Analysis for Challenge 4
Centrifugal force is given by
Force = mv2/r
But force is also given by
Force = ma
where a is acceleration. For purposes of this discussion, a = g. Then we can write
mg = mv2/r
We can cancel out the mass, and recognize that r must be the radius of the earth at the equator. r varies of course, but a quick look at Wikipedia indicates that the average radius of the earth is 3956.5 miles. g is about 32.2 ft/sec2. Thus, we can write
32.2 ft/sec2 = v2/(3956.5 miles)
Rearranging terms allows us to solve for v, as follows:
v = √[(32.2 ft/sec2)(3956.5 miles)]
v = √[(32.2 ft/sec2)(1mile/5280ft)(3956.5 miles)]
v = 4.91 miles/sec
This, of course, is equatorial velocity at Condition 0. We can now determine the length of a “day” at Condition 0 by dividing the circumference of the earth by equatorial velocity.
T = “day” = 2πr/v
T = “day” = 2π(3956.5 miles)/(4.91 miles/sec)
T = “day” = 5060.7 seconds
T = “day” ~ 1.41 hours
So, the circuitry must deliver a burst once every 1.41 hours. (Note: This time may be verified by doing an Internet search for “weightlessness at the equator”.)
The frequency, f, is given by
f = 1/T
f = (1/1.41H)(1H/3600s)
f = 197 µHz
Determining the DC Charging Current
v = (1/C)∫idt
In this case i will be a constant, Icharge, so we can write
v = (Icharge/C)t
Icharge = vC/t
v is known, C is known, and t is known, so we have
Icharge = (2.5V•0.001F/1.41Hr)(1Hr/3600s)
Icharge = 0.49μA
