Stellar Oscillations across Spiral Arms

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The way a comet can come to have enough energy or velocity to get into the gravity field of the sun is if it(the comet) came from a supernova event.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> It's not a question of whether interstellar comets can come <u>into</u> the Sun's gravity field; its whether they can lose sufficient energy to <u>remain</u> in the Sun's gravity field.

If Comet X is already traveling one kilometer/sec (relative to the Sun) when it enters the Oort Cloud, it will need to shed 1,000,000 Joule of energy per kilogram of mass; otherwise it will pass the Sun (as some very high speed*) and continue outward, leaving the Oort cloud behind at a speed of one kilometer/sec. Passing close to the left side of a planet (north being up), could transfer that much energy from the comet to the planet.

A Comet Y, blown out by a supernova might enter the Oort cloud at 100 km/s, so it would need to shed 10,000,000,000 Joule/kg in order to be captured by our solar system. Nothing less than a fiery collision would take away that much energy.

*You probably want to know how fast these comets will pass the Sun. Start by assuming some perihelion radius (from the Sun's center to the comet at perihelion); plug that value into the Escape Velocity Calculator . (In the box labeled "Radius (m)" plug in the perihelion radius, not the radius of the Sun. Actually, that calculator is not working for me, right now; I'll check it out and get back to you.) If you need to convert from stone-age USA units to metric, try the Conversion Calculator . Wait a minute; don't just add 1 km/sec to the escape velocity at perihelion. Instead, you have to calculate the square of the escape velocity at perihelion; then add 1,000,000 Joule/kg for Comet X or 10,000,000,000 Joule/kg for Comet Y; finally, take the square root of that sum. (Note: energy/mass is equivalent to velocity squared.)
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">...trash out-garbage in ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Aren't you glad we are so bad at recognizing trash when we see it. Otherwise, we'd have nothing to argue about. So, as the Ruskies say, "Trust but verify!"

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Actually, that calculator is not working for me, right now;... <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Not sure what I was doing wrong; it seems to work, now. More GIGO, I guess. If you have a slow internet connection, don't visit the homepage hosting that escape velocity calculator; you'll have to wait while it downloads background music---ten minutes of <i>PinkPanther.wav</i>, and no way to turn it off.

Here's a better source for Online Conversions . That other link doesn't include miles or kilometers; this one has just about everything you can think of.

Tom, you're the expert on orbital mechanics around these here parts. I'm having doubts about the accuracy of my earlier assertions that a comet gains energy by passing on the right side of a planet (north being up) and loses energy if it passes on the left. It didn't occur to me that some comets (especially interstellar ones) might have retrograde trajectories; i.e., pass on the left side of the Sun.

Now, I'm thinking it depends on whether the comet passes ahead of or behind the planet. For the sake of clarity, I'm talking about the comet's position relative to the Sun & planet at the point of closest approach to the planet. Consider a vector from the planet's center toward the comet at closest approach. If the eclyptic-plane component of that vector points to the right of the Sun (north being up), then the comet passes ahead of the planet.

If the comet passes ahead of the planet, it will tend to pull the planet foreward, so the planet gains energy; the planet's gain is the comet's loss; so the comet must then lose energy. If the comet passes behind the planet, then the comet gains energy. Have I got it right, this time? Am I even close?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by PhilJ</i>
<br />I'm having doubts about the accuracy of my earlier assertions that a comet gains energy by passing on the right side of a planet (north being up) and loses energy if it passes on the left. It didn't occur to me that some comets (especially interstellar ones) might have retrograde trajectories; i.e., pass on the left side of the Sun.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Not only that, but your "right" and "left" are undefined. Whose right or left? As seen from the Sun, or as viewed from far away in the direction the comet came from? You also neglected to specify if the comet was inbound or outbound. I assumed inbound. But it is equally likely to encounter a planet only on the outbound leg of its orbit.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Now, I'm thinking it depends on whether the comet passes ahead of or behind the planet.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That at least solves the right-left ambiguity.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If the ecliptic-plane component of that [closest approach] vector points to the right of the Sun (north being up), then the comet passes ahead of the planet.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This implies you are looking at an inbound comet from a deep space perspective. Is that true? Most astronomers would adopt a Sun perspective.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If the comet passes ahead of the planet, it will tend to pull the planet foreward, so the planet gains energy; the planet's gain is the comet's loss; so the comet must then lose energy. If the comet passes behind the planet, then the comet gains energy. Have I got it right, this time? Am I even close?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'd express it this way:

** If an inbound comet has its closest approach on the day side of a planet, or if an outbound comet has its closest approach on the night side of a planet, the comet receives a net increase in its solar orbital velocity and therefore a net energy gain relative to the Sun.

** If an inbound comet has its closest approach on the night side of a planet, or if an outbound comet has its closest approach on the day side of a planet, the comet receives a net decrease in its solar orbital velocity and therefore a net energy loss relative to the Sun.

Technically, the primary effect of the planet is usually on the comet's angular momentum. But we can ignore that and continue to speak of net energy for the sake of keeping things simple. We have also ignored transition cases where the planet switches a comet from prograde (direct) motion to retrograde, or vice versa. The impulse is assumed to be small.

Finally, note that there is an important asymmetry to be considered. Impulses to the comet are larger when the comet passes behind the planet because the comet then stays near the planet longer. That is why capture of Jupiter-family (low-inclination) comets is more common than capture of retrograde comets such as Halley.

I hope I did not overwhelm you with details. But nobody claims that orbital mechanics are very intuitive. -|Tom|-

Some of your response does run counter to my intuition. I'll need some time to digest it. If I counter respond too quickly, I'll only cause myself embarrassment.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by PhilJ</i>
<br />Some of your response does run counter to my intuition. I'll need some time to digest it. If I counter respond too quickly, I'll only cause myself embarrassment.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Here are some possible aids to dynamical intuition. The virial equation for 2-body orbits is: v^2 = GM (2/r-1/a), where v is velocity, G = gravitational constant, M = central mass (the Sun for us), r = solar distance, a = semi-major axis. (By Kepler's law, 4 pi^2 a^3 = GM P^2, where pi = 3.1416 and P = orbital period. So knowing P implies -a- and vice versa.)

The virial equation is an energy equation. v^2 is proportional to kinetic energy, -2GM/r is proportional to potential energy, and their sum -GM/a is the total energy. This last quantity is set to be zero for a parabolic orbit (always at escape velocity), negative for ellipses, and positive for hyperbolas. The key point to notice is that a comet's energy depends only on its distance and speed, but not on its direction of travel at that distance.

If a small planet perturbation enters the picture, the whole passage of the comet past the planet can be approximated as a single impulse at the point of closest approach. That impulse must be directed toward the planet. Such an impulse changes mainly the comet's speed. The change in solar distance at the instant of the impulse is usually negligible.

With these precepts, we can easily predict what the planet will do to the comet's energy. Any impulse that adds to the comet's net speed as it passes the planet will increase the comet's energy (ability to escape), and any impulse that decreases the comet's speed will decrease its energy. The examples I gave you then follow. -|Tom|-