You must assume circular orbits, as stated in another answer. The ground speed will also be variable due to inclination, unless it's equatorial. An extreme example is an east-west orbit versus a west-east orbit. In one case, the ground velocity is added, in the other case, it's subtracted.
But with the query limited to equatorial orbits, we can continue. Find the orbital velocity from the radius. Convert the orbital velocity to angular velocity, and then multiply by the Earth's radius. Last, subtract the Earth's movement from this.
$$ v_{gnd} = \sqrt{ \frac{ GM }{ r} } \frac{ 1 }{ r } R_e - 0.465 \frac{ km }{s} \\
= \left( 7.9 \frac{km}{s} \right) \left( \frac{ R_e }{ r } \right)^{3/2} - 0.465 \frac{ km }{s} $$
You can simply plot this. For length units, I'm using multiples of Earth's radius. So r=1 here is the surface.
For inclined orbits, the above relationship will be combined with some trigonometry, and change over the period of the orbit.