A satellite is not beyond gravity; it is falling continuously and moving sideways fast enough to keep missing. Set gravity's pull equal to the inward force a circle demands — $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ — cancel the satellite's own mass $m$ from both sides, and one line of algebra gives $v = \sqrt{\dfrac{GM}{r}}$: read aloud, v equals the square root of big G M over r.

The cast: $v$ is the orbital speed in m/s, $M$ is the central mass in kilograms — the planet, not the satellite — and $r$ is the orbital radius in metres, measured from the planet's CENTRE. That last word is where marks die. An orbit "400 km up" has $r = 6370 + 400 = 6770$ km, not 400: the altitude is a passenger, the radius is the quantity. Note too what is absent — the satellite's own mass cancelled, so a bolt and a space station on the same orbit fly at exactly the same speed. And push the units through: $\mathrm{m^3/(kg \cdot s^2)} \times \mathrm{kg} \div \mathrm{m}$ leaves $\mathrm{m^2/s^2}$, whose root is m/s. If your rearrangement cannot do that, it is wrong, no appeal — though units that do work out never prove you right.