# Celestial velocities

We normally think of our classroom or laboratory as being stationary when we are doing dynamics. But how valid is this assumption?

We will consider motion due to:

1. Spinning of the Earth about the North-South axis.
2. Orbit of the Earth about the Sun.
3. Rotation of the Sun around center the Milky Way.
4. Motion of the Milky Way through the universe.

As we will see below, some of these velocities are not small. Why is normally valid to assume that we are in an inertial reference frame?

• Elementary motions #rke

## Periods of rotation

The different types of motion have magnitudes roughly given in the following table. These motions are not all in the same direction, and may add to each other or act in opposite or even orthogonal directions.

Earth spin Earth orbit Milky Way Through CMB
period $T$ $24\rm\ h$ $365\rm\ d$ $200\rm\ My$
$8.64 \times 10^4\rm\ s$ $3.16 \times 10^7\rm\ s$ $6.31 \times 10^{15}\rm\ s$
radius $r$ $6370\rm\ km$ $1\rm\ AU$ $27.2\rm\ kly$
$6.37 \times 10^6\rm\ m$ $1.50 \times 10^{11}\rm\ m$ $2.57 \times 10^{20}\rm\ m$
ang. vel. $15.0^\circ/\rm h$ $0.986^\circ/\rm d$ $1.8^\circ/\rm My$
$\omega = 2\pi/T$ $7.27 \times 10^{-5}\rm\ rad/s$ $1.99 \times 10^{-7}\rm\ rad/s$ $9.96 \times 10^{-16}\rm\ rad/s$
velocity $1670\rm\ km/h$ $107\,000\rm\ km/h$ $922\,000\rm\ km/h$ $1\,990\,000\rm\ km/h$
$v = r\omega$ $4.63 \times 10^2\rm\ m/s$ $2.98 \times 10^4\rm\ m/s$ $2.56 \times 10^5\rm\ m/s$ $5.52 \times 10^5\rm\ m/s$
acceleration $0.343\%\ g$ $0.0605\%\ g$ $2.60 \times 10^{-9} \%\ g$
$a = r\omega^2$ $3.37 \times 10^{-2}\rm\ m/s^2$ $5.93 \times 10^{-3}\rm\ m/s^2$ $2.55 \times 10^{-10}\rm\ m/s^2$

The first people to have a rough idea of the radius of the Earth and the distance to the Sun were the ancient Greeks. Eratosthenes (276–195 BCE) computed the radius of the Earth to be 40 000 stadia (6800 km) by measuring the difference in Sun angle between Aswan and Alexandria.

Aristarchus of Samos (310–230 BCE) obtained the first estimates of the distance to the Sun by using observations of lunar eclipses and solar parallax. While his method was in principle correct, poor observational data meant that his computed Earth-Sun distance was quite inaccurate.

## Solar and sidereal time

We all know that one day is 24 hours long. But the period of the Earth's rotation is not 24 hours! This is because of the difference between solar time and sidereal time. Solar time is the time measured against the Sun, as we normally do. Sidereal time is measured against the stars, which is slightly different.

Schematic of the Earth's rotation about its own axis and about the sun, counting solar days and sidereal days. Here the Earth has just 8 solar days per year for better visualization.

As we can see above, the Earth rotates one more sidereal day each year than solar days. This means:

$\text{1 solar year} = \text{365 solar days} = \text{366 sidereal days}$

and so:

\begin{aligned} \text{sidereal day} &= \frac{365}{366}\ \text{solar day} \\ &= \frac{365}{366} \times 24\ {\rm h} \\ &= 23.93{\rm\ h} \\ &= 23{\rm\ h}\ 56{\rm\ min} \end{aligned}

The Earth's orbital angular velocity is thus actually $\omega = 360^\circ / (23.93{\rm\ h}) = 15.04^\circ/\rm h$.

The relationship between solar and sidereal days can also be computed by considering just a single day, as shown below.

Diagram of one solar day and one sidereal day for the Earth, not drawn to scale. The fact that $\omega_{\rm E}$ and $\omega_{\rm S}$ are in the same direction means that sidereal days are shorter than solar days (this is not a coincidence).

Just as for the definition of a day, there is a similar distinction between the synodic lunar month, which is the time between passes of the Moon between the Earth and the Sun, and the sidereal lunar month, which is the orbital period of the Moon in an inertial frame. In common usage, the phrase lunar month refers to the synodic month.

Just as sidereal days and solar days are different, the exact length of one year depends on how we define it. The sidereal year is the time for the Earth to complete one orbit relative to the fixed stars, and has length 365.256363 solar days. The tropical year is the time for the Earth to return to the same point in the seasons, which varies around a value of about 365.242189 solar days (about 20 minutes shorter than the sidereal year). These years are different because of the axial precession of the Earth.

In common usage the word “year” refers to the tropical year, as the seasons have historically been more important for people than the motion of the stars. Observe that: $365.242 \approx 365 + \frac{1}{4} - \frac{1}{100} + \frac{1}{400}.$ This is why leap years in our Gregorian calendar add an extra day every 4 years, unless the year is a whole century, except every 400 years. If our calendar used sidereal years instead of tropical, the fact that 365.256363 is slightly larger than 365.25 would mean that about every 200 years we would need to have a double leap year, when there would be two extra days (February 30?).

While leap years occur because the year is not an exact integer number of days, a similar problem is caused by the fact that the solar day is not exactly 24 × 60 × 60 seconds. The length of the day actually varies somewhat unpredictably due to tidal friction as well as climactic and geologic events such as glacier formation and mass redistribution in the mantle, both of which change the Earth's moment of inertia. To correct for these variations a leap second is occasionally added, in which the last minute of the day has 61 seconds.