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Thread: George's B-Day (238th)

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    George's B-Day (238th)

    Happy B-Day for "George's Star" (as named by its discoverer -- William Herschel)!

    ~ 40 years of discussion later it became Uranus, proving the need for discussion time limits.

    Since its 1781 discovery, it will have completed 3 orbits in ~ 15 more years. For those who enjoy something from nothing, 15 years is the amount of time Uranus would take if it stopped and fell directly into the Sun. Not a prophecy; just a numerical tease.
    We know time flies, we just can't see its wings.

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    Hmm, I'm pretty sure the community I live in is named after the same guy!
    Cum catapultae proscriptae erunt tum soli proscript catapultas habebunt.

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    Quote Originally Posted by George View Post
    Happy B-Day for "George's Star" (as named by its discoverer -- William Herschel)!

    ~ 40 years of discussion later it became Uranus, proving the need for discussion time limits.

    Since its 1781 discovery, it will have completed 3 orbits in ~ 15 more years. For those who enjoy something from nothing, 15 years is the amount of time Uranus would take if it stopped and fell directly into the Sun. Not a prophecy; just a numerical tease.
    I never thought about the outer planets falling into the sun before, that’s really interesting to contemplate.
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    Quote Originally Posted by George View Post
    Happy B-Day for "George's Star" (as named by its discoverer -- William Herschel)!

    ~ 40 years of discussion later it became Uranus, proving the need for discussion time limits.
    It's a very auspicious day as it is also Pi day and the anniversary of both Einstein's birth and Hawking's death.

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    Quote Originally Posted by Trebuchet View Post
    Hmm, I'm pretty sure the community I live in is named after the same guy!
    King George?
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    Quote Originally Posted by KaiYeves View Post
    I never thought about the outer planets falling into the sun before, that’s really interesting to contemplate.
    It was one of those forum coincidences. The Scholz's Star thread presented the idea of Oort objects coming our way, so I thought I would apply Newton's law and calculate the fall time for those objects, though knowing the slinging of those objects by the star was the real speed factor. What I learned is that it is very difficult to calculate free-fall time and involves what seems to be known as elliptical calculus, beyond my interest today in math problems.

    However, Kepler's law from the early 1600's offers a fine solution, surprisingly, by-passing the need for computers. Simply bump the eccentricity of the orbit to make a straight line, thus cutting the semi-major axis in half, use P2 = a3 [for our solar system], to get the orbital period, then 1/2 of this is the free fall time. I had to hunt it down by looking for someone who calculated the free-fall time for Earth (~ 65 days) and found this simple solution.
    Last edited by George; 2019-Mar-14 at 02:16 PM.
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    Quote Originally Posted by LaurieAG View Post
    It's a very auspicious day as it is also Pi day and the anniversary of both Einstein's birth and Hawking's death.
    That's impressive; how wise of Herschel?
    We know time flies, we just can't see its wings.

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    Quote Originally Posted by George View Post
    Simply bump the eccentricity of the orbit to make a straight line, thus cutting the semi-major axis in half, use P2 = a3 [for our solar system], to get the orbital period, then 1/2 of this is the free fall time. I had to hunt it down by looking for someone who calculated the free-fall time for Earth (~ 65 days) and found this simple solution.
    Yes, that is a very cute solution!
    Last edited by Ken G; 2019-Mar-15 at 01:02 PM.

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    Quote Originally Posted by Ken G View Post
    Yes, that is a very cute solution!
    Yep! I tried picking effective points to get average acceleration over short distances to allow iterations, and was still of 50%.

    I am fuzzy, however, on why changing the eccentricity to a mathematical straight line forces the semi-major axis to be cut in half? Wouldn't the Earth still take a year to go around for any revised eccentricity value due to things like conservation of ang. momentum? Why would the Earth have a new (a/2) experience?
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    Don't necessarily imagine the straight line, imagine a highly eccentric orbit that passed inside the radius of the Sun but never got to the center. That would still have the Earth crashing into the Sun, and it's clear the time it would take to do so. As for conservation of angular momentum, if you want th Earth to fall into the Sun, you'd have to reduce its angular momentum to almost zero, and that's what would cause it's new experience.

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    Why wouldn’t moving the Sun outward to achieve greater and greater eccentricity still produce its same 2a major axis?

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    Last edited by George; 2019-Mar-17 at 04:26 PM.
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    Quote Originally Posted by George View Post
    Yep! I tried picking effective points to get average acceleration over short distances to allow iterations, and was still of 50%.

    I am fuzzy, however, on why changing the eccentricity to a mathematical straight line forces the semi-major axis to be cut in half? Wouldn't the Earth still take a year to go around for any revised eccentricity value due to things like conservation of ang. momentum? Why would the Earth have a new (a/2) experience?

    Quote Originally Posted by George
    Why wouldn’t moving the Sun outward to achieve greater and greater eccentricity still produce its same 2a major axis?
    Those two quotes are describing two different hypothetical scenarios. In the first one we are leaving the aphelion unchanged while reducing the perihelion to zero or nearly so. That is what cuts the semimajor axis in half, as compared with a circular orbit of the same radius as the aphelion. The fall from a standstill at distance 2a takes the same time as half a circuit of a circular orbit of radius a.

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    Quote Originally Posted by Hornblower View Post
    Those two quotes are describing two different hypothetical scenarios. In the first one we are leaving the aphelion unchanged while reducing the perihelion to zero or nearly so. That is what cuts the semimajor axis in half, as compared with a circular orbit of the same radius as the aphelion.
    Yes, I clearly see that is what is being done, but I am unclear if that is correct physics. If a one-year period is independent of eccentricity for the Earth, then why make it otherwise? As soon as the semi-major axis is deemed to reduce, so too is its orbital period, but isn't there an orbital energy problem with this gedanken experiment? Cutting it in half means that the Earth would go from a one year period to ~ 130 days, so don't we need some bi-polar flows or MHD (hyperbole, of course ) to dump the orbital energy to achieve this new semi-major axis value?


    The fall from a standstill at distance 2a takes the same time as half a circuit of a circular orbit of radius a.
    But that would be 6 months, and that seems too long for a fall time since driving straight through town is faster than taking the loop around it, assuming no traffic. So I'm not doubting the 65 day result, but I'm having trouble grasping the physics of it.
    Last edited by George; 2019-Mar-17 at 06:07 PM.
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    Quote Originally Posted by Trebuchet View Post
    Hmm, I'm pretty sure the community I live in is named after the same guy!
    Quote Originally Posted by George View Post
    King George?
    The Third, at that. It's called Cape George, and the name was given to a point of land by yet another George, Vancouver.
    Cum catapultae proscriptae erunt tum soli proscript catapultas habebunt.

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    Quote Originally Posted by George View Post
    Yes, I clearly see that is what is being done, but I am unclear if that is correct physics. If a one-year period is independent of eccentricity for the Earth, then why make it otherwise? As soon as the semi-major axis is deemed to reduce, so too is its orbital period, but isn't there an orbital energy problem with this gedanken experiment? Cutting it in half means that the Earth would go from a one year period to ~ 130 days, so don't we need some bi-polar flows or MHD (hyperbole, of course ) to dump the orbital energy to achieve this new semi-major axis value?


    But that would be 6 months, and that seems too long for a fall time since driving straight through town is faster than taking the loop around it, assuming no traffic. So I'm not doubting the 65 day result, but I'm having trouble grasping the physics of it.
    The fall from 1 AU is the equivalent of aphelion to perihelion of an elongated ellipse with a major axis of 1 AU, which is half of Earth's major axis. I verified the time of 65 days from Kepler's formula. I wish I could remember how to integrate the straight in fall as a sanity check, but for now I trust the idea of treating this path as the limiting case of an extremely elongated elliptical orbit.

    To drag Earth to a halt would require a whopping retrograde slingshot action from a really close encounter with something. The corresponding action on a comet out in the Oort cloud, where the velocity is much lower for a circular orbit, would require only a nudge from a passing star.

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    Quote Originally Posted by Hornblower View Post
    The fall from 1 AU is the equivalent of aphelion to perihelion of an elongated ellipse with a major axis of 1 AU, …
    Ahhhh, yes. It's as simple as I thought it might be but I kept tripping myself up somehow. As you say, a stopped Earth must fall 1 AU and with an extreme eccentric orbit (pushed to a straight line) we also know the orbital time frame for this particular orbit with an a/2 semi-major axis. It's that simple. Nice.

    Thanks!

    To drag Earth to a halt would require a whopping retrograde slingshot action from a really close encounter with something. The corresponding action on a comet out in the Oort cloud, where the velocity is much lower for a circular orbit, would require only a nudge from a passing star.
    Yes. The free-fall time with an initial velocity from 0.92 AU [oops, lyr.] (~ Scholz's nearest approach) is almost 2.5 million years, hence the slinging velocity is the huge factor.
    Last edited by George; 2019-Mar-18 at 03:25 PM.
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    Quote Originally Posted by George View Post
    Ahhhh, yes. It's as simple as I thought it might be but I kept tripping myself up somehow. As you say, a stopped Earth must fall 1 AU and with an extreme eccentric orbit (pushed to a straight line) we also know the orbital time frame for this particular orbit with an a/2 semi-major axis. It's that simple. Nice.

    Thanks!
    I'm glad I was able to help.

    Yes. The free-fall time with an initial velocity from 0.92 AU (~ Scholz's nearest approach) is almost 2.5 million years, hence the slinging velocity is the huge factor.
    My bold. Don't you mean 0.92 light year?

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    Quote Originally Posted by Hornblower View Post
    My bold. Don't you mean 0.92 light year?
    Yep, fixed it.

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