Kepler's Third Law and Period Scaling
For a Keplerian ellipse of semi-major axis a around total gravitating mass M_tot, T^2 = 4 pi^2 a^3/(G M_tot); circular-orbit radius is the special case a=r.
Why this shows up in the exam
Planetary year scaling · Satellite radius-period ratios · Estimating central mass
Learn the idea
For the same central mass, orbital period squared scales with the cube of semi-major axis. Distant orbits are much slower: increasing orbital size by a factor k increases period by k^(3/2).
🧠 Memory hook: Period follows orbital size to the three-halves power.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- T² = 4 pi² a³/(G M_tot) — Kepler period relation
- T2/T1 = (a2/a1)^(3/2) — period ratio for unchanged central mass
How to approach it
- 1Identify the correct semi-major axis
- 2Check whether gravitating mass changes
- 3Take a ratio before numerical work
Common slip-ups that cost marks
- •Using altitude instead of orbital radius
- •Using ellipse minor axis instead of semi-major axis
- •Ignoring changed central or total mass
🌟 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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