Answer: Declination of the Sun - Spring Equinox

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Discussion Overview

The discussion centers around the declination of the Sun on the Spring equinox, particularly from the perspective of someone standing at the Tropic of Cancer. Participants explore the concepts of solar declination and altitude, and how these relate to the equinox.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Maria questions whether the declination of the Sun on the Spring equinox is 0 degrees or 23.5 degrees, expressing confusion about the definitions.
  • Tiny-Tim explains that at an equinox, the Sun is directly overhead at the equator, implying a declination of 0 degrees.
  • Maria reflects on the relationship between solar altitude and declination, suggesting that at the Tropics, the solar altitude would be 66.5 degrees.
  • Some participants clarify that declination is the same for all positions on Earth, reinforcing that the Sun's declination is 0 degrees at the equinox.
  • Marty confirms that at the Tropic of Cancer, the Sun culminates at an altitude of 66.5 degrees at local noon, with a zenith distance of 23.5 degrees.
  • Maria seeks clarification on how to determine the altitude of a star without a telescope, exploring practical methods such as using a ruler held at arm's length.

Areas of Agreement / Disagreement

Participants generally agree that the declination of the Sun is 0 degrees at the equinox, but there is some confusion regarding the relationship between declination and altitude, particularly from the perspective of different latitudes. The discussion remains somewhat unresolved as participants clarify their understanding without reaching a definitive consensus on all points.

Contextual Notes

There are limitations in the discussion regarding the definitions of declination and altitude, as well as the assumptions made about the observer's location and the implications of solar position at different latitudes.

Maria76
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Hi,

This might sound like a simple question, but it's not for me.

I'm standing at the Tropic of Cancer - what would be the declination of the Sun on the Spring equinox. I believe it should be zero degrees, but my brother says it's 23.5 degrees. Can anybody shed light?

Thanks in advance.

Maria
 
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Welcome to PF!

Maria76 said:
I'm standing at the Tropic of Cancer - what would be the declination of the Sun on the Spring equinox. I believe it should be zero degrees, but my brother says it's 23.5 degrees. Can anybody shed light?

Hi Maria! Welcome to PF! :smile:

At an equinox, the Sun is overhead (at midday) at the equator. :wink:
 
Hi Tiny-Tim,

Thanks for your answer.

Now I'm having a little trouble visualizing the situation ...

If the Sun is overhead at midday at an equinox at the equator, it means that the solar altitude is 90 degrees. At the Tropics, which are at latitude 23.5 degrees, the solar altitude would be at 66.5 degrees (90 - 23.5) ... is that right? If I'm visualizing it properly, it means that the Sun's declination is 23.5 degrees at the Tropics at an equinox, and 0 degrees at the equator.

But then I'm thinking that at the equinox, the Earth's rotation axis is perpendicular to the line of intersection between the ecliptic and the equator of the celestial sphere ... so that's what makes me think that the Sun's declination is 0 degrees, because the Sun is also perpendicular to that line of intersection.

Have I totally misunderstood? I think I may have confused the Sun's declination with its altitude.

Thanks again.

Maria
 
hmm … looking at http://en.wikipedia.org/wiki/Declination" , declination of a star is the amount north or south of the celestial equator, and so is the same for all positions on the Earth.
 
Last edited by a moderator:
Maria76 said:
Hi,

This might sound like a simple question, but it's not for me.

I'm standing at the Tropic of Cancer - what would be the declination of the Sun on the Spring equinox. I believe it should be zero degrees, but my brother says it's 23.5 degrees. Can anybody shed light?

Thanks in advance.

Maria

Declination of the Sun on the celestial sphere is 0 degrees at the March Equinox.

If you are standing at the Tropic of Cancer you are 23.5 degrees north of the equator. The Sun will culminate (highest) at local noon at a zenith distance 23.5 degrees or an altitude above northern horizon at 66.5 degees

Marty
 
Marty-0750 said:
Declination of the Sun on the celestial sphere is 0 degrees at the March Equinox.

If you are standing at the Tropic of Cancer you are 23.5 degrees north of the equator. The Sun will culminate (highest) at local noon at a zenith distance 23.5 degrees or an altitude above northern horizon at 66.5 degees

Marty

Hi Marty & Tiny-Tim.

Let me see if I have this right. As Tiny-Tim says, declination of a star is the amount north or south of the celestial equator, and so is the same for all positions on the Earth. In that case, if I am standing at the Tropic of Cancer, the declination of the Sun is 0 degrees at the equinox. But it has an altitude of 66.5 degrees above the horizon. Do I have it right?

Thanks again.

Maria
 
Maria76 said:
Hi Marty & Tiny-Tim.

Let me see if I have this right. As Tiny-Tim says, declination of a star is the amount north or south of the celestial equator, and so is the same for all positions on the Earth. In that case, if I am standing at the Tropic of Cancer, the declination of the Sun is 0 degrees at the equinox. But it has an altitude of 66.5 degrees above the horizon. Do I have it right?

Thanks again.

Maria

Yep.

Marty
 
Welcome to PF!

Yep. :smile:

and also … Welcome to PF, Marty! :biggrin:
 
Hi again,

Just out of curiosity, given that the altitude of a star is its height above the horizon, how would I be able to determine the altitude of a star without a telescope? I'd probably need something like a sextant or an astrolabe? Or is there a more simple way?

Thanks again.

Maria
 
  • #10
Maria76 said:
Hi again,

Just out of curiosity, given that the altitude of a star is its height above the horizon, how would I be able to determine the altitude of a star without a telescope? I'd probably need something like a sextant or an astrolabe? Or is there a more simple way?

Thanks again.

Maria

For a metric ruler held at arms length, 1 cm is close to 1 degree across the sky.

Hold it out vertically and with one eye aligning the botton with 0 cm on horizon.

Align star with edge of ruler.

The cm reading is its approximate altitude +/- 1 degree

For refinment due to somewhat varying arm lengths for different people, measure length (L cm) from eye to ruler when you are holding ruler vertically.

The the angle (degrees) 1 cm subtends will be:

1 / L * 57.3

Marty
 

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