Vai's Uni Notes

Orbital Period

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The orbit in time, part 1: how long one revolution takes (Curtis section 2.5).

Area Swept Out

  • From Kepler’s second law, the area swept out in a given time interval is
    • 𝐴𝑡1,𝑡2=12ℎ(𝑡2−𝑡1)
  • Over one full orbital period 𝒫︀ the whole ellipse is swept out:
    • 𝐴0,𝒫︀=12ℎ𝒫︀=𝜋𝑎𝑏

Period of an Ellipse

  • Solve for 𝒫︀, using 𝑏=𝑎√1−𝑒2 (ellipse geometry), 𝑝=ℎ2𝜇⇒ℎ=√𝜇𝑝, and 𝑝=𝑎(1−𝑒2):
    • 𝒫︀=2𝜋𝑎𝑏ℎ=2𝜋𝑎2√1−𝑒2√𝜇𝑝=2𝜋𝑎2√1−𝑒2√𝜇√𝑎(1−𝑒2)=2𝜋√𝜇𝑎32
    • units check: [𝑇]

Orbital Period

  • 𝒫︀=2𝜋√𝜇𝑎32
  • Curtis (2.83)
  • Depends only on the semimajor axis, not on the eccentricity

Kepler’s Third Law

  • Square the period:
    • 𝒫︀2𝑎3=(2𝜋)2𝜇
  • The same constant for every orbit around the same central body, which is Kepler’s third law (Kepler’s Laws: 𝒫︀21𝑎31=𝒫︀22𝑎32)
  • Next: where on the orbit is the satellite at a given time? → Mean Anomaly

Graph View

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