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Confused on Orbits: Ellipses vs Circles

  1. Aug 29, 2005 #1
    As someone who only studied first year physics and maths and have taken no interest in it since than I was rather surprised to wake up one morning and realise along with 99.5% of the population that I really had no idea how the planets orbits worked beyond the vague word ellipse which I didn't really understand.

    http://en.wikipedia.org/wiki/Keplerian_harmonic_law#Kepler.27s_first_law

    Is Kepler's Law correct? Or rather is it roughly that simple with very minor perterbutions due to other planets' gravity?

    From my memory the 17th century observations revolved around comets that had heavily ellipitical orbits with the Sun foci very much as one end.

    Equally, the Earth appears to go around the Sun at at least an approximately circular orbit with no easily observable perigee or apogee.

    Are the other planets like this? Or do they have large recognisable elipse shapes?

    I thought that celestial mechanics had been worked out in the 18th century with some relatively simple and robust equations that allowed us (or them) to work out the position of the planets at every point in history.

    I have briefly browsed this forum and seems that there are some highly knowledgable people here and I am sure this is not the first time this question has been asked.

    Hopefully an answer or directions to a thread that explain this simply for a non-expert might be available?
     
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  3. Aug 29, 2005 #2

    SpaceTiger

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    I like your choice of username, given the question. :smile:


    The latter, but that doesn't really mean Kepler's law is incorrect, just that it should be viewed as an approximation.


    The ellipse has two foci and the sun is sitting at one of them.


    It depends on how you define "easily", but this is basically correct. Another way to say it is that the ellipse's eccentricity is small.


    Pretty much, except Pluto's orbit, which is highly elliptical.


    That's pretty much right, though Einstein later modified the equations very close to the sun. Also, we can now do numerical simulations that achieve much higher accuracy than would have been reasonable to achieve by hand.


    You seem to be on the right track already and I'm not sure that I would expect even the knowledge you already have from the average person. Anyway, I hope that I answered your questions clearly enough.
     
  4. Aug 29, 2005 #3

    selfAdjoint

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    Maybe just a quick review of conic sections would help?
     
  5. Aug 29, 2005 #4
    Bring back the Epicycle, that I could understand.

    So I guess my basic confusion is the ellipse that the earth does around the sun so close to circular to be not noticed in normal life?

    But presumably the earth's orbit is well known and pretending we are in the two body system (which for approximation is true, the sun being so much larger).

    What is the distance between the two foci of the ellipse, the sun and its foci pair?
    What point during the calender is the closest approach to the sun and what point the furtherest away? How far to away from the sun are these two points?

    What are the acceleration effects that we would feel as we slow down and speed up during the elliptical orbit?

    Or is our elliptical orbit a special kind of ellipse where the two foci are almost overlapping? (usually known as a circle?)
     
  6. Aug 29, 2005 #5

    LURCH

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    Much info here .

    The eccentricity of the Earth's orbit is very slight, as you can see. 146 million km minimum, 152 max.
     
  7. Aug 30, 2005 #6

    tony873004

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    No effect at all, even if Earth's orbit were very eccentric. The Earth and anything on it would be subject to the same gravitational pull from the Sun.

    Yes about the almost-overlapping foci.
     
  8. Aug 30, 2005 #7

    Astronuc

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    The is a nice simple discussion "Vector Treatment of Classical Orbit Theory", in "Mathematics of Classical and Quantum Physics", by Frederick W. Byron, Jr. and Robert W. Fuller.

    It is discusses Kepler's second law and the so-called Runge-Lenz vector, with it's implication with respect to orbits.
     
  9. Aug 30, 2005 #8

    BobG

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    The Earth is closest to the Sun about the first week of January and furthest from the Sun about the first week of July. (The reason for the seasons is the tilt in the Earth's axis relative to the Earth's orbit, not the relative distance from the Sun - otherwise both the Northern and Southern hemisphere would experience the same seasons at the same time.)

    The distance between the closest and furthest points is equal to the major axis (or about 146 million km + 152 million km).

    Celestial mechanics weren't always worked out so nicely, even in the 18th century. Huygens determined the approximate distance between the Earth and Sun. He did this by looking at Venus when it was 90 degrees ahead or behind the Earth (he figured this out by watching Venus's phases). He then had a right triangle and could visually measure the angle between the Sun and Venus for a second angle. Then he only needed the distance between Earth and Venus to figure out both how far the Earth was from the Sun and how far Venus was from the Sun.

    That creates the whole new problem of figuring out how far Venus is from the Earth. He could estimate the change in size and brightness of Venus at different points in its orbit to get some idea of the variation in the distance of Earth-Venus - but he needed the size of Venus to supply the last bit of needed data. Through numerology and mysticism, Huygens figured out that Venus must be the same size as Earth :uhh: . By coincidence, Venus is about 95% as big as the Earth - close enough for Huygens numbers to turn out fairly accurate.

    Cassini later used a better method to figure out the Earth-Sun distance. He used Mars and parallax to determine the Earth-Sun distance. In fact, I think Cassini is normally credited with figuring out the distance between the Earth and Sun, since it would be kind of embarrassing to credit Huygens.
     
  10. Aug 30, 2005 #9
    .

    So if I understand my elliptical orbits correctly the Earth should be travelling it's fastest when it is closest to the Sun and at its slowest when furtherest away from the Sun. At least that is the case with comet ellipitical orbits.

    Although the difference is only about 4% (between closest and furthest points, I don't know how this translates in terms of speed) should any acceleration and deacceleration effects be detectable, even if only be sensitive instruments?

    Also there ought to be some differences with the Sidereal day depending on the point in the orbit?
     
  11. Aug 30, 2005 #10

    tony873004

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    30.28 km/s fastest
    29.31 km/s slowest

    No. Being on an accelerating Earth is not the same as being in an accelerating car.

    In the car, the drivetrain accelerates the car, and the car seat accelates you. So you feel the acceleration as you are pushed back in your seat. Drop a set of keys in your car while you are accelerating and they will not fall straight down. They will fall backwards as the car accelerates and they do not. They can't accelerate in the horizontal direction until they make contact with the car.

    The acceleration due to the Sun's gravity is different. It pulls on both the Earth and you. You do not need to be in contact with the accelerating Earth for you to accelerate as well. You and the Earth accelerate together, therefore Earth does not have the opportunity to accelerate you.
     
  12. Aug 30, 2005 #11

    BobG

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    You're close on differences in the sidereal day. Actually, the sidereal day is almost constant, but the solar day varies. Astronomers measure the solar day from noon to noon. At perihelion, not only is the Earth's speed greater, but its angular velocity is greater. It sweeps out a bigger angle each day, meaning the Earth has to rotate further to get to the next local noon.

    Of course, it's not quite that simple. Having the tilt axis perpendicular to the Earth's radius vector makes for a shorter day, as well, meaning the summer and winter solstice would be the longest days if the Earth was in a circular orbit. The longest solar day of the year (from noon to noon) winds up occurring between the winter solstices and perihelion (it occurs on Christmas this year).

    The Royal Observatory of Greenwich has a web page on this.
     
  13. Aug 31, 2005 #12
    .

    Stressing that I am completely non-expert but your

    Sounds suspect. I think acceleration is acceleration.

    But the acceleration from 29 km/s to 30 km/s spread over a 6 month period is not going to be detectable except via instrumentation.

    Thank you for all your links.

    I think we should have stuck with the epicycles.
     
  14. Aug 31, 2005 #13

    SpaceTiger

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    There may be instruments/experiments that could detect the acceleration directly, but it's easier to detect the eccentricity of the earth's orbit via the motions of the sun and other planets across the sky. If they were to detect the acceleration, they would have to be very sensitive instruments. Why? Consider first that the modulation you're describing is a small correction to the centripetal acceleration one expects in a circular orbit. If the earth does a full circle around the sun in a year, then its speed is just:

    [tex]v = \frac{2\pi d_{sun}}{P_{orbit}}=\frac{2\pi (1~AU)}{1~year} \simeq 30,000~m/s[/tex]

    Then given, the usual formula for centripetal acceleration, you get:

    [tex]a=\frac{v^2}{r}=\frac{(30,000~m/s)^2}{1~AU}\simeq0.01~m/s^2[/tex]

    But that's not all. There's another motion to consider -- the rotation of the earth. If you're standing at the equator, then you're moving at a speed of

    [tex]v = \frac{2\pi R_{earth}}{P_{earth}}=\frac{2\pi (6400~km)}{1~day} \simeq 500~m/s[/tex]

    Again, the acceleration is given by

    [tex]a=\frac{v^2}{r}=\frac{(500~km/s)^2}{6400~km}\simeq0.03~m/s^2[/tex]

    Finally, the acceleration due to gravity, as you may remember from high school physics, is:

    [tex]g=9.8~m/s^2[/tex]

    The effect you're talking about is around four orders of magnitude (factors of 10) smaller than the acceleration due to gravity, so don't expect to be able to measure it in your living room.
     
    Last edited: Aug 31, 2005
  15. Aug 31, 2005 #14

    LURCH

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    It may be a little hard to swallow, but it really is true; there really is no way of "feeling" acceleration caused by gravity. No instrumentation, no matter how sensitive, could detect the Earth's acceleration, because there is no "force" of acceleration to detect. The Earth is in free fall, both when it is at its furthest point from the sun and traveling at 29.31 km/s and when it is at its closest point and traveling 30.28 km/s. And at every point in between.
     
  16. Aug 31, 2005 #15

    pervect

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    As Bob G points out, there are subtle but definitely measurable differences in the length of the day depending on the point in the orbit.

    The difference, however, is in the solar day, not the Sidereal day. See

    http://en.wikipedia.org/wiki/Solar_time

    You can consider that to a high degree of approximation, the Earth rotates with a constant angular velocity with respect to the distant stars. This is what makes the sidereal day essentially constant. The variation of the Earth's orbit makes the solar day non-constant. (The tilt of the Earth also has an effect, this mostly affects the length of the daylight period rather than the length of the day).
     
  17. Aug 31, 2005 #16

    SpaceTiger

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    The laws of physics are the same in that frame, but that doesn't prevent you from measuring the acceleration relative to some other frame (in this case, the sun would be the most sensible choice). This is why we can measure orbital parameters by looking at the modulations of the positions of objects in the sky. You're right, however, that we won't feel it or see the effects locally. We could measure the effects locally if we take the orbital approximation to the next order and consider precession of the ellipse by planetary perturbations, but that would be an even smaller effect.
     
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