Question

If $\theta$ and $\delta$ be the latitude of an observer and declination of a heavenly body respectively, the upper culmination of the body will be south of zenith, if its zenith distance is

• A. $\delta - \theta$
• B. $\theta - \delta$
• C. $\theta + \delta$
• D. $(1/2) (\theta - \delta)$

Hint:

The zenith distance ($z$) of a celestial body at upper culmination depends on the observer's latitude ($\theta$) and the body's declination ($\delta$).

• If culmination is between the zenith and the celestial pole (north of zenith for Northern Hemisphere observer, i.e., $\delta>\theta$): $z = \delta - \theta$.
• If culmination is between the zenith and the celestial equator (south of zenith for Northern Hemisphere observer, i.e., $\theta>\delta$): $z = \theta - \delta$.
• If $\theta = \delta$, the body culminates at the zenith, and $z=0$.

The sign convention is crucial for determining the direction of culmination relative to the zenith.

Answer 1

The Correct Option is B

Step 1: Understand astronomical coordinates and zenith distance.

• Latitude ($\theta$): The angular distance of a place north or south of the Earth's equator. For an observer, it determines the elevation of the celestial pole.
• Declination ($\delta$): The angular distance of a point north or south of the celestial equator. It's analogous to latitude on Earth for a celestial body.
• Zenith: The point directly overhead an observer.
• Upper Culmination: The point at which a celestial body reaches its highest altitude in the observer's sky, crossing the local meridian.
• Zenith Distance ($z$): The angular distance of a celestial body from the zenith.

Step 2: Relate zenith distance, latitude, and declination at upper culmination.
The general formula for the zenith distance ($z$) of a celestial body at its upper culmination (when it crosses the local meridian) is given by: $z = |\text{observer's latitude} - \text{celestial body's declination}|$ However, the specific direction (north or south of zenith) provides the exact form. Consider an observer in the Northern Hemisphere (latitude $\theta$ is positive, measured north from the equator).

• The zenith is at an altitude of $90^\circ$. Its declination (celestial latitude) is equal to the observer's latitude, $\theta$.
• The celestial equator has a declination of $0^\circ$. Its altitude is $90^\circ - \theta$.
• The celestial pole (North Celestial Pole for Northern Hemisphere) has a declination of $90^\circ$. Its altitude is $\theta$.

For the upper culmination of the body to be south of the zenith, it means the celestial body's declination ($\delta$) is less than the observer's latitude ($\theta$), and the body is located between the zenith and the celestial equator (assuming both $\theta$ and $\delta$ are positive, i.e., in the same hemisphere). In this geometric configuration, the angular distance from the zenith to the body is the difference between the zenith's declination (which is $\theta$) and the body's declination ($\delta$). Therefore, the zenith distance ($z$) is: $z = \theta - \delta$
Step 3: Select the correct option.
Given the condition that the upper culmination is south of zenith, the zenith distance is $\theta - \delta$. $$\boxed{\theta - \delta}$$