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at ground level g is 9.8m/s^2 suppose the earth starts to increas its angular velocity how long would a day be when people on t...Asked by Alamir
At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people on the equator were just 'thrown off'? Why is the expression 'thrown off' a bad one?
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Answered by
Damon
Ac = v^2/R = omega^2 R
when Ac = g =about 9.8 m/s^2, gravity no longer is strong enough to hold your feet on the ground. You would not be thrown, just weightless :)
so
omega^2 = 9.8 / R
R is about 6.38*10^6 meters
so
omega^2 = (9.8/6.8) * 10^-6
omega = 1.2 * 10^-3 radians/second
how many seconds to go 2 pi radians?
T = (2 pi / 1.2) 10^3 = 5.23 *10^3 seconds
there are 3600 seconds in an hour
so
5230 /3600 = 1.45 hour long day
Not even time for lunch.
when Ac = g =about 9.8 m/s^2, gravity no longer is strong enough to hold your feet on the ground. You would not be thrown, just weightless :)
so
omega^2 = 9.8 / R
R is about 6.38*10^6 meters
so
omega^2 = (9.8/6.8) * 10^-6
omega = 1.2 * 10^-3 radians/second
how many seconds to go 2 pi radians?
T = (2 pi / 1.2) 10^3 = 5.23 *10^3 seconds
there are 3600 seconds in an hour
so
5230 /3600 = 1.45 hour long day
Not even time for lunch.
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