Central-Force Angular Momentum and Kepler's Second Law
For a central force tau = r cross F = 0, hence L=m r cross v is constant and dA/dt=L/(2m); at apsides this gives r_p v_p = r_a v_a.
Why this shows up in the exam
Equal-area timing · Comet speeds at apsides · Elliptical-orbit speed ratios
Learn the idea
Zero torque about the force centre conserves angular momentum and makes areal velocity constant. A planet sweeps equal areas in equal times, so it moves fastest when nearest the attracting focus.
🧠 Memory hook: Closer means faster because r times tangential speed stays fixed.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- dL/dt = tau = 0 — angular-momentum conservation for a central force
- dA/dt = L/(2m) — constant areal velocity
- r_p v_p = r_a v_a — speed relation at periapsis and apoapsis
How to approach it
- 1Locate the force centre and apsides
- 2Use torque or equal-area reasoning
- 3Apply angular momentum with vector directions intact
Common slip-ups that cost marks
- •Assuming speed is constant in an ellipse
- •Using distance from ellipse centre instead of focus
- •Applying r v conservation away from purely tangential apsides
🌟 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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