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Mercurys orbit and Brian Cox

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  • 16-07-2019 8:50am
    #1
    TomSweeney
    Registered Users Posts: 3,193 ✭✭✭


    I just watched the first episode of the fantastic BBC series "The Planets".
    In the first episode Brian Cox was talking about Mercury and it's synced orbit, so for every 2 orbits around the sun, it rotates 3 times on it's axis, he then went on to say "So it's day is twice as long as it's year"

    So ..
    2y = 3d
    d = 2/3y

    So one day is 2/3 of it's year surely ?

    If you Google Mercurys day/year you get

    Orbital period = 88 earth days = 2,112 hours
    Mercury Day = 1407.6 hours

    Which is roughly 2/3 of it's year .... but I just can't believe BBC and Cox are wrong ...


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    #2
    Peregrinus
    Registered Users Posts: 26,056 ✭✭✭✭Peregrinus


    No. He's right, more or less.

    A day, remember, is not the time it takes a planet to rotate once on its axis. It's the time it takes for the sun to move through the sky back to the (apparent) point from which it started - say, noon to noon. These are not the same thing although, on Earth, they are very nearly the same thing.

    Imagine Mercury did not revolve around the sun at all - was stationery in space. (I know that's impossible, but imagine it, to illustrate the point.) Mercury's day would then be defined entirely by its rotation around it's own axis. From the point of view of an observer on the eqautor of mercury, after exactly one rotation, the sun would be in exactly the same position overhead.

    But now imagine that Mercury revolves around the sun very fast, completing one revolution in exactly the same time as it rotates once around its own axis. From the point of view of an observer on the surface of the planet, the sun would not appear to move at all. Any apparent westward motion resulting from the rotation of Mercury would be exactly cancelled by an apparent eastward motion resulting from the orbit around the sun. For the observer on Mercury, the sun would remain at a fixed point in the heavens. The day would be infinitely long - unless you're on the wrong side of the planet, in which case the night would be infinitely long.

    What this shows is that the length of the day is actually affected both by a planet's rotation and by its orbital period. This hardly matters on earth, where in one rotation - 24 hours - the planet has only moved 1/365th of the distance around the sun, shifting the apparent position of the sun in the sky by just one degree. But it matters a lot on Mercury, because after one rotation Mercury has moved two-thirds of the way around the sun, shifting the position of the sun in the sky by about 240 degrees. So for the observer on the planet the day is not yet complete - the sun has not got back to the point in the sky where it appeared to be one rotation ago.

    When you allow for this, Cox is saying, the sun won't actually return to its apparent position overhead until Mercury has completed two full orbits.


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    #3
    TomSweeney
    Registered Users Posts: 3,193 ✭✭✭TomSweeney


    Peregrinus wrote: »
    No. He's right, more or less.

    A day, remember, is not the time it takes a planet to rotate once on its axis. It's the time it takes for the sun to move through the sky back to the (apparent) point from which it started - say, noon to noon. These are not the same thing although, on Earth, they are very nearly the same thing.

    Imagine Mercury did not revolve around the sun at all - was stationery in space. (I know that's impossible, but imagine it, to illustrate the point.) Mercury's day would then be defined entirely by its rotation around it's own axis. From the point of view of an observer on the eqautor of mercury, after exactly one rotation, the sun would be in exactly the same position overhead.

    But now imagine that Mercury revolves around the sun very fast, completing one revolution in exactly the same time as it rotates once around its own axis. From the point of view of an observer on the surface of the planet, the sun would not appear to move at all. Any apparent westward motion resulting from the rotation of Mercury would be exactly cancelled by an apparent eastward motion resulting from the orbit around the sun. For the observer on Mercury, the sun would remain at a fixed point in the heavens. The day would be infinitely long - unless you're on the wrong side of the planet, in which case the night would be infinitely long.

    What this shows is that the length of the day is actually affected both by a planet's rotation and by its orbital period. This hardly matters on earth, where in one rotation - 24 hours - the planet has only moved 1/365th of the distance around the sun, shifting the apparent position of the sun in the sky by just one degree. But it matters a lot on Mercury, because after one rotation Mercury has moved two-thirds of the way around the sun, shifting the position of the sun in the sky by about 240 degrees. So for the observer on the planet the day is not yet complete - the sun has not got back to the point in the sky where it appeared to be one rotation ago.

    When you allow for this, Cox is saying, the sun won't actually return to its apparent position overhead until Mercury has completed two full orbits.


    Wow! thanks for that - nice one !

    Makes perfect sense now, but just to be clear, my maths above is correct basing days solely on axial rotations ?

    Did you know this before or work it out yourself ? cos it is easy now to understand, but I'd imagine it's difficult to come up with it.

    In fact now that I think of it, Cox's very next line on the show was "If I walk at a contant speec of 2/3mph to the west, the sun will just stay in the same position of the sky" ..


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    #4
    Peregrinus
    Registered Users Posts: 26,056 ✭✭✭✭Peregrinus


    I started from the assumption that Brian Cox Is Always Right, and then asked myself how he might be right in this instance. The rest followed.


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    #5
    Twenty Grand
    Closed Accounts Posts: 1,452 ✭✭✭Twenty Grand


    Well you learn something new every day..


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    #6
    TomSweeney
    Registered Users Posts: 3,193 ✭✭✭TomSweeney


    Peregrinus wrote: »
    I started from the assumption that Brian Cox Is Always Right, and then asked myself how he might be right in this instance. The rest followed.


    He did screw up on relativity once, I must dig it up ...


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    #7
    SophieLockhart
    Registered Users Posts: 203 ✭✭SophieLockhart


    The time from noon to noon is called a solar day, the time to complete one full rotation is called a sidereal day.


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    #8
    ps200306
    Registered Users Posts: 1,589 ✭✭✭ps200306


    If you think Mercury's strange, check out the days on Uranus (no juvenile jokes please ;) ).

    Uranus is lying pretty much on its side. At certain times in its year it points its poles at the Sun, other times its equator. It takes 17 hours to rotate on its axis and 84 earth years to orbit the Sun. In 2029 it's north pole will face the Sun and the Sun will appear almost stationary. Much of the northern hemisphere will have continuous daylight for (earth) years. The Sun will trace an ever-widening spiral in the sky until eventually it will be rising and setting again. Then it gets lower in the sky each day until it disappears for years on end.


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    #9
    Capt'n Midnight
    Moderators, Recreation & Hobbies Moderators, Science, Health & Environment Moderators, Technology & Internet Moderators Posts: 90,700 Mod ✭✭✭✭Capt'n Midnight


    Perfect planet if you are Scottish

    https://en.wikipedia.org/wiki/Hogmanay
    Hogmanay (Scots: [ˌhɔɡməˈneː];[1] English: /ˌhɒɡməˈneɪ/ HOG-mə-NAY[2]) is the Scots word for the last day of the year and is synonymous with the celebration of the New Year (Gregorian calendar) in the Scottish manner. It is normally followed by further celebration on the morning of New Year's Day


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