B Interstellar asteroids

[OmCheeto]

TL;DR Summary

Has anyone solved a similar problem?

I came up with this problem several years ago, and decided I was too old and stupid to solve it. Just curious if anyone has seen such a problem.

 

[Ibix]

Quick approximation: zero. It falls the same distance towards the binary system as it climbs back out, so there's no net change in its energy.

A more detailed calculation depends at least on the orientation of the plane of the binary components' orbits, and probably their eccentricities and the initial speed of the asteroid. It may well need a numerical integrator in general, unless anyone else knows more about test particle orbits in binary systems than I do.

 

[renormalize]

Quick approximation: zero. It falls the same distance towards the binary system as it climbs back out, so there's no net change in its energy.

Are you sure? As I read their (confusing?) diagram, the asteroid starts on the left at a distance of $3\times 10^{13}\,\text{m}$ but is evaluated on the right at $6\times 10^{13}\,\text{m}$ (twice the distance on right as on the left).

 

[Ibix]

Are you sure?

I read it as the blue line in the bottom left being $3\times 10^{13}\mathrm{m}$ and the green arrow across the middle being the twice that length. That's consistent with the total distance travelled being 20 times the distance between the stars, and adds that the distance is supposed to be symmetric about the barycentre. I think my answer is correct on that interpretation, but I agree other interpretations are possible and I would be wrong on yours.

 

[jbriggs444]

A more detailed calculation depends at least on the orientation of the plane of the binary components' orbits, and probably their eccentricities and the initial speed of the asteroid. It may well need a numerical integrator in general, unless anyone else knows more about test particle orbits in binary systems than I do.

We are given initial speed and position. We can assume a perfect circular orbit of the stars about each other. A quick look at the numbers says that the orbital velocities of the two stars exceed the initial velocity of the asteroid by a factor of three. The stars will have revolved more than once before the encounter gets interesting. This breaks the apparent symmetry. The asteroid will be deflected by whichever star it encounters first.

Without running the simulation, both a temporary capture or a quick ejection at an unknown angle with an unknown speed seem plausible.

 

[OmCheeto]

Quick approximation: zero. It falls the same distance towards the binary system as it climbs back out, so there's no net change in its energy.

What about the momentum? Shouldn't the stars both be attracted to the asteroid, move to the left and inward initially and then inward and to the right after the asteroid passes through the origin, O?

A more detailed calculation depends at least on the orientation of the plane of the binary components' orbits, and probably their eccentricities and the initial speed of the asteroid. It may well need a numerical integrator in general, unless anyone else knows more about test particle orbits in binary systems than I do.

The stars orbit each other in a perfect circle. They also orbit each other such one star moves into the page while the other moves out. Thus the asteroid is always on a linear path.

I read it as the blue line in the bottom left being 3×1013m and the green arrow across the middle being the twice that length.

Correct.

 

[OmCheeto]

...the orbital velocities of the two stars exceed the initial velocity of the asteroid by a factor of three

Sorry to laugh, but my calculations said that the asteroid's initial velocity exceeded the two stars velocities by a factor of 6.

Not only was I off by a factor of two, but the reciprocal to boot.

 

[TensorCalculus]

I saw a very, very similar problem in a textbook once: I think it was Ryden (but don't quote me on that). If I can find it I'll paste a picture here.

These sorts of problems certainly aren't uncommon when preparing for astrophysics olympiads...

 

[jbriggs444]

The stars orbit each other in a perfect circle. They also orbit each other such one star moves into the page while the other moves out. Thus the asteroid is always on a linear path.

Ahhh, my mistake. I visualized all motion in the plane of the page. Your way restores the symmetry.

 

[renormalize]

What about the momentum? Shouldn't the stars both be attracted to the asteroid, move to the left and inward initially and then inward and to the right after the asteroid passes through the origin, O?

At only $1\text{ kg}\,$, the asteroid is a negligible test body compared to the $2\times 10^{30}\text{ kg}$ star masses, so it's effect on the stars' motions is nil.

 

[OmCheeto]

At only $1\text{ kg}\,$, the asteroid is a negligible test body compared to the $2\times 10^{30}\text{ kg}$ star masses, so it's effect on the stars' motions is nil.

Negligible, but not zero. I believe this is the crux of the problem. How much the non-zero change of momentum of the stars affects the asteroid on its path.

 

[Ibix]

What about the momentum? Shouldn't the stars both be attracted to the asteroid, move to the left and inward initially and then inward and to the right after the asteroid passes through the origin, O?

I'd still say zero.

A more precise calculation, given that the asteroid is on axis, would be

1. In the initial setup, calculate the position of the center of mass of the three bodies, which will be on axis and slightly to the left of the stars.
2. The center of mass cannot move, so now you can calculate the position of the stars in the final setup, which will be slightly to the left of the center of mass.
3. Compute the change in gravitational potential energy of the asteroid from the different distances. This gives you the velocity change.

You will likely find that you will need to compute $\left((x+\delta x)^2+y^2\right)^{-1/2}$ where $\delta x$ is much, much smaller than either $x$ or $y$. Neglecting the $\delta x^2$ term$$(x+\delta x)^2+y^2\approx r^2\left(1+\frac{2x\delta x}{r^2}\right)$$so a Taylor expansion to first order gives$$\frac{1}{\sqrt{(x+\delta x)^2+y^2}}\approx\frac{1-x\delta x/r^2}{r^2}$$
Note that this is still neglecting any effect that the asteroid has on the orbits of the stars, but I expect that would be negligible compared to this already tiny number.

Edit: forgot to put the square root in the maths, now corrected.

 

[Ibix]

The center of mass cannot move,

This isn't quite true - I forgot the asteroid wasn't initially at rest. The center of mass is moving to the right, which ends up meaning that you need to know how long it takes the asteroid to fly the distance, which is a messier calculation.

 

[sophiecentaur]

At only 1 kg, the asteroid is a negligible test body compared to the 2×1030 kg star masses, so it's effect on the stars' motions is nil.

Gravity assist (slingshot) orbits are achieved for tiny objects going past a vast planet. The 1kg mass does not prohibit momentum transfer from one or both stars. The asteroid's momentum is what counts. You'd need to do the calculation to make sure that the fly through would take a very small part of the binary orbital period. Otherwise, the path of a star could take it close to the asteroid - slingshot situation.

B ut the question appears to be more about GPE and KE at 20 X distance

 

[jbriggs444]

Gravity assist (slingshot) orbits are achieved for tiny objects going past a vast planet. The 1kg mass does not prohibit momentum transfer from one or both stars.

For a transit that follows the rotational axis of the two star system, there is no possibility of a slingshot. We are told that this is the intended scenario.

For a transit in the orbital plane, a slingshot would be a high probability.

But yes, the small mass of the test object does not preclude a slingshot.

 

[renormalize]

Gravity assist (slingshot) orbits are achieved for tiny objects going past a vast planet. The 1kg mass does not prohibit momentum transfer from one or both stars. The asteroid's momentum is what counts.

Yes, but @OmCheeto asked: "Shouldn't the stars both be attracted to the asteroid, move to the left and inward initially and then inward and to the right after the asteroid passes through the origin O?" and I replied regarding the asteroid that "it's effect on the stars' motions is nil" (emphasis added). Do you disagree with that statement?

 

[OmCheeto]

Yes, but @OmCheeto asked: "Shouldn't the stars both be attracted to the asteroid, move to the left and inward initially and then inward and to the right after the asteroid passes through the origin O?" and I replied regarding the asteroid that "it's effect on the stars' motions is nil" (emphasis added). Do you disagree with that statement?

I would guess that it depends on whether you define 'nil' as 'zero' or 'negligible'.

Perhaps I'll analyse the problem with various sets of star masses, much like I did when I found it VERY peculiar that people claimed that regardless of somethings mass, it ALWAYS accelerated towards earth at the same rate. While the maths worked out that EVERYONE but me was correct, it was also very interesting looking at the numbers plotted on a log-log scale. From dirt clods through Sgr A*. Simply fascinating.

 

[Bandersnatch]

It shouldn't matter, should it? You can always analyse the problem in the rest frame of the binary barycentre, and from there the asteroid comes and leaves with the same energy. Similar to how it works with slingshot manoeuvres.

 

[OmCheeto]

It shouldn't matter, should it?

I think it should.

You can always analyse the problem in the rest frame of the binary barycentre, and from there the asteroid comes and leaves with the same energy. Similar to how it works with slingshot manoeuvres.

Exaggerated a bazillion times, while the binary barycenter of stars may remain the same, the component attraction of each star on the asteroid changes, from left to right.

hmmm.... This reminds me of that bizarre thing I learned in university, of how outside a hollow sphere you're gravitationally attracted to said sphere, but once inside: NOTHING!

Boogie man science*, IMHO.

-----
*Yes, I know. The maths worked out for that too. (maybe.......)

 

[sophiecentaur]

I would guess that it depends on whether you define 'nil' as 'zero' or 'negligible'.

You're not wrong. But orbital physics and 'rocket science neglect these things.

This reminds me of that bizarre thing I learned in university, of how outside a hollow sphere you're gravitationally attracted to said sphere, but once inside: NOTHING!

For a solid sphere the force is only nothing at the centre of mass . On the way in, the g is due to the mass of a planet beneath your feet. i.e. the force towards the centre is proportional to the distance to the centre. That's the same rule as for a spring where the restoring force is proportional to the extension. Passing through the mid point of the pair, there would be zero force.

For a transit that follows the rotational axis of the two star system, there is no possibility of a slingshot. We are told that this is the intended scenario.

For a transit in the orbital plane, a slingshot would be a high probability.

But yes, the small mass of the test object does not preclude a slingshot.

I can't find wording in the original problem which was definitely either scenario. I was assuming the transit was in the orbital plane; the other scenario is much easier and probably solvable analytically(?).

 

[jbriggs444]

I can't find wording in the original problem which was definitely either scenario. I was assuming the transit was in the orbital plane; the other scenario is much easier and probably solvable analytically(?).

That clarification was in #6:

The stars orbit each other in a perfect circle. They also orbit each other such one star moves into the page while the other moves out. Thus the asteroid is always on a linear path.

 

[OmCheeto]

... the orbital velocities of the two stars exceed the initial velocity of the asteroid by a factor of three.
...

Would you be willing to double check that? I double checked a second source, and I'm still getting my initial answer.

 

[jbriggs444]

Would you be willing to double check that? I double checked a second source, and I'm still getting my initial answer.

Sure can. I have no reason to be very confident in my numbers.

We have $F = \frac{mv^2}{r} = \frac{Gm^2}{r^2}$ where $F$ is the attractive force of the stars for each other, $r$ is the stellar orbital radius, $m$ is the stellar mass and $v$ is the stellar orbital velocity.

Edit: One can see that I've screwed up this calculation. If $r$ is the orbital radius, the relevant distance for purposes of gravitational force is $2r$. So we need to modify the above equation to:$$F = \frac{mv^2}{r} = \frac{Gm^2}{4r^2}$$
We can solve [wrongly] for $v = \sqrt{\frac{Gm}{r}}$ and take $r = 1.5 \times 10^{12}$ while $m = 2 \times 10^{30}$ and $G = 6.7 \times 10^{-11}$

Edit: The corrected solution $$v = \sqrt{\frac{Gm}{4r}}$$

Double checking units -- kilograms, meters and seconds. No obvious problems there.

I get 9500 m/s this time.

Edit: Corrected to 4250 m/s.

Meanwhile the object is moving at 30000 m/s. So my revised estimate is that the stars are slower than the asteroid by a factor of three six. Not faster as I'd initially suggested.