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### Earth's Rotational Speed

```
Date: 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/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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