LOG#221. Kepler+Cosmic speeds.

Johannes Kepler
Kopie eines verlorengegangenen Originals von 1610

Hi, there! Today some Kepler third law stuff plus cosmic speed calculations and formulae. Cosmic speed sounds cool…But first…

From High School, you surely calculated how fast is A PLANET moving around the sun (or any star, indeed). To simplify things, take units with $G_N=4\pi^2$ for a moment (nasty trick that works!). Kepler third law reads

$T^2=\dfrac{a^3}{M}$

where $M=M_\star+M_p$ is the total mass, sum of the star mass and the planet star, $a$ is the major semiaxis of the ellipse and $T$ is the period of the motion. Note that for binary systems like binary stars, you can not neglect the $M_p$ term since it is comparable to the higher mass. Simple calculus, let you obtain

$v_P=2\pi\left(\dfrac{M}{T}\right)^{1/3}$

Have you ever asked yourself what is the STAR speed? Generally speaking, the star is NOT static either in gravitation! So, forget that picture on your head telling you that the sun is fixed, it also moves. What is the star speed? It shows that you can compute easily the star speed with the aid of the conservation of linear momentum. Linear momentum $p=mv$ is conserved, due to translational invariance in 3d space, and thus,

$P_\star+P_p=constant$

Set the constant to $0$ , and take the modulus, so you can see now that

$V_\star=\dfrac{m_pv_p}{M_\star}\approx\dfrac{m_pv_p}{M}$

Then, you get

$V_\star=2\pi\left(\dfrac{M}{T}\right)^{1/3}\dfrac{m_p}{M}$

or equivalently

$V_\star=2\pi\left(\dfrac{1}{M}\right)^{2/3}\dfrac{m_p}{T^{1/3}}$

In our units with $G_N=4\pi^2$ , speeds are indeed measured in $AU/yr$ ! Now, you can not only calculate the speed of Earth around the sun, you can indeed calculate the speed of the sun around the Earth. You can extend this argument and calculate the difference between the planet speed, the star speed and the center of mass speed. It is a quite pedagogical exercise! In fact, there are two extra corrections to the abouve formulae in the general setting of celestial mechanics: you must include the effect of eccentricity and the inclination of the system with respect to the observer. The above formulae suppose you look perpendicular to the system, and the eccentricity is small or zero. If you use standard $G_N$ SI units, you would get instead

$v_{p}=\left(\dfrac{2\pi GM}{T}\right)^{1/3}$

$V_\star=\dfrac{m_pv_p}{M_\star}=\dfrac{m_p}{M_\star}\sqrt[3]{2\pi}\left(\dfrac{GM}{T}\right)^{1/3}$

and generally

$P_{CM}=\dfrac{m_pv_p+M_\star V_\star}{M}$

and normally you choose $P_{CM}=0$ for convenience, but it can also be calculated with respect to the planet or star frames!

Angular speed in Kepler law (or velocities/speeds) are related to the space dimension in space-time! Thus, in $D=d+1$ space-time you would get angular speeds

$\Omega^2=\dfrac{G_{d+1}M}{R^{d}}$

and periods would scale as $R^{d/2}$ . Moreover, if you are orbiting a star but a GR rotating object, it is described better by a Kerr metric. In Kerr spacetimes, Kepler third law gets generalized into ( $G_N=c=1$ )

$\Omega=\pm\dfrac{M^{1/2}}{r^{3/2}\pm aM^{1/2}}$

where $a$ is now the Kepler parameter. Reintroduce units to get instead

$\Omega=\pm\dfrac{\sqrt{GM}}{r^{3/2}\pm\chi\left(\dfrac{\sqrt{GM}}{c}\right)^3}$

Kepler third law can also be extended, for instance,in Finsler-like general relativity. There, you could get

$\dfrac{T^2}{R^3}=\dfrac{4\pi^2}{GM}\left(1-\dfrac{A(R)}{R^4}\right)^{-1}$

or even stranger formulae in other gravitationally modified theories of gravity are even possible. Therefore, if you modify gravity with extra dimensions or more general theories, you obtain corrections to the Kepler third law. Even simple rotating black holes provide a generalized Kepler third law (the above formula is for ecuatorial orbits only!) for orbitating bodies! Thus, observations on orbital patterns could provide you hints on modified gravity. Unfortunately, no observation is yet giving you a MOG (MOdified Gravity) or extended theory of gravity confirmation. It implies strong bounds on the possible sizes of these corrections or discard them till now!

To end this post, I will review the so-called cosmic speeds:

First cosmic speed. Namely, the orbital speed. For usual spacetime dimensions read

$V_1=\sqrt{\dfrac{GM}{R}}$

Second cosmic speed. It is the escape velocity/speed. It yields

$V_2=\sqrt{\dfrac{2GM}{R}}=\sqrt{2}V_1$

Third cosmic speed. It is the escape velocity from the solar system. A naive calculation for Earth third cosmic speed gives

$V_3=\sqrt{\dfrac{2GM_\odot}{R_E}}=42km/s$

but the fact that Earth is also moving, let us reduce this value to a lower number, since $V(oS)=V_S-V_o=12.3km/s$ , where $V_o=29.8km/s$ is the orbital Earth speed, such as

$\dfrac{1}{2}mV_3^3-\dfrac{GM_Em}{r_E}=\dfrac{1}{2}mV_{oS}^2$

$V_3=16.7km/s$

Fourth cosmic speed. You need naively $V_4'=350km/s$ for escape from the Milky way, but as the solary system is moving with respect to it, you can easily show up (exercise for you!) that you would need only $V_4=130km/s$ to escape, similarly to the case of the third cosmic speed.

Remember the instantaneous speed is:

$v=\sqrt{GM\left(\dfrac{2}{r}-\dfrac{1}{R}\right)}$

Let me remark a final two constants (the hidden secret constant will be the topic of a future blog post about the full Kepler problem and its generalizations) formulae for the Kepler reduced 2-body problem:

$E_t=-\dfrac{GM\mu}{2R}$

$L=\mu\sqrt{GMR(1-e^2)}$

where $E_t, L$ are the total energy and angular momentum $M$ is the total mass, $\mu=(M_1+M_2)/M$ the reduced mass, and R the orbital major semiaxis, and $e$ is the orbital eccentricity. Are you eccentric today? Problem: What would be the escape velocity from our Universe?

A summary table:

See you in a future new blog post!

P.S.: Some earthling speeds are

i) Rotational speed of Earth on the equator is about 1670km/h or 0.46km/s.

ii) Rotational speed of Earth around the sun is about 107000km/h, or about 30 km/s.

iii) Rotational speed of the solar system (the sun) around the Milky Way is about 220km/s (828000km/h). The Milky Way spins at about 270 km/s with respect to its center.

iv) Milky Way speed towards the Big Attractor is about 611km/s, or about 2.2 million km/s.

v) Milky Way speed with respect to the CMB is about 2268000km/h or about 630 km/s.

vi) What about the fifth cosmic velocity? Yes, it can be defined. The fifth cosmic speed is the velocity necessary to escape from this Universe. Take $R_U, M_U$ as the radius and mass of this Universe. You can calculate the 5th cosmic speed as follows

$v_5=\sqrt{\dfrac{GM_U}{R_U}}$

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LOG#221. Kepler+Cosmic speeds.

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