Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Earth's rotational speed in the past.

  1. Jun 20, 2008 #1
    I was hoping to discover how the current rotational speed of the earth differs from the past. I'm also curious about the moon's relative distance and orbital speed in the past. And can anyone tell me how fast I am currently moving through space, given all of the variables involved.
     
  2. jcsd
  3. Jun 20, 2008 #2

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    At present, the Earth's period of rotation slows at a rate of 1.5 milliseconds per century.
    At present, the Moon is receding from the Earth at 3.8 cm per year. You can find the change in orbital speed by applying Kepler's Laws to this.
    Relative to what reference?
     
  4. Jun 20, 2008 #3
    As Janus pointed out that question is actually an incredibly important detail and the foundation of Einstein's relativity theory. But I'm going to assume the answer you're looking for is how fast is the earth moving around the sun if we consider the sun to be stationary (correct me if I'm wrong). In which case we have:

    [tex]\frac{mv^2}{r}=\frac{GMm}{r^2} \rightarrow v = \sqrt{\frac{GM}{r}}[/tex] where M is the mass of the sun (approximately 2*10^30) and r is approximately one AU (Astronomical unit = 149*10^9 meters). Therefore, the earth's orbital speed relative to a stationary sun is approximately:

    [tex]\sqrt{\frac{(6.67\times10^{-11})(2\times10^{30})}{149\times10^9}} \approx 30 000 m/s = 108 000 km/h[/tex]

    So 108,000 km/h. We can double check this by seeing that:

    [tex]\omega = \frac{v}{r} = \frac{30 000}{149\times10^9} \approx 2\times10^{-7}[/tex]

    and since [tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{2\times10^{-7}} \approx 3.12\times10^7 s[/tex]

    Therefore, by the velocity we calculated it takes 3.12*10^7 seconds for the earth to revolve around the sun once which is approximately 361 days (remember we rounded pretty much everything a fair bit).
     
  5. Jun 20, 2008 #4
    Thank you for the information! My curiosity as to my relative speed includes the sun's orbit in its solar cluster, the galaxy arm's relative speed and the galaxy's motion itself. My relative speed in mph. all things considered. Thanks!
     
  6. Jun 20, 2008 #5
    Well a cosmic year (the approximate time it takes our solar system to rotate once around the milky way galaxy) is approximately 225 million years (= 1.971*10^12 hours) and the earth's approximate distance from the center of the milky way is 28,000 light years (2.649*10^17 km) so:

    [tex]T=\frac{2\pi}{\omega} = \frac{2\pi r}{v} \rightarrow v = \frac{2\pi r}{T} = \frac{2\pi (2.649\times10^{17})}{1.97\times10^{12}}=845 000 km/h[/tex]

    So the earth is orbiting the milkyway of a speed that is roughly 845 000km/hr
     
  7. Jun 20, 2008 #6

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    A good way to remember the motions was described by the noted cosmologist E Idle.

    Just remember that you're standing on a planet that's evolving
    And revolving at nine hundred miles an hour,
    That's orbiting at nineteen miles a second, so it's reckoned,
    A sun that is the source of all our power.
    The sun and you and me and all the stars that we can see
    Are moving at a million miles a day
    In an outer spiral arm, at forty thousand miles an hour,
    Of the galaxy we call the "Milky Way".

    It doesn't descibe the motion of the galaxy through the local virgo supercluster - either this wasn't well known in the 70s or just didn't ryhme!
     
    Last edited: Jun 20, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Earth's rotational speed in the past.
  1. Earths Rotation (Replies: 3)

  2. Earth rotation (Replies: 2)

Loading...