\surd\kappa$. This simplifies to $\cos^{-1}\surd\kappa$ via the $\cos^{-1}x+sin^{-1}x = \pi/2$. To evaluate the sine term I used $\cos(\sin^{-1}x) = \sqrt{1-x^2}$ for $x\in[0,\pi/2]$.

 

[rahuldandekar]

I got it till the integration, and then made another substitution which didn't seem to yield a right answer... that teaches me to neglect trigonometric substitutions! ;)

That methods of yours really is a gem, DH. Especially the way you handled the phis and the kappas ;).

 

[D H]

That methods of yours really is a gem, DH. Especially the way you handled the phis and the kappas ;).

Thanks. 'Twas a fun little problem.