The Two-Body Problem

Newtonian Gravitation and Kepler's Third Law

Aerospace EngineeringOrbital MechanicsFree preview
⏱️ About 25 min
The Two-Body Problem — illustration
Decorative illustration.

If Earth's gravity pulls inward, how does the Moon stay suspended in the sky instead of crashing into us?

💡
The big idea: By treating a spacecraft and Earth as two point masses, we derive the fundamental equation of motion that defines all closed orbits.
🎯 By the end, you'll be able to
  • State the equation of motion for the restricted two-body problem.
  • Calculate circular orbit speed using \(v = \sqrt{\mu/r}\).
  • Compute the orbital period using Kepler's Third Law.
📎 Helpful to know first
  • A basic understanding of classical Newtonian mechanics and algebra.

Newton's Law of Universal Gravitation

The foundation of orbital mechanics lies in Newton's law of universal gravitation, which states that two masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. In the restricted two-body problem, we assume the primary body (Earth) is fixed and the secondary body (satellite) has negligible mass.

Equation of Motion

Applying Newton's second law to the gravitational force gives us the fundamental differential equation governing orbital motion. The gravitational parameter \(\mu\) is the product of the gravitational constant and the mass of Earth, \(\mu = 398600\) km\(^3\)/s\(^2\).

\[ \ddot{\vec r} = -\dfrac{\mu}{r^3}\,\vec r \]

Circular Orbits

For a circular orbit, the radius \(r\) is constant. The velocity required to maintain this orbit is perpendicular to the radius vector. For a circular orbit, the semi-major axis \(a\) is simply the radius \(r\).

\[ T = 2\pi\sqrt{\dfrac{a^3}{\mu}} \]
🔑 Kepler's Third Law

Kepler's Third Law states that the square of the orbital period is directly proportional to the cube of the semi-major axis (\(T^2 \propto a^3\)). This law holds for all closed orbits under an inverse-square gravitational field.

🎮 Two-Body Orbit Visualizer LIVE
Predict first: Adjust the orbital radius to see how it affects orbital speed and period.
A simulation of a satellite orbiting a primary body under ideal two-body assumptions.
📝 Worked example: A satellite is in a circular Earth orbit with radius \(a = 7000\) km. Calculate its orbital velocity and period.
  1. 1. Given \(\mu = 398600\) km\(^3\)/s\(^2\) and \(a = 7000\) km.
  2. 2. Orbital velocity \(v = \sqrt{\mu/a} = \sqrt{398600/7000} = 7.546\) km/s.
  3. 3. Orbital period \(T = 2\pi\sqrt{a^3/\mu} = 2\pi\sqrt{7000^3/398600} = 5828.5\) s.
  4. 4. Convert to minutes: \(T = 5828.5 / 60 = 97.14\) min.
✓ The orbital velocity is about 7.546 km/s and the period is about 97.1 minutes.
✏️ Practice: Calculate the orbital velocity and period for a circular orbit with radius \(a = 8000\) km.
Solution
  1. Hint: Use v=sqrt(mu/r) and T=2 pi sqrt(a^3/mu).
  2. Answer: v = 7.059 km/s; T = 118.7 min.

Check your understanding

1. Which parameter represents Earth's gravitational parameter?
\(\mu = GM\) is the standard gravitational parameter, combining the gravitational constant and Earth's mass.
2. If the orbital radius of a circular orbit doubles, what happens to the period?
By Kepler's Third Law, \(T \propto r^{3/2}\), so doubling the radius increases the period by \(2^{1.5} \approx 2.828\).
🔎 Related lessons
Next: Conic Sections & Vis-Viva
✅ Key takeaways
  • The restricted two-body problem reduces orbital motion to a single ordinary differential equation.
  • Circular orbit velocity is found by balancing gravitational force and centripetal acceleration.
  • Kepler's Third Law links the orbital period directly to the semi-major axis.
➡️ Next, we will generalize beyond circular orbits to ellipses, parabolas, and hyperbolas using the vis-viva equation.
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