Orbital elements

1. Newton's law of gravitation

Why it matters: satellites for weather, GPS and phone signals, and every space mission, depend on a single law of gravity plus Newton's laws of motion. Every mass attracts every other with a force along the line joining them ($G$ is the gravitational constant, $M_\oplus$ the mass of the Earth, $r$ the distance between centres):

$$\vec F = -\frac{G M m}{r^2}\,\hat r, \qquad \mu \equiv GM_\oplus = 3.986\times10^{14}\ \text{m}^3\,\text{s}^{-2}$$

For a satellite the mass $m$ cancels, so its acceleration is $\ddot{\vec r} = -\mu\,\vec r/r^3$.

2. Circular orbits and escape

For a circle, gravity supplies exactly the centripetal force, $\mu m/r^2 = m v^2/r$:

$$v_c = \sqrt{\frac{\mu}{r}}, \qquad T = 2\pi\sqrt{\frac{r^3}{\mu}}, \qquad v_{esc} = \sqrt{\frac{2\mu}{r}} = \sqrt2\,v_c$$

An object in orbit is falling all the time; it just moves sideways fast enough that the ground curves away beneath it. At the escape speed the total energy is zero and the path becomes a parabola.

3. Conic sections and orbital elements

Every two-body orbit is a conic with the planet at one focus:

$$\begin{gathered}r(\nu) = \frac{p}{1 + e\cos\nu}\\ p = \frac{h^2}{\mu}\\ \vec e = \frac{(v^2 - \mu/r)\,\vec r - (\vec r\cdot\vec v)\,\vec v}{\mu}\end{gathered}$$

Here $h = |\vec r\times\vec v|$ is the specific angular momentum and $\nu$ the true anomaly (angle from periapsis, the point nearest the planet; the farthest point is apoapsis). $a$ is the semi-major axis (half the longest diameter of the ellipse). The eccentricity sets the shape: $e = 0$ circle, $0 < e < 1$ ellipse, $e = 1$ parabola, $e > 1$ hyperbola. The specific orbital energy fixes the semi-major axis:

$$\varepsilon = \frac{v^2}{2} - \frac{\mu}{r} = -\frac{\mu}{2a}, \qquad r_p = a(1-e), \quad r_a = a(1+e)$$

4. The vis-viva equation

Rearranging the energy equation gives the speed anywhere on an orbit, the most useful single equation in astrodynamics:

$$v^2 = \mu\left(\frac{2}{r} - \frac{1}{a}\right)$$

5. Kepler's laws

  1. Ellipses. Planets move on ellipses with the Sun at one focus (generally: conics).
  2. Equal areas in equal times. The line from the Sun sweeps area at the constant rate $dA/dt = h/2$; this is conservation of angular momentum. The coloured sectors in the sandbox each take the same time, so they have the same area: long and thin near apoapsis, short and fat near periapsis.
  3. Harmonic law. $T^2 = \dfrac{4\pi^2}{\mu}a^3$, so $T^2/a^3$ is the same for everything orbiting the same body (1 yr²/AU³ for the Sun's planets).

6. Changing orbits: the Hohmann transfer

Burning prograde at one point raises the opposite side of the orbit. The most fuel-efficient two-burn transfer between circular orbits $r_1$ and $r_2$ uses an ellipse that touches both (semi-major axis $a_t = (r_1 + r_2)/2$):

$$\begin{gathered}\Delta v_1 = \sqrt{\frac{\mu}{r_1}}\left(\sqrt{\frac{2r_2}{r_1 + r_2}} - 1\right)\\ \Delta v_2 = \sqrt{\frac{\mu}{r_2}}\left(1 - \sqrt{\frac{2r_1}{r_1 + r_2}}\right)\\ t_{transfer} = \pi\sqrt{\frac{a_t^3}{\mu}}\end{gathered}$$

7. The rocket equation and getting to orbit

A rocket gains speed by throwing mass backwards at exhaust speed $v_e = I_{sp}g_0$. Integrating momentum conservation gives Tsiolkovsky's equation for each stage:

$$\Delta v = I_{sp}\,g_0 \ln\frac{m_0}{m_f}, \qquad \dot m = \frac{F}{I_{sp}g_0}$$

Low Earth orbit needs about 7.8 km/s of orbital speed, but a real launch needs roughly 9.3–9.5 km/s of Δv because of gravity losses (thrust spent holding the rocket up, about $g\sin\gamma$ per second where $\gamma$ is the flight-path angle) and drag losses. A gravity turn minimises both: after a short vertical climb the rocket is tipped slightly east (the pitch-over) and then simply follows its velocity vector, so gravity bends the path towards horizontal with no sideways loads. Staging throws away empty tank mass so later stages accelerate less dead weight.

Why launch from the equator? Earth's surface at the equator moves east at $\omega_\oplus R_\oplus = 465$ m/s, a free boost for eastward launches, and an equatorial site can reach a 0° inclination orbit (such as geostationary) without an expensive plane change. Kenya hosts a real example: the Luigi Broglio Space Centre (San Marco platforms) off the coast near Malindi at about 2.9° S, operated by Italy. Scout rockets launched from it between 1967 and 1988, including NASA's Uhuru X-ray satellite in December 1970, named in honour of Kenya's independence.

How the lab computes

The lab integrates the equation of motion $\ddot{\vec r} = -\mu\,\vec r/r^3$ (section 1) with the velocity Verlet method (a symplectic integrator: it does not let the energy drift away systematically, so orbits stay closed over many revolutions).

$$\begin{gathered}\vec v_{n+1/2} = \vec v_n + \tfrac12\vec a(\vec r_n)\,\Delta t, \qquad \vec r_{n+1} = \vec r_n + \vec v_{n+1/2}\,\Delta t\\ \vec v_{n+1} = \vec v_{n+1/2} + \tfrac12\vec a(\vec r_{n+1})\,\Delta t\end{gathered}$$
Model limits. All motion is in a single plane. Earth is a perfect sphere (no $J_2$ oblateness), the atmosphere is exponential with an 8.5 km scale height, drag uses a fixed $C_d$, burns in the sandbox are instantaneous, and the solar-system view uses Keplerian ellipses from mean J2000 elements ignoring inclinations and planetary perturbations, so planet positions are approximate.

Example 1: the ISS orbit

The ISS orbits about 420 km above Earth ($R_\oplus = 6371$ km). Find its speed and period.

  1. $r = 6371 + 420 = 6791$ km $= 6.791\times10^6$ m.
  2. $v_c = \sqrt{3.986\times10^{14}/6.791\times10^6} = 7661$ m/s $\approx 7.66$ km/s (about 27 600 km/h).
  3. $T = 2\pi r/v = 2\pi(6.791\times10^6)/7661 = 5570$ s $= 92.8$ minutes: about 15.5 orbits a day.

Example 2: escape speed

  1. At Earth's surface: $v_{esc} = \sqrt{2\mu/R_\oplus} = \sqrt{2(3.986\times10^{14})/6.371\times10^6} = 11.19$ km/s.
  2. From a 12 000 km circular orbit, $v_c = 5.763$ km/s and $v_{esc} = 8.151$ km/s, so the escape burn is $(\sqrt2 - 1)v_c = 2.387$ km/s.

Example 3: Hohmann transfer to geostationary orbit

Transfer from a circular orbit of radius $r_1 = 12\,000$ km to GEO, $r_2 = 42\,164$ km.

  1. $v_1 = \sqrt{\mu/r_1} = 5763$ m/s. Transfer ellipse $a_t = 27\,082$ km. Vis-viva at $r_1$: $v_p = \sqrt{\mu(2/r_1 - 1/a_t)} = 7191$ m/s, so $\Delta v_1 = 1428$ m/s.
  2. At $r_2$: $v_a = \sqrt{\mu(2/r_2 - 1/a_t)} = 2047$ m/s, while $v_2 = \sqrt{\mu/r_2} = 3075$ m/s, so $\Delta v_2 = 1028$ m/s.
  3. Total $\Delta v = 2456$ m/s. Transfer time $= \pi\sqrt{a_t^3/\mu} = 6.16$ h. Check the GEO period: $2\pi\sqrt{r_2^3/\mu} = 23.93$ h, one sidereal day.
  4. In the lab: choose Circular, set Δv to 1428 m/s and burn prograde, then arm a 1028 m/s burn at apoapsis.

Example 4: the rocket equation for the upper stage

  1. Stage 2 with a 500 kg payload: $m_0 = 900 + 9000 + 500 = 10\,400$ kg, $m_f = 1400$ kg.
  2. $\Delta v = 345 \times 9.807 \times \ln(10\,400/1400) = 3383 \times 2.005 = 6785$ m/s.
  3. Stage 1 carries the rest: $m_0 = 54\,400$ kg, $m_f = 14\,400$ kg, $\Delta v = 311 \times 9.807 \times \ln 3.778 = 4054$ m/s (vacuum; less at sea level).
  4. Total about 10.8 km/s, plus 0.46 km/s from Earth's rotation, against roughly 9.4 km/s needed. Doubling the payload costs about 1 km/s.

Example 5: the equatorial bonus

Eastward surface speed $= \omega_\oplus R_\oplus\cos\phi$ with $\omega_\oplus = 7.292\times10^{-5}$ rad/s and $R_\oplus = 6378$ km (equatorial):

SiteLatitudeSurface speed
Malindi, Kenya (San Marco)2.9° S464 m/s
Kourou, French Guiana5.2° N463 m/s
Cape Canaveral, USA28.5° N409 m/s
Baikonur, Kazakhstan45.6° N325 m/s

Challenges are checked automatically.

Open experiments

  1. Prograde at periapsis. On an ellipse, compare a 200 m/s prograde burn at periapsis with the same burn at apoapsis. Which raises the energy more? (This is the Oberth effect: $\Delta\varepsilon \approx v\,\Delta v$.)
  2. Radial burns. A radial burn rotates the ellipse rather than enlarging it much. Try it and watch the periapsis marker move.
  3. Gravity turn. Launch with pitch-over angles of 2°, 6° and 12°. Compare maximum dynamic pressure, how much stage 2 propellant is left and whether you reach orbit at all.
  4. Inner and outer planets. Watch for a year in the Inner view, then switch to All. Use the planet panel to confirm that $T^2/a^3$ is the same for every planet.

Kepler's third law with real planetary data

Periods $T$ (in years) and semi-major axes $a$ (in AU). The final column is very close to 1 for every planet. The small departures for the giant planets come from their own masses (Jupiter has 0.1% of the Sun's mass, and strictly $T^2 = 4\pi^2a^3/G(M_\odot + m)$) and from the planets perturbing one another, which shifts their averaged orbital elements slightly.

Vertical axis: $\log_{10} T$ (years). On log–log axes $T \propto a^{3/2}$ is a straight line of gradient 1.5.