A space satellite problem:

You are to compute the altitude and velocity of a 1000 kilogram
communication satellite that is to orbit above a location on the
Earths equator. All values are in metric units (meter, kilogram, second).

Constants you need to know:
The mass of the Earth is  Me = 5.97 10^24 kilogram
The universal gravitational constant G = 6.674 10^-11
The radius of the Earth is R = 6.378 10^6 meter.
The circumference of the Earth at the equator is 40.07 10^6 meter.
The location on the Earths equator is 6.378 10^6 meter from the Earth center
The Earth rotates in 24 hours = 1440 minutes = 86,400 seconds
Thus the location on the equator has a velocity of Ve=463.7 meter per second
  (40.07 10^6 / 86,400)
The mass of the satellite Ms = 1000 kilogram
The satellite will be orbiting at distance d meter from center of Earth.

The satellite must rotate above the location on the equator at
distance  d meters from center of Earth for orbiting
velocity  v meters per second.
The force on the satellite due to gravity is F
F = G * Ms * Me / d*d
The centripetal force of the orbiting satellite Fs at distance d, velocity v
Fs = Ms * v^2 / d
F must equal Fs for a stable orbit
For the satellite to stay above the location on the equator, equal ratios
v/d = Ve/R

Your problem is to find the numerical values of v and d to at least
three significant digits.

This can be expressed as a nonlinear minimization problem:
Minimize err to about zero
err = | F - Fs | + | v/d - Ve/R |

err = abs(G*Ms*Me/(d*d) = Ms*v*v/d) + abs(v/d - Ve/R)

Start   d > 4*R  compute v to make second part zero,
iterate d to larger values to reduce err toward zero.
(ok to drive F-Fs toward zero)

Another problem is to determine the launch trajectory of the rocket
that is to place the satellite at distance d from center of earth,
traveling at velocity v parallel to the equator and over the
location on the equator.
(left to students to solve, remember rocket problem)