I thought this might be of some interest, as it pretains to weather and I just finished a term paper on it.
We will use Kepler's third law

rs3/Ts2 = rm3/Tm2

where Ts = peroid of orbit of Satellite = 1 day
Tm = period of orbit of Moon = 27.3 days
rm = distance from center of Moon to center of Earth = 3.84 x 108 m
and rs = distance from Satellite to center of Earth

Solving: rs3/Ts2 = rm3/Tm2

for rs gives us

rs = rm (Tm/Ts)2/3

or rs = 0.11 rm

hence

(a) the satellite is situated about
1/10 th of the distance to the moon

(b) rs = 0.11 rm = 42.35 x 106 m
and since R = 6.38 x 106 m
then the satellite is 6.64 earth radii from
the earth's centre or 5.64 earth radii above
the ground at the equator.

(c) its height above the ground is 5.64 R
or 5.64 6.38 x 106 m
or 35,970 km above the earth at the equator.
For those that don't feel that my artimetic is right, 2/3 means square it and then divide by the cube root of the squared. Looks pretty hard, but it isn't!