Deriving orbital speed v = √(GM/r)
Setting gravity equal to the centripetal force, GMm/r² = mv²/r, m cancels, and rearranging gives v = √(GM/r) = √(gr). So a larger orbit means a slower satellite, independent of the satellite's mass.
v = √(GM/r) = √(gr) · T = 2πr/v = 2π√(r³/GM)What this actually means
Four lines, done. Write GMm/r² = mv²/r. Cancel m to get GM/r² = v²/r. Multiply through by r to get v² = GM/r. Square root. Show every step, because 'derive' means the marker wants to see the cancellation.
Because g = GM/r², you can also write v = √(gr), where g is the field strength AT THE ORBIT, not at the surface. That distinction bites in questions about satellites 'orbiting close to the surface', where the two happen to coincide.
The counterintuitive bit is that v falls as r grows. Push a satellite further out and it moves SLOWER, not faster. Students expect the opposite because the orbit is longer, but the period grows faster than the circumference does, so the speed drops.
m cancelling is the headline result. Two satellites of wildly different masses at the same radius have identical speeds and identical periods. Ask what happens to the orbit if the satellite's mass doubles, and the answer is nothing at all.
Feed v back into T = 2πr/v and you get T = 2π√(r³/GM), which squared is Kepler's third law. So orbital speed and Kepler III are the same physics arriving in different currency.
Sanity numbers worth memorising: low Earth orbit is about 7.9 km s⁻¹ with a period near 90 minutes, geostationary is about 3.1 km s⁻¹ with a period of 24 h. If your answer is nowhere near those, check your radius.
Assuming a bigger orbit means a faster satellite, or using surface g instead of g at the orbital radius.
Prove it — watch it be true
- Drag the orbit radius slider outwards and watch the orbital speed readout drop.
- Check a specific radius against √(GM/r) using the values the panel displays.
- Change the satellite mass and confirm the speed and period readouts do not budge.