What is true purpose of polar aligning? Last night, I looked at Mars. It was moving fast. I still need to adjust RA axis to track fast-moving Mars. Why the telescope didn't track it by itself?

  • 1
    $\begingroup$ The proper motion of Mars isn't that large. I suspect your scope wasn't aligned correctly to the equatorial plane. Try tracking a star & see if you need to adjust the RA. $\endgroup$
    – PM 2Ring
    Oct 7, 2018 at 2:45

2 Answers 2


The Earth's rotation makes the stars and planets appear to move westward across the eyepiece field of a stationary telescope at about 900 arcseconds per minute. An equatorial mount can counter this by moving only on its polar axis at a constant rate. An altitude-azimuth mount must move on both axes at variable rates in order to track.

In early October 2018, Mars appeared to move about 1.1 arcseconds per minute relative to the stars. In an equatorially mounted telescope, properly aligned and tracking at the sidereal rate, Mars would take about 13 minutes to move through its own apparent diameter of 15 arcseconds. Without a tracking motor, you'd have to move the RA axis by hand, but you could leave the declination axis alone for a while.

If the mount has a tracking motor designed to be usable in either hemisphere, it may have a switch to reverse direction. If that switch is in the wrong position, stars appear to move westward twice as fast as if the motor were switched off.

If an equatorial mount is misaligned, stars appear to drift in declination, depending on their position in the sky, up to 15 arcseconds per minute per degree of polar alignment error. For example, if you're in the northern hemisphere and the mount's polar axis is 5 degrees too far to the east, objects near the meridian appear to drift southward at 75 arcseconds per minute, whereas objects 4 hours off the meridian drift only half that fast. If the polar axis is more than 30 degrees off the celestial pole, it might as well be an alt-az mount.


There are two main coordinate systems used in astronomy: equatorial and horizontal. The equatorial system imagines a sphere around the earth on which stars occupy (for our purposes) fixed positions. (In fact, the stars do move slightly, but not with a speed that matters for casual amateur observation.) The point is that this system works well for objects (particularly, stars) with negligible amounts of motion. We use declination and right ascension to describe positions on the sphere, just as we use latitude and longitude to describe a position on the sphere of the earth. The point of polar aligning an equatorial mount is to find a convenient starting point, which is the North Celestial Pole: it's how your telescope gets aligned with this grid on the sky. Once your telescope "knows" where the pole is, you can manually move it using calibrated mechamisms to a particular position in RA & Dec. The second coordinate system is the horizontal one, which uses the observer's position as its starting point (it is location-dependent), and measures objects using altitude (from the observer's horizon) & azimuth (true bearing from the observer's position). An alt-az telescope mount can move freely up and down, left and right. Sometimes it has markings for horizontal coordinates.

An equatorial mount was not designed to track planets, but it can do so. The advantage of using an equatorial mount is that you can finely adjust the scope to account for the movement of the planet during the (limited) time you're observing.

Why didn't the telescope track it by itself?

Do you have a GoTo system?

  • $\begingroup$ aha... "An equatorial mount was not designed to track planets, but it can do so. " This was crucial information. Planets move much faster for tracking by the mount, I guess. $\endgroup$
    – KawaiKx
    Oct 8, 2018 at 8:43
  • $\begingroup$ @KawaiKx Mars was moving about 0.1% slower. $\endgroup$
    – Mike G
    Oct 8, 2018 at 11:31
  • $\begingroup$ @KawaiKx I'm skeptical that the movement of Mars against the stellar background over a time period of a few hours is anywhere near sufficient to exit your telescope's field of view. Did the stars leave the FoV as well? $\endgroup$ Oct 8, 2018 at 18:08
  • $\begingroup$ @KawaiKx The path of the planet isn't parallel to the celestial equator. For short time-spans, it doesn't matter. $\endgroup$
    – Alphecca
    Oct 8, 2018 at 19:44

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