|A practical application for cube root|
Message #1 Posted by Karl Schneider on 25 Jan 2007, 3:31 a.m.
HP Forum readers --
Many low-end calculators have a built-in cube root function, which was added to the HP-33S as well. Consensus is that practical applications for the cube root function are limited.
There's certainly one use for it -- determining the edge of a cube having a given volume. Aside from that, what else might there be?
Here's one -- the height of a geostationary satellite, which stays over the same equatorial spot on the earth's surface as it orbits the planet in the direction of the earth's rotation. How high is the orbit? Too low, and the satellite must have a higher angular speed to apparently "travel eastbound"; too high, and it slips behind to apparently "travel westbound". (Away from the equator, and the component of gravity not needed for centripetal acceleration pulls the satellite toward the equator.)
Assume a perfectly-spherical Earth that stays at a fixed point, with a satellite in a circular orbit at altitude "h", in whose plane the Equator lies.
R (radius of Earth) 6.37 x 106 meter
w (angular speed of Earth) 7.2921 x 10-5 rad/s
M (mass of earth) 5.97 x 1024 kilogram
G (Gravitational Constant) 6.67259 x 10-11 m3*kg-1*s-2
h (height of satellite)
m (mass of satellite) -- irrelevant to the answer
The attractive force "FG" between the two objects:
G * M * m
FG = ---------
(R + h)2
This equals the required centripetal force "FC":
m * [w * (R + h)]2
FC = -----------------
(R + h)
G * M * m
--------- = m * w2 * (R + h)
(R + h)2
Canceling "m" and combining,
G * M = w2 * (R + h)3
Finally isolating for "h",
h = cbrt ( G * M / w2) - R
= 3.578 x 107 meter
= 22,236 miles!
NOTE: "w" is 2*pi radians divided by the sidereal day (23 hours, 56 minutes, 4.091 seconds) converted to seconds.
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Edited: 27 Jan 2007, 5:43 p.m. after one or more responses were posted