**Analysis for Challenge 4**

Centrifugal force is given by

Force = mv^{2}/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 = mv^{2}/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/sec^{2}. Thus, we can write

32.2 ft/sec^{2} = v^{2}/(3956.5 miles)

Rearranging terms allows us to solve for v, as follows:

v = √[(32.2 ft/sec^{2})(3956.5 miles)]

v = √[(32.2 ft/sec^{2})(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, I_{charge}, so we can write

v = (I_{charge}/C)t

I_{charge} = vC/t

v is known, C is known, and t is known, so we have

I_{charge} = (2.5V•0.001F/1.41Hr)(1Hr/3600s)

I_{charge} = 0.49μA