This is essentially what Redbelly posted, but I jumped straight to acceleration rather than force:
Start with what we know:
1. Distance = velocity * time
2. Distance in an orbit is 2*pi*a (the circumference formula), where a is semi-major axis, same as radius for a circular orbit
Set them equal to each other:
(1) v_{circ} \cdot t = 2\pi a
Use the circular orbital velocity formula
(2) v_{circ} = \sqrt {\frac{{GM}}{a}}
If you'd like to derive this, you start with formula for centrepital acceleration
(3) \frac{{v^2 }}{a}
and Newton's gravitational formula for acceleration
(4) \frac{{GM}}{{a^2 }}
and set them (3) & (4) equal to each other, then use algebra to isolate v
<br />
\begin{array}{l}<br />
\frac{{v^2 }}{a} = \frac{{GM}}{{a^2 }}\,\,\,\, \Rightarrow \,\,\,\, \\ <br />
\\ <br />
v^2 r^2 = GMr\,\,\,\, \Rightarrow \,\,\,\, \\ <br />
\\ <br />
v^2 r = GM\,\,\,\, \Rightarrow \,\,\,\, \\ <br />
\\ <br />
v^2 = \frac{{GM}}{a}\,\,\,\, \Rightarrow \,\,\,\, \\ <br />
\\ <br />
v = \sqrt {\frac{{GM}}{a}} \\ <br />
\end{array}<br />
Re-write formula (1), substituting the circular velocity formula for velocity
(4) \sqrt {\frac{{GM}}{a}} \cdot t = 2\pi a
Use algebra to isolate a
<br />
\begin{array}{l}<br />
\sqrt {\frac{{GM}}{a}} = \frac{{2\pi a}}{t} \\ <br />
\\ <br />
\frac{{GM}}{a} = \left( {\frac{{2\pi a}}{t}} \right)^2 \\ <br />
\\ <br />
\frac{{GM}}{a} = \frac{{4\pi ^2 a^2 }}{{t^2 }} \\ <br />
\\ <br />
GMt^2 = 4\pi ^2 a^3 \\ <br />
\\ <br />
a^3 = \frac{{GMt^2 }}{{4\pi ^2 }} \\ <br />
\\ <br />
a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} \\ <br />
\end{array}<br />
And since you want altitude, subtract Earth's radius (r)
(5) a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} - r
Plugging in numbers (using meters, kilograms, seconds)
(6.6725985E-11*5.97369125232006E+24*86400^2/(4*pi^2))^(1/3)=42241095.3597612
I used 86400 for t, which is 24 hours. You can get a little more exact if you look up the length of a sideral day (~23h 56m) in seconds
