Circular-Orbit Speed, Period, and Angular Momentum
For a satellite mass m in a circular orbit of radius r about mass M with m much smaller than M, GMm/r^2 = mv^2/r, so v=sqrt(GM/r), T=2 pi sqrt(r^3/GM), and L=m sqrt(GMr).
Why this shows up in the exam
Satellite speed comparisons · Orbital angular momentum · Weightlessness and free fall
Learn the idea
In a circular gravitational orbit, gravity supplies centripetal acceleration and fixes speed for each radius. A satellite cannot choose any speed at a chosen circular radius; balancing inward gravity selects one value.
🧠 Memory hook: Circular orbit means gravity equals centripetal need.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v_c = sqrt(GM/r) — circular orbital speed
- T = 2 pi sqrt(r³/GM) — circular orbital period
- L = m sqrt(GMr) — circular orbital angular momentum
How to approach it
- 1Set gravity equal to centripetal force
- 2Solve symbolically for the requested quantity
- 3Check that larger radius gives lower speed and longer period
Common slip-ups that cost marks
- •Using satellite mass in orbital speed
- •Using altitude instead of centre radius
- •Interpreting weightlessness as zero gravity
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A satellite moves in a circular orbit where GM = 4 x 10^14 m^3/s^2 and orbital radius is 2 x 10^6 m. Find its orbital speed.
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