From: Marc Afifi (mafifi@redshift.com)
Date: Sun Nov 07 1999 - 08:04:42 PST
Message-ID: <3825A316.5F56@redshift.com> Date: Sun, 07 Nov 1999 16:04:42 +0000 From: Marc Afifi <mafifi@redshift.com> Subject: Re: Pinhole Digest #283 - 11/07/99
Sorry David and Art, I already wrote you in private but I'm home now so
I'll post my reply.
Here's how I interpret the problem:
The centripetal force holding an object in orbit around the Earth is the
force due to gravity. Therefore
mg = mv^2/r
notice that the mass drops out of this equation and the relevant
relationship is
v = root (r*g) (g varies with distance by the law of gravitation)
This equation tells the speed required for orbit at any radius (don't
forget to recalculate g at different distances)
If an object has a velocity greater than root(r*g) then it will not be
able to maintain orbit.
Now let's make some calculations:
At the equator, the radius is about 6400 km and g is 9.81 m/s^2.
Therefore, the speed required for orbit at the Earth's surface is
root(6.4x10^6 * 9.81) which is about 7900 m/s.
If you now calculate the tangential speed that an object has by virture
of the Earth's rotation, then the speed of the object is about 460 m/s.
(circumference is about 40 000 km and it takes 24 h to complete one
revolution, you do the math)
Notice that the tangential speed of 460m/s is substantially less than
the 7900m/s needed for orbit and therefore the object cannot "fly off
the Earth." It is not even capable of mainting orbit. Indeed, if the
Earth's surface wasn't there, the object would spiral into the center.
If on the other hand, the Earth's day was only 1.4 hours long, then the
tangential velocity would exceed the velocity required for orbit and
everything at the Earth's surface would indeed fly off into space.
What a great question!
-- Marc Afifi Physics, Chemistry and Marine Science Pacific Grove High School 615 Sunset Dr. Pacific Grove, CA, 93950 http://www.pghs.org/staff/afifi/d5hp.html(831) 646-6590 ext. 223
Favorite Oakland Raiders Motto: "Just When Baby?"
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