Star Culmination and Altitude Calculation

The question asks us to find the altitude of a star at its lower culmination, given its declination and the fact that it culminates at the zenith.

Let's break down the terms:

• Declination ($\delta$): This is the celestial equivalent of latitude on Earth, measuring a star's angular distance north or south of the celestial equator. In this case, the star has a declination of 60° N.
• Culmination: This is when a celestial object crosses the local meridian. There are two types: upper culmination (highest point in the sky) and lower culmination (lowest point in the sky for circumpolar objects).
• Zenith: This is the point directly overhead the observer, with an altitude of 90°.

The problem states that the star culminates at the zenith. For a star to culminate exactly at the zenith, the observer's latitude ($\phi$) must be equal to the star's declination ($\delta$) and both must be in the same hemisphere. Since the declination is 60° N, the observer's latitude must be 60° N.

So, we have:

• Observer's Latitude ($\phi$) = 60° N
• Star's Declination ($\delta$) = 60° N

Calculating Star Altitude at Lower Culmination

For an observer in the northern hemisphere ($\phi$ N) and a star with northern declination ($\delta$ N), the altitude ($a$) of the star when it crosses the meridian is given by:

• At Upper Culmination: $a_{uc} = 90° - (\phi - \delta)$ if $\phi \ge \delta$
• At Upper Culmination: $a_{uc} = 90° + (\phi - \delta)$ if $\phi < \delta$
• At Lower Culmination: $a_{lc} = \phi + \delta - 90°$ (if the star is circumpolar)

The fact that the star culminates at the zenith ($a_{uc} = 90°$) confirms our latitude calculation:

If $\phi = \delta = 60° N$, then $a_{uc} = 90° - (60° - 60°) = 90° - 0° = 90°$. This matches the zenith culmination.

Now, let's find the altitude at lower culmination. We use the formula for lower culmination altitude for a northern star viewed from the northern hemisphere:

\begin{equation*} a_{lc} = \phi + \delta - 90° \end{equation*}

Substitute the values $\phi = 60° N$ and $\delta = 60° N$:

\begin{equation*} a_{lc} = 60° + 60° - 90° \end{equation*}

\begin{equation*} a_{lc} = 120° - 90° \end{equation*}

\begin{equation*} a_{lc} = 30° \end{equation*}

This formula is valid because $\phi + \delta = 60° + 60° = 120° > 90°$, which means the star is circumpolar from this latitude (it never sets below the horizon).

Therefore, the star's altitude at the lower culmination is 30°.