Astronomy 217

 
 

Prof. Andrew W. Steiner

 
 
Sep. 10, 2021
 
 

TA James Ternullo

Last Time

  • Newton's Laws
  • More on Vectors
  • Gravitational Force
  • Conservation
  • Angular Momentum
  • Centripetal Force
  • Inverse Square Laws

Today

  • Revisiting Kepler's Third Law
  • Ellipses in terms of \( L \)
  • Barycenter
  • Kinetic and Potential Energy
  • Orbit classification
  • Orbital velocity
  • Transfer orbits
  • Stellar Parallax

Kepler's Third Law, Revisited

  • A relationship between the period \( P \) and the semi-major axis \( a \) $$ P^2 = \frac{4 \pi^2 a^3}{G M} $$
  • Check units! $$ \mathrm{s}^2 = \frac{\mathrm{m}^3} {(\mathrm{N}~\mathrm{m}^2/\mathrm{kg}^2)~\mathrm{kg}} $$

Mass of a Planet from Moon's Motion, Revisited

  • Use Newton's laws plus circular motion, where \( m \) is the moon's mass, \( M \) is the planet's mass, and \( r \) is the distance between the planet and the moon $$ F = \frac{m v^2}{r} = \frac{G M m}{r^2} $$ where $$ v = \frac{2 \pi r}{t_{\mathrm{orbit}}} $$ and \( t_{\mathrm{orbit}} \) is the time it takes for the moon to orbit the planet
  • Solving for \( M \) $$ M = \frac{4 \pi^2 r^3}{G t_{\mathrm{orbit}}^2} $$

Ellipse, Revisited

  • Instead of using the semi-major axis for the ellipse $$ r = \frac{a(1-e^2)}{1+e \cos \theta} $$
  • We can now write it in terms of the angular momentum $$ r = \frac{L^2}{G M m^2 (1+e \cos \theta)} $$ using $$ a = \frac{L^2}{G M m^2 (1-e^2)} $$

Solar System Barycenter

  • Small correction to Kepler’s 3rd law for the more massive planets.
  • The origin: Newton’s 3rd law. Force of the Sun on the planet is equal to the force of the planet on the Sun
  • Both planet and Sun follow ellipses around the common center of mass or "barycenter".
  • Effect unimportant when \( m \ll M \)

Kinetic and Potential Energy

  • Kinetic energy: $$ K = \frac{1}{2} m v^2 = \frac{p^2}{2 m} $$
  • Potential energy is the integral of the force (for a central force) $$ U_G = \int_R^{\infty} - \frac{G M m}{r^2}~dr = - \frac{G M m}{R} $$
  • Energy is conserved, what does this mean for motion around an ellipse?

Kinetic and Potential Energy, in terms of \( L \)

  • Total energy $$ E = K + U = \frac{1}{2} m v^2 - \frac{G M m}{r} $$
  • We can write in terms of \( L \) and the eccentricity \( e \): $$ K = \frac{1}{2} m^3 \left( \frac{G M}{L} \right)^2 \left( 1 + 2 e \cos \theta + e^2 \right) $$ $$ U = - m^3 \left( \frac{G M}{L} \right)^2 (1+e \cos \theta) $$ the sum is $$ E = K + U = \frac{m^3}{2} \left( \frac{G M}{L} \right)^2 \left( e^2 - 1 \right) $$
  • This is less than zero if \( e < 1 \)
  • What if \( e \geq 1 \)?

Classification of orbits

  • Four different types of orbits
  • \( e=0 \) : circular orbit (minimum energy)
  • \( 0<e<1 \) : elliptic orbit
  • \( e=1 \) : marginally bound parabolic orbit $$ v = v_{\mathrm{esc}} = \sqrt{2 G M}{r} $$
  • \( e>1 \) : Hyperbolic orbit

Orbital Velocity

  • Going back to the kinetic energy $$ v^2 = \left( \frac{G M m}{L} \right)^2 \left( 1 + 2 e \cos \theta + e^2 \right) $$
  • And using $$ L = G M m^2 a ( 1 - e^2) $$
  • Gives the "vis viva" ("living force") equation $$ v^2 = G M \left( \frac{2}{r} - \frac{1}{a} \right) = \frac{2 \pi a}{P} \left( \frac{2 a}{r} -1 \right)^{1/2} $$

Hohmann Transfer Orbit

  • Hohmann transfer orbit, an ellipse whose perihelion is the inner planet's orbit and whose aphelion is the outer planet's orbit.
  • The transfer orbit from Earth to Mars has a semimajor axis $$ a_{\mathrm{TO}} = 0.5 ( a_{\oplus} + a_{\mathrm{Mars}} ) $$ and period P to = 1.41 years, thus perihelion velocity $$ v_{\mathrm{pe,TO}} = \frac{2 \pi a_{\mathrm{TO}}}{P_{\mathrm{TO}}} \left( \frac{2 a_{\mathrm{TO}}}{a_{\mathrm{\oplus}}} - 1\right)^{1/2} $$ is \( 32.9~\mathrm{km}/\mathrm{s} \), Since \( v_{\oplus} = 29.8~\mathrm{km}/\mathrm{s} \) a 3.1 km/s boost is needed.
  • Similarly \( v_{\mathrm{ap,TO}} = 2.17~\mathrm{km}/\mathrm{s} \), a further boost of 2.3 km/s is needed to match \( v_{\mathrm{Mars}}=24.0~\mathrm{km}/\mathrm{s} \)

Stellar Parallax

  • The Earth’s rotation has a base line of 6371 km.
  • Using the Earth’s orbit provides a longer base line of \( 1.5 \times 10^{8}~\mathrm{km} \)
  • Diurnal Parallax: $$ 11.4~\mathrm{ly} = 1.08\times 10^{14}~\mathrm{km} $$ $$ \pi = \tan^{-1} \left[6371/(1.08 \times 10^{14}) \right] $$ or $$ (3.39 \times 10^{-9})^{\circ} = 12~\mu\mathrm{as} $$
  • Annual Parallax: $$ \pi = \tan^{-1} \left[ 1.5 \times 10^{8} / (1.08 \times 10^{14}) \right] $$ or $$ (7.94 \times 10^{-5})^{\circ} = 0.286~\mathrm{as} $$