Earth's Rotational SpeedDate: 07/25/99 at 21:19:38 From: R.Lambkin Subject: Earth's Rotational Speed If you were standing at the equator at sea level, how fast would you be travelling (to the nearest 1/2 mile) in relation to the center of the earth? Date: 07/25/99 at 22:02:22 From: Doctor Jeremiah Subject: Re: Earth's Rotational Speed There are 24 hours/day and the earth rotates all the way around each day. The earth is about 25000 miles around at the equation at sea level so going all the way around means about 25000 miles/day. 25000 miles/day / 24 hours/day = 1041.66 miles/hour The reason I keep saying "about" is because these numbers are kind of close but they are not exactly right. According to NASA's web site at: http://windows.ivv.nasa.gov/earth/statistics.html the Earth's diameter is 7,926 miles and thus the circumference is 24900.26 miles. 25900.26 miles/day / 24 hours/day = 1037.5 miles/hour - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 09/13/2001 at 13:09:29 From: John Gatti Subject: Rotational speed of the earth Dear Dr. Math, Sometimes we discuss that what is obviously correct is not correct. One simple question that has an obvious answer is how long it takes the earth to make one revolution on its axis. The obvious answer that is incorrect is 24 hours. The next factor people realize is that the is the leap year issue, but this issue has less of an effect that the fact that one of our years' days is due to the orbit of out earth around the Sun. I think that subtracts a day rather than adds a day. So we would have 365 24 hour days / 366 revolutions x 24 hours = hours per revolution = 23.934 hours, or 23 hours 56 minutes. From the point of view of centripetal forces, which are necessary to understand for satellite orbits, ocean tides, etc., what is the rotational rate of the Earth? Thanks for your consideration, John Date: 09/13/2001 at 14:17:53 From: Doctor Rick Subject: Re: Rotational speed of the earth Hi, John, thanks for writing to Ask Dr. Math. You are correct that we should really use the sidereal day (the period of one rotation of the earth relative to the stars) rather than the familiar solar day (the period from one noon to the next). The solar day gives the rotation of the earth relative to a reference frame (the line joining the sun and the earth) that rotates once a year. It makes more sense to use an inertial reference frame, and that is what the sidereal day gives us. What is the sidereal day? NASA says: http://liftoff.msfc.nasa.gov/academy/rocket_sci/satellites/sidereal.html The Length of a Day A sidereal (pronounced sigh-dear'-real) day refers to the rotation of the Earth measured relative to the stars. It is the time it takes the Earth to rotate 360 degrees and is equal to 23 hours, 56 minutes and 4 seconds. You are correct. If you had included more digits of precision, you would have gotten the 4 seconds as well (actually, 3.93). Strictly speaking, we should use the sidereal year (365.2564 days) instead of 365 days, but this makes a tiny correction, giving 4.10 seconds. The above is answering the question, how fast does a point on the equator move, to the nearest 1/2 mph? The speed arrived at in the answer, 1037.5 mph, is off by 0.28%, or almost 3 miles per hour. This error is well over the requested accuracy of 0.5 mph. The correct speed should be 24900.26 miles/day / 23.934 hours/day = 1040.37 miles/hour - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/