Numbers, units and time

Quote a result so that its last digit is the first uncertain one. Round the uncertainty to one or two significant figures, then round the value to the same decimal place:

• $P_B = 1.12570 \pm 0.052$ d becomes $P_B = (1.13\pm0.05)$ d.
• $r = (2.7389\pm0.19)\times10^{-3}$ au becomes $(2.74\pm0.19)\times10^{-3}$ au.

Keep more digits in intermediate steps than in the final answer, especially when you subtract nearly equal numbers. The 2016 Moon problem finds the Sun's angle as $180^\circ - 36.2^\circ - 143.71^\circ = 0.09^\circ$: rounding $143.71^\circ$ to $144^\circ$ would make it negative.

Work in units that keep numbers near 1, and convert once at the end.

• Angles: $1\ {\rm rad} = 206\,265''$; $1'' = 1000$ mas; $1^\circ = 3600''$.
• Distances: $1\ {\rm pc} = 206\,265$ au $= 3.086\times10^{16}$ m; $1\ {\rm au} = 1.496\times10^{11}$ m; $R_\odot = 6.96\times10^8$ m.
• Galaxy dynamics: $G = 4.30\times10^{-6}\ {\rm kpc\,(km/s)^2}\,M_\odot^{-1}$. With $r$ in kpc and $v$ in km/s, $M = rv^2/G$ comes out directly in $M_\odot$.
• Velocities: $1\ {\rm km/s}\approx1.02$ pc/Myr; $1\ {\rm au/yr} = 4.74$ km/s, which is why tangential velocity is $v_t = 4.74\,\mu d$ ($\mu$ in arcsec/yr, $d$ in pc).
• Flux densities: $1\ {\rm Jy} = 10^{-26}\ {\rm W\,m^{-2}\,Hz^{-1}}$.
• A handy shortcut: an object of radius $R$ (in $R_\odot$) at distance $D$ (in kpc) has angular diameter $\theta = 9.30\times10^{-3}\,R/D$ mas, because $2R_\odot/1\,{\rm au} = 9.30\times10^{-3}$.

Astronomical times are continuous day counts, so intervals are a subtraction.

• Julian Date (JD): days since noon on 1 January 4713 BC. JD 2451545.0 is 2000 January 1, 12:00 TT.
• Modified Julian Date: $\mathrm{MJD} = \mathrm{JD} - 2\,400\,000.5$. It starts at midnight, not noon.
• Truncated dates: papers often subtract a further constant, such as 2016's tMJD (days since MJD 2 440 000). Check the definition every time.
• HJD and BJD: times corrected to the Sun or the Solar System barycentre. The correction is up to $\pm8.3$ minutes (the light travel time across 1 au), which matters for timing problems.

When counting days by hand across years, list the leap years: every fourth year, except century years not divisible by 400 (1900 was not a leap year, 2000 was). The 2007 Moon problem spans 1879 to 2006 and needs exactly 31 of them.

Remember the different "years" and "months": sidereal year 365.2564 d, tropical year 365.2422 d, sidereal month 27.3217 d, synodic month 29.5306 d.

Many data rounds give a figure instead of a table: a light curve, a calibration relation, an image of an eclipse.

• Read to about one fifth of the smallest grid division, and state that as your reading uncertainty.
• For a trend through scattered points, draw the line first, then read off the line, not the nearest point (the 2015 IC 342 Cepheids).
• To find a time when a curve crosses a level ($t_2$, the time to fade by 2 mag; $\Delta t_0$ for a supernova), interpolate linearly between the two bracketing points.
• To measure a circle that is only partly visible (the Earth's umbra on an eclipsed Moon), draw two chords and their perpendicular bisectors. They meet at the centre.