4
$\begingroup$

I'm trying to solve this task:

Tula and Moscow are located on the same meridian. In which city length of the day is bigger on 22 of June? 22 of December?

I was thinking like that: we need to calculate the height of the sun above the horizon and where it will be bigger there the length of a day is bigger. The formula of height is:

$$\text{h = 90 - latitude + declination}.$$

On 22 June the declination of Sun is maximum and equals 23,5. Latitude of Tula - 54,2 , latitude of Moscow - 55,3. It seems like in Tula the height of the Sun is bigger than in Moscow, so bigger is the length of the day. But in reality it is the opposite. The further from equator a city is located the bigger the length of the day in this city.

Could you please tell me where am I wrong and how to solve this task correctly?

$\endgroup$
3
  • $\begingroup$ the premise that the max height of the sun determines the length of the day is wrong. On 21Mar all Earth has the same day length, but only in the equator the sun in Zenith. $\endgroup$
    – d_e
    Jul 13 at 9:32
  • $\begingroup$ @d_e That is not true because the length of day is defined as the interval between when the top of the Sun first appears above the horizon and when the top of the Sun finally disappears below the horizon. An extreme example is the South Pole, where it takes several 24 hour days for the Sun to rise around September 22, and later it takes several 24 hour days for the Sun to set around March 21. $\endgroup$ Jul 13 at 14:21
  • $\begingroup$ @DavidHammen, I agree with you. Not sure what is the that that "is not true"? Other than that, my comment might not be 100% accurate, but basically it is, I think, and demonstrate that the length of the day is not a function of the max height of the sun in the that day, when comparing different locations. $\endgroup$
    – d_e
    Jul 13 at 14:36
6
$\begingroup$

Note that the Sun here is point-like and there is no refraction.

What have I done wrong?

Your supposition that the altitude of the Sun is directly related to the length of the day is wrong. Proof with counterexample: observe the altitude of the Sun on the equator and in the Moscow on the equinox; but still, the lengths are the same.


And what is the correct way of dealing with it?

My images have helped to many people, I hope they will help you as well:

Figure 1 where the first arabic number corresponds to the day of the month and the second roman number is the month (I for January, II for February ...)

Note that the altitude of the Sun on the June 21 is really larger in Tula than in Moscow, but the length of the Sun's trajectory across the sky in Tula is smaller than in Moscow. So, this is a short graphical solution.

But astronomers always want the numerical solutions

so here you have it: Such length is the so called hour angle $H$, which is the angular distance between the celestial meridian and the setting Sun [derivation]. The equation is

$$\cos H=-\tan\delta\tan\phi$$ where $\delta$ is the declination of Sun on date and $\phi$ is the latitude. So, with quick calculations for June 21 ($\delta=23.5°$),

  • $H_{Tula}=127.1°$
  • $H_{Moscow}=128.9°$

So, really, the hour angle in Moscow is larger, which corresponds to longer day.


What about December 21?

You can easily calculate it as well using $\delta=-23.5°$.

$\endgroup$
2
  • 1
    $\begingroup$ I believe that 21 is the day and VI is month, so 21.VI means the 21 of June. $\endgroup$
    – ALiCe P.
    Jul 14 at 5:58
  • 1
    $\begingroup$ @uhoh Oh, sorry, I edited it. (That's how it is shorted in central part of Europe.) $\endgroup$
    – User123
    Jul 14 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.