Celestial Body moves along Parallel of Declination

Theorem

Consider the celestial sphere $C$ whose observer is $O$.

Let $B$ be a celestial body on $C$.

Let $\delta$ be the declination of $B$.

Let $P_\delta$ be the parallel of declination whose declination is $\delta$.

Then $B$ appears to move along the path of $P_\delta$, in the direction from the north to east to south to west.

It takes $1$ sidereal day for $B$ to travel around $P_\delta$.

The apparent movement of $C$ is caused by the rotation of Earth about its axis.

This rotation takes $1$ sidereal day, by definition, for the principal vertical circle to be in the same position again.

As the Earth rotates, the observer $O$ moves in an easterly direction.

Hence the celestial sphere $C$ appears to move in the opposite direction, so from north to east and so on.

$B$ remains at the same distance from the north celestial pole as it did.

Hence its declination $\delta$ stays the same.

Hence the result.

$\blacksquare$

1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text {II}$. The Celestial Sphere: $19$. Declination and hour angle.