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 $$(7.94 \times 10^{-5})^{\circ} = 0.286~\mathrm{as}$$