Falling to a Star With Varying Acceleration

[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.

 

[Janus]

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

Where x is the radius of your neutron star
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.)

 

[Tom MS]

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

Where x is the radius of your neutron star
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.)

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

 

[PeroK]

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

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.

 

[Tom MS]

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.

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?

 

[PeroK]

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?

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)

 

[houlahound]

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

 

[PeroK]

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

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

 

[Tom MS]

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)

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