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2017 Eclipse Blog / FAQ



How fast is the shadow moving across the US during the eclipse?


A great question! The Moon does indeed “carry” its shadow along with it, as it moves in its orbit around the Earth.

shadow_scrape

And whenever that shadow happens to scrape across the Earth, we have a total eclipse!  (Of course, you have to be in the Path of Totality to see the total eclipse, and you have to use eye protection whenever the eclipse is not total – regardless of whether you’re in the path or not!)

a_rev

Because of the geometry of the Earth’s shape, the shadow will travel faster across its surface and the ends of the eclipse path, and slowest right in the middle.  By using one of the best eclipse calculators out there (thanks to Xavier Jubier!), we can see that the Moon’s shadow (also called the “Umbra”) is moving:

  • 2410mph in Western Oregon
  • 1747mph in central Nebraska
  • 1462mph in Western Kentucky
  • 1502mph near Charleston SC

Make sure you are in the path on eclipse day!  And we’ll see you… in the shadow!

39 Responses to “How fast is the shadow moving across the US during the eclipse?”

  1. Thank you for the info on the speed of the moon’s shadow during the eclipse. I live in southeast Missouri, and will probably watch the eclipse from Farmington.

    • Admin says:

      Excellent! Just make sure you are IN the path, and have good weather on eclipse day!

      Dan

      • James Bass says:

        Hi, can you post the mathematics that was used in the calculator that clocks the speed of the shadow of the moon moving across the earth?

        • Admin says:

          It is a bit more complicated than what can be posted in a blog. 🙂 I will give you some references though!

          Explanatory Supplement to the Astronomical Ephemeris and the American Ephemeris and Nautical Almanac (1961). London: H.M. Stationery Office.

          Chauvenet, W. (1960). A Manual of Spherical and Practical Astronomy. New York: Dover (reprint).

          Meeus, J.; Grosjean, C.C.; Vanderleen, W. Canon of solar eclipses (1966). Oxford/New York: Pergamon Press.

          Meeus, J. Elements of Solar Eclipses 1951-2200 (1989). Richmond, VA: Willman-Bell.

          Comrie, L.J. The Computation of Total Solar Eclipses (1933). Monthly Notices of the Royal Astronomical Society, Vol. 93, p.175.

    • Jack says:

      Farmington will get at least 2 minutes of total eclipse, so you have a front row seat (weather permitting)!

  2. Ron Hadfield says:

    I will be in Driggs, ID on August 21st and I believe it is in the centerline. Is there a way for me to determine the speed of the shadow in Driggs?

  3. Jim Boyle says:

    What day and time will the eclipse appear over the various Cities, ie Casper Wy?

  4. Randy P. says:

    Thank you for the cool info on the eclipse!

  5. Andrea L. says:

    So here is a great math problem: If I am in a vehicle going 65 mph under the solar eclipse, how long will I be in the dark? As a point of reference… Driving south from Columbia, SC to Charleston, SC on US I 26, on August 21st… Barring any traffic impasses.

  6. Tom says:

    This page says the umbra moves at 2955 mph in Western Oregon. I don’t think that’s right. Xavier’s eclipse calculator for Lincoln City, OR (on the coast) says 2410 mph. Should this be corrected or am I missing something? I love your website! Thanks.

  7. Harry J MacNeil says:

    Hi,
    I’ll be in Beatrice Nebraska.
    Harry

  8. Tom Gordon says:

    I will be in Beatrice NE. Taking the day off work to travel an hour south from Omaha so I can be in the path.

  9. Ben says:

    Hi I’ll be in garden of the gods illinois.the cliffs there are quite a bit higher elevation than the adjacent valley. Is the shadow of the moon something tjat could be visibly discerned approaching me. Thinking that would he really neat video footage if so!

    • Admin says:

      If you have a clear view to the western horizon, the sight is spectacular. It’s like a big black curtain rushing toward you … from the West! 😉

  10. Gene Bahnsmath, not calculus. says:

    No worry, your Learjet can fly in the shadow. It is simple math, not calculus!

    Google for GPS coordinates:
    Madras, OR =44.6N, 121.1W Time Eclipse begins = 10:19 AM
    Paducah, KY=39.0N, 84.7W Time Eclipse begins = 1:22 PM

    Length of an arc = radius x angle in radians (1.0 radian = 57.3 deg approx)
    Radius of Earth = 3959 (somewhere). I used 3960.

    N-S displacement = 3960 x (44.6-39)/57.3 = 390 mi.
    Radius for E-W displacement = distance to polar axis (take an average elevation)
    So Avg elevation = (44.6+37.1)/2=40.85 deg N
    EW radius = 3960 x cosine(40.85 deg) = 3960 x 0.756 = 2994 mi.
    E-W displacement= 2994 x (121.1-84.7)/57.3 = 1902 mi.
    Approx Total Displacement = sqrt( 390^2 + 1902^2) = 1942 mi.

    Speed of Shadow = 1942/(13h 22m – 10h 19m) = 1942/(3+3/60)= 637 mph
    Speed of sound at 20,000 feet = 707 mph (Google again)
    637/707 = 0.9 Mach Number
    So the Learjet might fall behind. Better take an F-16.

    • Gene Bahnsmath, not calculus. says:

      Correction: Paducah KY is at 37.1 n, 88.6 W (I confused it with Covington)

      So, not bothering to redo the math right now, I believe it would correct to about 680 mph. Supersonic now, so definitely need the F-16.

      Gene B

      • Gene Bahnsmath, not calculus. says:

        Correction 2: I was too hasty. Forgot that 88 deg W Lat ia WEST of 84 deg W Lat. So it would be 45 mph less, not more.

        So rough estimate is about 592 mph, or 0.83 Mach at 20,000 ft, The Learjet might come close.

        Gene B

    • Ben Prather says:

      I think you forgot to consider the two hour time zone changes from Oregon to Kentucky.

      Thus it should be 1942/(1+3/60)~1850, or Mach 2.6. Thus you will need a SR-71 blackbird, MiG 25, or some top secret aircraft we don’t know about. (Several others might be able to be pushed beyond there specs to reach this speed.)

      http://www.aerospaceweb.org/question/performance/q0030b.shtml

      🙂

  11. Jaco Breytenbach says:

    Oh, terrible, these mph’s! Why don’t you work in kilometers per hour like all scientifically educated people?

    • Admin says:

      Because most Americans are more comfortable thinking in mph, and the site is not made for scientists – it’s made for everyday people who want to enjoy the eclipse.

      Otherwise, we could give the speeds in furlongs per fortnight! 🙂

  12. Love the site. Wish I knew about it earlier in the year.

    Thanks for your dedication.

  13. Norm Norm says:

    Assuming 1,747 mph is correct for the speed of shadow movement in Grand Island, Nebraska, it should take the shadow about two seconds to cover a mile.

  14. Adam says:

    Hi,
    I live just outside of Portland Oregon , went to Salem. Got me almost 2 minutes in the shadow. Very amazing!

    Just wanted to say thank you for providing the information to scratch that Curiosity itch I had.

  15. Tex says:

    I raced it….the shadow beat me a little on the straightway

  16. Charles Naumann says:

    Ground speed of shadow = width of totality / length of time in totality at center

    If the totality band is 70 miles and you have 2.5 minutes of totality at the center, the shadow is moving about 70miles / 2.5 minutes = 1600 miles per hour

    • Admin says:

      The shadow’s speed for this eclipse varied from about 2400mph over Oregon to about 1500mph over S Carolina. In general, your thumbnail calculation is accurate, but only for locations where the path outline on the Earth’s surface is reasonably circular.

  17. Chris Leuthold says:

    What is it about the geometry of the earth’s shape that causes the shadow to change speeds?

    • Admin says:

      Because the Earth is round. When the Moon’s shadow first touches down on the Earth, it is basically tangent to the surface and at that moment mathematically has an infinite speed. As the shadow moves across the face of the Earth, the outline of the shadow on the Earth becomes more and more circular as the curvature of the Earth causes the umbra to hit it more and more perpendicularly. We have a VERY crude diagram that demonstrates this. (It is not to scale, but the underlying logic is not changed). When you watch this, note how the leading edge of the shadow appears to trace out its path more and more slowly as the shadow makes its way toward the point where the earth is most bulged out.

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