4
$\begingroup$

From several texts I have read, I have learned that right ascension is measured on a scale of 24 hours. I understand that Earth rotates almost exactly 360 degrees relative to distant objects, which is what RA is used for. The 24 hour system should only apply to the 361 degree rotation period (solar day). Why then do we split the sidereal day of the earth into 24 hours, when if it were split into 23 hours and 56 minutes we could know where an object would be x hours away after x hours?

More (possibly) relevant information: I understand the Hour Angle HR is the measure of how far Aries is from the local meridian (RA of current meridian of longitude). This way you could calculate how many hours away the object is, but once again, this would not be exactly correct as 4 minutes has to be subtracted from the 24 hour scale, multiplying the whole time to something like 99.7% (14.44 degrees per hour), correct?

Thanks, this is really confusing me and I would appreciate some help. I did see a similar question but that had nothing to do with the sidereal day v. solar day issue I'm having.

Improve this question
$\endgroup$
4
  • $\begingroup$ My guess: it's because right ascension was originally defined in analogue to (or even by) longitude used for navigation. If you're measuring longitude using measurements of local time compared to a clock, it's most convenient to use a 24 hour, 60 min, 60 sec system. Then, when you're measuring the position of stars, the spherical trigonometry tools and tables used for navigation become convenient. $\endgroup$
    – Sean Lake
    Sep 21, 2017 at 6:05
  • 1
    $\begingroup$ While your complaint is technically valid, mathematical purists such as myself use radians for everything :) $\endgroup$
    – user21
    Sep 21, 2017 at 15:56
  • 2
    $\begingroup$ So if someone is specifying right ascension in degrees rather than hour/min/sec, the celestial equator should not be divided into 360 degrees? $\endgroup$
    – DJohnM
    Sep 21, 2017 at 21:20
  • $\begingroup$ No, it should be divided into 360 degrees, as the earth rotates that much with respect to the vernal equinox. However, when we convert that to hours, I claim it should be not 24 hours but rather the time it takes the vernal equinox to appear at the same spot in the sky (about 4 minutes under 24 hours). $\endgroup$
    – Rithwik Sudharsan
    Sep 23, 2017 at 2:10

3 Answers 3

Reset to default
5
$\begingroup$

Right ascension is a historical oddity.

To specify a point in the sky you need a coordinate system, the one which we have come to use has it's origin at the Point of Aries on the Equator and the Ecliptic (a reasonable choice), It uses the Equator for one axis, and the meridian through the point of Aries for the other, again these are convenient and reasonable.

The coordinate system also requires a unit to be chosen for measuring angular distance. For measuring declension, we use degrees. The more mathematically pure may prefer radians, but degrees are a commonly used unit for measuring angles.

For measuring the angular distance around the equator we have come to use "hours", by dividing the full turn into 24 parts. This is a historical oddity. It has a few conveniences: The sidereal day is only slightly shorter than 24 hours, so a star with RA of 1hr will be due south about 1 hour after a star with a RA of 0hr (not exactly but close enough for a rule of thumb). A telescope with an equatorial mount that is set to make a full turn in 24 hours will track stars well enough for a human observer.

Dividing the equator into (23+56/60) parts would be inconvenient since its not a whole number. For these reasons Flamsteed used Hours for his catalogue in 1712, and we have followed his tradition.

So the use of "hours" may be roughly related to the apparent motion of the stars, but it is really just a way of specifying an angle, and using 24 is easier than (23+56/60)

Improve this answer
$\endgroup$
1
  • $\begingroup$ As JohnHoltz has pointed out, if you waited for a star to move 6 hours solar time while the RA is measured in sidereal time, then the star would have moved 1 minute ahead. But over many days this would cause it to be very far from your line of sight, correct? If you agree, telescopes should be set to complete a full turn in one sidereal day rather than a solar one, and there would be no true conveniences? $\endgroup$
    – Rithwik Sudharsan
    Sep 23, 2017 at 2:21
3
$\begingroup$

The problem is not that right ascension goes from 0 to 24 hours (or 0 to any other number). The problem is that the Earth is revolving around the Sun, and our time system is based on the Sun. Because of the motion around the Sun, one solar day (24 hours 0 minutes 0 seconds) is different than one sidereal day (23 hours 56 minutes 4 seconds). So you need two clocks: one for the sun, and one for the stars.

An example may help. Let's change right ascension to be from 0 to 100. At midnight (00:00) on day 1, let's define that 0 right ascension is on the meridian. At 23:56, the Earth has made one complete rotation relative to the stars, so 0 right ascension is back on the meridian. At 24:00, the Earth has made one complete rotation relative to the Sun (one day), but the right ascension has increased to 0.28 during the extra 4 minutes.

Day 2. At 23:52, the Earth has completed a second full rotation relative to the stars, so the right ascension is again 0. But after two complete days (24:00 on day 2), the right ascension is now 0.56.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you for making my thoughts clearer, this did help me understand the problem more clearly. However, once again, that is my question: If we still use our 24 hour system, then shouldn't the sidereal rotation of 360 degrees be 4 minutes less? I doubt any non-astronomers would be dealing with such numbers anyways, so why not do this instead of calculating in 24 hours sidereal time (which is obviously 23.9 hours solar time). $\endgroup$
    – Rithwik Sudharsan
    Sep 23, 2017 at 2:16
  • $\begingroup$ Yes, the Earth's rotation makes the sun rotate 360 degrees in 1440 minutes (24 hours), and the stars rotate 360 degrees in 1436 minutes. I think you are trying to make 1 hour of right ascension (RA) equal 1 hour of clock time, but there is no advantage to do that. In fact there are disadvantages to do that such as 360 degrees would equal 23:56:04 RA. $\endgroup$
    – JohnHoltz
    Sep 23, 2017 at 13:36
  • $\begingroup$ That exactly what I am doing, why is that last bit a disadvantage? One would have to convert sidereal time to solar time nevertheless, for observational purposes at least $\endgroup$
    – Rithwik Sudharsan
    Sep 23, 2017 at 19:58
  • $\begingroup$ If you still need a formula to convert solar time to sidereal time based on the day of the year and the time, what is the advantage to make 1 hr RA equal to 1 hr of time? You are replacing 1 equation with another. $\endgroup$
    – JohnHoltz
    Sep 24, 2017 at 1:10
  • $\begingroup$ well it's take your hour angle and just add it to your local time, but now I understand that you would rather use local sidereal time and would end up converting that too anyways. $\endgroup$
    – Rithwik Sudharsan
    Sep 25, 2017 at 3:29
2
$\begingroup$

It's 23 hours, 56 minutes, and 4.1 seconds, more or less. That's one sidereal day expressed in solar time. Solar time is neither a particularly convenient nor particularly useful time scale when looking at the stars.

Alternatively, one could use sidereal time, which is what hour angle refers to. There are 24 sidereal hours in a sidereal day, 60 sidereal minutes in a sidereal hour, and 60 sidereal seconds in a sidereal minute.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Why the random drive by downvote? This answer is correct. Hour angle refers to sidereal time, not solar time. $\endgroup$
    – David Hammen
    Sep 22, 2017 at 0:46
  • $\begingroup$ Thanks for your answer. Does this mean that astronomers did/do this simply for roundness? I assume one would have to convert to solar time for any real word adjustments for ease of recognizability. $\endgroup$
    – Rithwik Sudharsan
    Sep 23, 2017 at 2:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .