Problem 1

Problem 1: Orbital Period and Orbital Radius

Derivation of Kepler's Third Law (Circular Orbits)

For a small body of mass \(m\) orbiting a much larger body of mass \(M\), we assume a circular orbit.

Newton's Law of Gravitation:

\(F_g = \frac{G M m}{r^2}\)

Centripetal Force:

\(F_c = \frac{m v^2}{r}\)

Set \( F_g = F_c \):

\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)

Cancel \(m\), simplify:

\(v^2 = \frac{G M}{r}\)

Orbital period \(T\) is:

\(T = \frac{2\pi r}{v}\)

Substitute \(v\):

\(T = \frac{2\pi r}{\sqrt{\frac{G M}{r}}} = 2\pi \sqrt{\frac{r^3}{G M}}\)

Final Form (Kepler’s Third Law):

\(T^2 = \frac{4\pi^2}{G M} r^3\)

Implications for Astronomy

  1. Mass Calculations:

\(M = \frac{4\pi^2 r^3}{G T^2}\)

  1. Planetary Distances: Given \(T\) and \(M\), solve for \(r\).

  2. Satellite Design: Engineers use this to set satellite altitudes and periods (e.g., GPS, geostationary).

Real-World Example: Moon Around Earth

  • \(r = 3.84 \times 10^8 \, \text{m}\)
  • \(T = 2.36 \times 10^6 \, \text{s}\)

Use:

\(M = \frac{4\pi^2 r^3}{G T^2}\)

To compute Earth's mass.

colab

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Visualizing Circular Orbits

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Extension to Elliptical Orbits

Kepler’s Third Law holds for elliptical orbits as well, where the orbital radius \(r\) is replaced by the semi-major axis \(a\):

\(T^2 \propto a^3\)

This means the square of the orbital period is proportional to the cube of the semi-major axis, even if the orbit is not a perfect circle.

This applies to:

  • Elliptical planetary orbits (e.g., Mars, which has a noticeable eccentricity)
  • Binary star systems, where two stars orbit their common center of mass
  • Asteroids and comets, which often have highly eccentric elliptical orbits

The generalized form of Kepler’s Third Law becomes:

\(\frac{T^2}{a^3} = \frac{4\pi^2}{G(M_1 + M_2)}\)

Where: - \(T\) = orbital period - \(a\) = semi-major axis of the orbit - \(M_1\), \(M_2\) = masses of the two orbiting bodies

colab simulation.