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, 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