I Falling to a Star With Varying Acceleration

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1. Apr 9, 2016

Tom MS

Take an object of negligible mass that is dropped from rest 2 kilometers away from a neutron star of mass 1.989*10^30 kilograms (1 solar mass) and radius 7,802 meters. How long will it take the object to reach the surface of the neutron star?

I'm not terrible at calculus, but I know for a fact that a problem such as this involves some integration that I haven't been able to pin down.

2. Apr 9, 2016

Janus

Staff Emeritus
$$t = \frac{ \cos^{-1} \left ( \sqrt{\frac{x}{r}} \right ) + \sqrt{\frac{x}{r} \left ( 1-\frac{x}{r} \right ) }}{\sqrt{2 \mu}} r^{\frac{3}{2}}$$

r is the distance of the object from the center of the neutron star
$\mu$ is equal to G(M+m) with M+m being the sum of the neutron star and object masses. (if m is really small with respect to M, you can ignore it without sacrificing accuracy too much.)

3. Apr 9, 2016

Tom MS

Thank you! Do you have a derivation for this that I could look at?

4. Apr 9, 2016

PeroK

There was one on here but it was a long time ago and I can't find it. I've got the solution written down if you are interested.

How much maths do you know? It's really just a solution to a differential equation. It's not too hard but involves a trick or two.

5. Apr 9, 2016

Tom MS

I'm taking high school level AP calculus right now, but I'd be happy to see the solution! In terms of differential equations, I only know how to solve simple linear differential equations and separable differential equations.

Also, I just realized. Why couldn't I just use conservation of energy?

6. Apr 9, 2016

PeroK

Yes, you can use conservation of energy, but that's not all you need. I'll show you both methods. I've got slightly different notation:

$M$ is the mass of the large body, $R$ is the radius of the large body, $r_0$ is the starting position of the falling mass, $r_1$ is the final position of the falling mass, where $r_1 \ge R$, and $r$ is the variable position of the falling mass. And, $r'(0) = 0$ meaning the falling mass starts from rest.

First, you have the differential equation for $r$:

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

The first trick is to multiply both sides by the integrating factor $2r'$ to give:

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

$\frac{d}{dt}(r')^2 = \frac{d}{dt}(\frac{2GM}{r})$

$(r')^2 = \frac{2GM}{r} + C = \frac{2GM}{r} - \frac{2GM}{r_0} = 2GM(\frac{r_0 - r}{rr_0})$ (using the initial position to find $C$)

This gives us the first equation:

$(r')^2 = 2GM(\frac{r_0 - r}{rr_0})$ Equation (1)

The alternative, as you suggested, is to use conservation of energy:

$-\frac{GMm}{r_0} = -\frac{GMm}{r} + \frac{1}{2}m(r')^2$

Which gives you equation (1) more easily.

The main trick is a non-obvious substitution:

Let $r = r_0 cos^2\theta$ hence $r' = -2r_0(cos\theta sin\theta)\theta'$

Substituting this into (1) gives:

$4r_0^2(sin^2 \theta cos^2 \theta)(\theta')^2 = 2GMr_0(\frac{1 - cos^2 \theta}{r_0^2 cos^2 \theta}) = \frac{2GMsin^2 \theta}{r_0 cos^2 \theta}$

$cos^4 \theta (\theta')^2 = \frac{GM}{2r_0^3}$

$cos^2 \theta (\theta') = \sqrt{\frac{GM}{2r_0^3}}$ Equation (2)

Now we integrate this with respect to $t$, noting that at $t = 0, r = r_0, \theta = 0$ and at $t = t_1, r = r_1, \theta = \theta_1$

$\int_0^{\theta_1} cos^2 \theta d \theta = \sqrt{\frac{GM}{2r_0^3}}t_1 + C$

$\frac{1}{4}[2\theta_1 + sin2\theta_1] = \sqrt{\frac{GM}{2r_0^3}}t_1$ (as $C = 0$)

This gives us essentially an intermediate solution for $t_1$:

$t_1 = \sqrt{\frac{r_0^3}{8GM}}[2\theta_1 + sin2\theta_1]$ Equation (3)

Where $r_1 = r_0 cos^2 \theta_1$ hence $\theta_1 = cos^{-1} \sqrt{\frac{r_1}{r_0}}$

Now, of course, you can use trig identities to give $sin 2\theta_1$ in terms of $r_0, r_1$:

$sin2\theta_1 = 2cos\theta_1 sin\theta_1 = 2 cos\theta_1 \sqrt{1- cos^2\theta_1} = 2 \sqrt{\frac{r_1}{r_0}} \sqrt{1- \frac{r_1}{r_0}}$

This gives you the formula that Janus posted:

$t_1 = \sqrt{\frac{r_0^3}{2GM}}[cos^{-1} \sqrt{\frac{r_1}{r_0}} + \sqrt{\frac{r_1}{r_0}(1- \frac{r_1}{r_0})}]$ Equation (4)

Although, this is actually a more general formula for time to fall from any distance $r_0$ to any other distance $r_1$.

There is a special case worth mentioning. If the object falls from a long way away and you model the large mass as a point mass, then $r_1 = 0$ and you get a good estimate for the time it takes the two objects to collide. The exact equation would, of course, have $r_1 = R$:

$t_1 = \sqrt{\frac{r_0^3}{2GM}}[\frac{\pi}{2} + 0] = \pi \sqrt{\frac{r_0^3}{8GM}}$ Equation (5)

Last edited: Apr 9, 2016
7. Apr 9, 2016

houlahound

Is it valid to divide the distance r by t above to get the speed of entry?

8. Apr 9, 2016

PeroK

No. You can get the speed at any point simply from conservation of energy.

9. Apr 9, 2016

Tom MS

Wow. Thank you I actually understand that. That is really awesome.