What is orientation of a needle shaped spacecraft?

[sayebms]

this is a small part of a problem on tidal forces and I wasn't sure what the question asks as it seems to me that more information is needed. Am I right or is there something I am missing? the question goes as:

"A spacecraft approaches a neutron star of radius 10 km and mass 1.5 times mass of sun. what is its likely orientation at its closes approach of 20 km from the center of the star?"

I appreciate someone clarify what the question is asking.
Thank you.

 

[DrStupid]

I appreciate someone clarify what the question is asking.

The question seems quite obvoius: Will the spaceship be orientated horizontally or vertically ore something between at its closes approach? However, I do not know if this question can be answered without the starting orientation and angular momentum.

 

[Dale]

it seems to me that more information is needed.

No, actually much less information is needed. None of the details of the mass or distance matter. The question could be simplified:

"A spacecraft approaches close to a neutron star, what is its likely orientation at its closest approach?"

Think about what makes one state more likely than another state.

 

[jbriggs444]

No, actually much less information is needed. None of the details of the mass or distance matter. The question could be simplified:

"A spacecraft approaches close to a neutron star, what is its likely orientation at its closest approach?"

There are two ways of answering this that give opposing answers. Is the rotation of the approaching spacecraft damped in any way? [Larry Niven had to admit that Beowulf Schaeffer does not survive].
Please ignore the above. My recollection of the sequence of events was erroneous. There is an unambiguous answer.

 

[Bystander]

[Larry Niven had to admit that Beowulf Schaeffer does not survive].

Huh? PM me w' the explanation ... we don't need to give away answers.

Edit: Well, shucks ... trying to figure out what I'd missed in the Fleet of Worlds n-volume whatever it is.

 

[jbriggs444]

Edit: Well, shucks ... trying to figure out what I'd missed in the Fleet of Worlds n-volume whatever it is.

Tried to PM, failed: "You may not start a conversation with the following recipients: Bystander."

 

[sayebms]

After much thinking I have found that we need to use the equation for tidal forces acting on the spaceship as it approaches the star. and it appears that there is a big difference (a few hundred g's if i remember correctly) in gravitational acceleration between tip of the spaceship and its tail hence the ships orientations is most likely radial towards the center of the star.

Thank you.

 

[Dale]

hence the ships orientations is most likely radial towards the center of the star.

That is correct. You can also come to the same realization by considering potential energy. The radial orientation has a lower PE.

 

[DrStupid]

You can also come to the same realization by considering potential energy. The radial orientation has a lower PE.

How does that lead to the maximum propability for the vertical position independent from the starting conditions? The the lower potential energy results in an increased kinetic energy and therefore in an increased angular velocity compared to the horizontal position. Shouldn't that result in an increased propability for the horizontal position in case of a rotating ship?

 

[Dale]

How does that lead to the maximum propability for the vertical position independent from the starting conditions?

It is just an observation that a system is most likely to be found in the lowest energy state. For a gravitational system that is the tidally locked state with the long axis pointed towards the primary.

 

[DrStupid]

It is just an observation that a system is most likely to be found in the lowest energy state.

The gravitational potential energy of a planet in an elliptic orbit is a simple counter example. What makes you sure that the potential energy resulting from the tidal forces acting on the rocket is not another exception?

For a gravitational system that is the tidally locked state with the long axis pointed towards the primary.

Is the rocket tidally locked during the closest approach?

 

[Dale]

The gravitational potential energy of a planet in an elliptic orbit is a simple counter example

It isn't a counter example. Even for a planet in an elliptical orbit the most likely configuration is tidally locked. It certainly doesn't mean that all planets in elliptical orbits are tidally locked, just that it is more likely than any other configuration.

Is the rocket tidally locked during the closest approach?

We don't know, but it is more likely than any other possibility.

 

[rcgldr]

Assuming there's no internal friction related to rotation of the spacecraft , would it tend to oscillate or be out of sync with respect to its path, similar to the moons libration (the moon rotates about about the same rate, but it's orbit is slightly elliptical).

 

[DrStupid]

Even for a planet in an elliptical orbit the most likely configuration is tidally locked.

That's not what I was talking about. It is more likely to find the planet in a position with a higher potential energy - no matter if it is tidally locked or not.

We don't know, but it is more likely than any other possibility.

I would like to see your calculation.

 

[jbriggs444]

Consider a pendulum in free fall in an elevator shaft. We do not know its initial orientation. Assume that its initial angular velocity is small. We apply a small upward force on the axis about which the pendulum is free to swing. The pendulum begins swinging back and forth. As time passes we gradually increase the upward force that is applied. What effect does this have on the amplitude of the swings of the pendulum (measured in displacement from the "down" orientation)?

What is the likelihood of finding the pendulum at 90 degrees from straight down after the upward force has been increased by a factor of 1000 from its starting value?

 

[Dale]

That's not what I was talking about. It is more likely to find the planet in a position with a higher potential energy - no matter if it is tidally locked or not.

OK, but that isn't relevant to the thread. The thread was asking about the orientation at the position of closest approach, which is most likely to be "vertical" because the tidally locked configuration is the most likely.

I would like to see your calculation

Well, in the solar system there are 173 natural satellites. Of those, 34 are tidally locked, so that is just under 20% likelihood. So the orientation at closest approach for all of those is vertical. All other orientations have less than 1% likelihood. So at least in our solar system it is the most likely orientation empirically.

 

[Delta Kilo]

What is the likelihood of finding the pendulum at 90 degrees from straight down after the upward force has been increased by a factor of 1000 from its starting value?

Apparently, not likely at all. Just type x''(t) = -(g+k t) x(t) into wolfram alpha and it will give you the answer which turns out to be a pair of Airy functions which look like this, except there is a minus sign in the argument so time goes towards negative x on this graph:

The amplitude gradually decreases while the frequency increases indefinitely (or rather until the whole thing shatters to pieces)

 

[DrStupid]

What is the likelihood of finding the pendulum at 90 degrees from straight down after the upward force has been increased by a factor of 1000 from its starting value?

As the pendulum will get closer to its stable equilibrium (hanging straight down) the likehood is very low. This is indeed a good point. It suggests that strongly increasing tidal forces will result in an oscillation close around the vertical orientation of the ship during the closest approach. However, this argumentation requires that the period of the oscillation is short compared to the time of the flyby and therefore also depends on the starting conditions. Without a corresponding calculation (or at least rough estimation) I cannot say if this is the case - especially if the process starts near the instable equilibrium (horizontal orientation of the ship or standing upside-down in case of the pendulum analogy).

OK, but that isn't relevant to the thread.

It might be irrelevant for the OP but not for the thread. The example shows that the "observation that a system is most likely to be found in the lowest energy state" can't be generalized to the potential energy of any system. Thus you need to show that it applies to the OP if you want to use it in your argumentation.

The thread was asking about the orientation at the position of closest approach, which is most likely to be "vertical" because the tidally locked configuration is the most likely.

In contrast to jbriggs444 you did not yet provided a proper justification for this claim. As you do not refer to the given details regarding mass and distance (which are important for his argumentation) you need another explanation.

So at least in our solar system it is the most likely orientation empirically.

The most likely orientation of bullets with hyperbolic trajectories would be different (just to give you a single counter example). Why need a spacecraft approaching a neutron star needs to behave like a natural sattellite in the solar system and not like the bullets?

 

[Dale]

The example shows that the "observation that a system is most likely to be found in the lowest energy state" can't be generalized to the potential energy of any system. T

I agree with you on this point now. I think that I have overstated it above.

What we can say is something along the lines that the tidally locked state is the most likely state for gravitational system. There might need to be some further qualifications also, but I am not certain of that.

you need another explanation.

I don't know why. I observe the fact that the tidally locked orientation is the most likely orientation. From that fact I can immediately answer the OP. The question isn't asking which orientation it actually has, just which is most likely.

The most likely orientation of bullets with hyperbolic trajectories would be different (just to give you a single counter example)

I don't think that is a counterexample. What makes you believe that the most likely orientation for hyperbolic trajectories would be different than for elliptical trajectories?

 

[DrStupid]

What we can say is something along the lines that the tidally locked state is the most likely state for gravitational system.

I'm still missing a proper justification for this claim. A statistic based on a non-representative selection of objects is not sufficient. You need to show it for any kind of gravitational systems or at least for the system we are talking about.

What makes you believe that the most likely orientation for hyperbolic trajectories would be different than for elliptical trajectories?

Because there is not anough time to change the initial orientation and angular momentum of the bullet during a single flyby in the gravitational field of a body in the solar system.

 

[Dale]

A statistic based on a non-representative selection of objects is not sufficient.

What makes it non representative? The same gravitational laws apply. Tidal forces still exist. The objects still don't have perfect spherical symmetry. There is a difference in scale, certainly, but not such a difference that we would expect any new quantum effects. So on what basis is it non representative.

Because there is not anough time to change the initial orientation and angular momentum of the bullet during a single flyby in the gravitational field of a body in the solar system.

I understand your point, but again, it is not asking for which orientation it will definitely have, but just which is most likely. There will be tidal forces, even during a single flyby. With such a short time it may not be >20 times as likely, but it should still be ever so slightly more likely than any other orientation.

 

[DrStupid]

What makes it non representative?

It consists of objects in close orbits only.

There will be tidal forces, even during a single flyby. With such a short time it may not be >20 times as likely, but it should still be ever so slightly more likely than any other orientation.

Why?

 

[Bystander]

Why?

This happens if the spaceship distorts/warps in any way under the influence of the tidal force(s); no distortion means no energy/work, and no lock.

 

[DrStupid]

This happens if the spaceship distorts/warps in any way under the influence of the tidal force(s); no distortion means no energy/work, and no lock.

Distortion is neither required (see jbriggs444's argumentation with the pendulum) nor sufficient for tidal locking (Earth is distorted but not locked).

 

[Dale]

Why?

Because the tidal forces are still present, even in a single flyby.

 

[jbriggs444]

I am somewhat lost trying to follow the discussion. My best guess is that Dale is making a kind of argument from thermodynamic equilibrium -- that the lowest potential energy state is the highest probability state. That argument seems patently unsound. We have an example in astronomy where the lowest potential energy state (the low point in an orbit) is the lowest probability. Satellites spend more time high than low.

If the object were rotating rapidly as it hits its low point near the neutron star then it would tend to spend less time vertical (spinning more rapidly) and more time horizontal (spinning less rapidly). However, it is pretty clear that the intended interpretation of the problem does not involve insanely high initial rotation rates.

Nonetheless, I would refer you to https://en.wikipedia.org/wiki/Neutron_Star_(short_story):

• In the "Afterthoughts" section of the Tales of Known Space collection, Niven writes: "I keep meeting people who have done mathematical treatments of the problem raised in the short story 'Neutron Star', ... Alas and dammit, Shaeffer can't survive. It turns out that his ship leaves the star spinning, and keeps the spin."

 

[Dale]

that the lowest potential energy state is the highest probability state. That argument seems patently unsound.

Yes, I agree that it is unsound and I backed off from that.

Nonetheless, it is empirically clear that a tidally locked configuration is more probable than any other configuration.

 

[DrStupid]

However, this argumentation requires that the period of the oscillation is short compared to the time of the flyby and therefore also depends on the starting conditions. Without a corresponding calculation (or at least rough estimation) I cannot say if this is the case - especially if the process starts near the instable equilibrium (horizontal orientation of the ship or standing upside-down in case of the pendulum analogy).

For the case that the ship is small compared to the distance, the angular acceleration can be estimated with the quadrupole moment of the gravitational field:

$\ddot \varphi = - 3 \cdot M \cdot G \cdot \frac{{\left( {r \times R} \right) \cdot \left( {r \cdot R} \right)}}{{r^2 \cdot \left| R \right|^5 }}$

The minimum period of the resulting oscillation is

$T < 2 \cdot \pi \cdot \sqrt {\frac{{\left| R \right|^3 }}{{3 \cdot M \cdot G}}}$

As this is only by the factor sqrt(3) below the duration of a full circular orbit in the same distance, the requirement for the pendulum analogy is not fulfilled. The ship doesn't oscillate during the flyby.

As I do not have another idea how to answer this question without calculation, I tried to solve the equations of motion numerically. But all tested integrators failed to solve the three dimensional problem (x,y,orientation) for the conditions given by the OP (I didn't even try to simulate rotations around more then one axis). Thus I turned to polar coordinates and used the equation

$\left| R \right| = R_{\min } \cdot \frac{{1 + \varepsilon }}{{1 + \varepsilon \cdot \cos \vartheta }}$

for the trajectory in order to reduce the problem to the orientation only. Then I eliminated the time to avoid the numerical problems resulting from the high velocities and accelerations during the close flyby. The resulting differential equation was a surprise (at least for me):

$\varphi '' = \frac{{\varphi ' \cdot 2 \cdot \varepsilon \cdot \sin \vartheta - 3 \cdot \left( {\cos \varphi \cdot \sin \vartheta - \sin \varphi \cdot \cos \vartheta } \right) \cdot \left( {\cos \varphi \cdot \cos \vartheta + \sin \varphi \cdot \sin \vartheta } \right)}}{{1 + \varepsilon \cdot \cos \vartheta }}$

I didn't expected it to be independent from the mass of the star and the periapsis distance. However, it still depends on the starting conditions. As we don't have any information about possible starting conditions and their probabilities, no general answer can be given. Thus I tried to solve the problem at least for the following special starting conditions:

- Start in Apoapsis or infinity
- No starting angular velocity
- Equally distributed starting orientations

The result was even more surprising:

There actually are preferred orientations for eccentricities below 2, but they are not vertically as expected in most of the posts above. Instead the ship is most likely turned out of the vertical orientation by at least 18°. This is so very counterintuitive that I initially expected an error in my calculation. Thus I repeated it several times with different derivations and numerical methods but I always get the same result.

 

[Dale]

Instead the ship is most likely turned out of the vertical orientation by at least 18°. This is so very counterintuitive that I initially expected an error in my calculation. Thus I repeated it several times with different derivations and numerical methods but I always get the same result.

Hmm, that is very interesting.

 

[anorlunda]

Great thread. I just wish that the problem was stated more clearly so that I could be sure that all the posters were commenting on the same scenario.

My initial read of the OP was that there would be no orbit but rather a direct hit with the spacecraft headed toward the COM of the star. And if the question was multiple choice, the choices would be

1. Mostly horizontal
2. Mostly vertical
3. No preference

 

[DrStupid]

I just wish that the problem was stated more clearly

Yes, that was also my first thought.

but rather a direct hit with the spacecraft headed toward the COM of the star.

In this case I would expect the word "impact" instead of "closes approach". But you are right, this is a possible interpretation.