When you look at a nearby star against the distant stellar background, the foreground star moves. Not much, mind you. But it is detectable. In fact, a nearby star may change position by as much as several arc seconds a year as a result of its velocity relative to the solar system.

But things get a tad more complicated than that, because the Earth itself is a moving platform. As a matter of fact, the unit of distance that is preferred by astronomers, the parsec, is actually defined as the distance at which an object appears to be displaced by one arc second when the Earth is displaced by one astronomical unit (the Earth-Sun distance, i.e., the approximate radius of the Earth's orbit.)

So how do you calculate the exact sky position of a star at a given time?

Star positions are usually given using a form of spherical coordinates: equatorial coordinates centered on the Earth, with the sky position of the star given as its right ascension $\alpha$ and declination $\delta$. The proper motion of a star is given using the corresponding angular components $\mu_\alpha$ and $\mu_\delta$, typically expressed as milliarcseconds per year. It is to this proper motion that we must also add the parallax contribution due to Earth's orbital motion.

The first step is to convert the star's coordinates to equatorial Cartesian coordinates. For this, we also need the distance $r$ to the star:

\begin{align}

\begin{pmatrix}x_{\tt eq}\\y_{\tt eq}\\z_{\tt eq}\end{pmatrix}=\begin{pmatrix}\cos\delta\cos\alpha\\\cos\delta\sin\alpha\\\sin\delta\end{pmatrix}r.\tag{1}

\end{align}

Next, we perform the transformation from equatorial to geocentric ecliptic coordinates, with $\epsilon\sim 23.5^\circ$ representing the inclination of the ecliptic plane with respect to the celestial equator:

\begin{align}

\begin{pmatrix}x_{\tt ec}\\y_{\tt ec}\\z_{\tt ec}\end{pmatrix}=

\begin{pmatrix}

1&0&0\\0&\phantom{-}\cos\epsilon&\sin\epsilon\\0&-\sin\epsilon&\cos\epsilon

\end{pmatrix}

\begin{pmatrix}x_{\tt eq}\\y_{\tt eq}\\z_{\tt eq}\end{pmatrix}.\tag{2}

\end{align}

Next, we apply the apparent motion of the Earth (assuming a circular orbit of radius $R=1$ AU with orbital period $T=1$ year):

\begin{align}

\begin{pmatrix}x'_{\tt ec}\\y'_{\tt ec}\\z'_{\tt ec}\end{pmatrix}=

\begin{pmatrix}x_{\tt ec}\\y_{\tt ec}\\z_{\tt ec}\end{pmatrix}+

\begin{pmatrix}

\phantom{-}R\sin 2\pi t/T,\\

-R\cos 2\pi t/T,\\

0

\end{pmatrix}.\tag{3}

\end{align}

We can now transform this back into the equatorial coordinate system:

\begin{align}

\begin{pmatrix}x'_{\tt eq}\\y'_{\tt eq}\\z'_{\tt eq}\end{pmatrix}=

\begin{pmatrix}

1&0&0\\0&\cos\epsilon&-\sin\epsilon\\0&\sin\epsilon&\phantom{-}\cos\epsilon

\end{pmatrix}

\begin{pmatrix}x'_{\tt ec}\\y'_{\tt ec}\\z'_{\tt ec}\end{pmatrix}.\tag{4}

\end{align}

Finally, we can convert back into right ascension and declination:

\begin{align}

\begin{pmatrix}\alpha'\\\delta'\end{pmatrix}=

\begin{pmatrix}

\arctan y'_{\tt eq}/x'_{\tt eq}\\

\arctan z'_{\tt eq}/\sqrt{{x'}^2_{\tt eq}+{y'}^2_{\tt eq}}.

\end{pmatrix}.\tag{5}

\end{align}

The parallax motion of the star, then, due to the Earth's orbital motion is the time derivative:

\begin{align}

\begin{pmatrix}

\mu'_\alpha\\\mu'_\delta

\end{pmatrix}

=

\frac{d}{dt}

\begin{pmatrix}\alpha'\\\delta'\end{pmatrix}.\tag{6}

\end{align}

Last, the apparent motion of the star can be obtained by adding these values to its proper motion $\mu$:

\begin{align}

\begin{pmatrix}

\mu^{\tt app}_\alpha\\\mu^{\tt app}_\delta

\end{pmatrix}=

\begin{pmatrix}

\mu'_\alpha\\\mu'_\delta

\end{pmatrix}+

\begin{pmatrix}

\mu_\alpha\\\mu_\delta

\end{pmatrix}.\tag{7}

\end{align}

These steps are not terribly complicated (especially when using, never even mind computer algebra, a spreadsheet will do just fine) but the actual expressions can get messy, so there's no point trying to express the final result as a closed form expression.

It is worthwhile to observe that the radius of the Earth's orbit figures only in Eq. (3), which could be rewritten using $R/r$ in place of $R$ in the second matrix on the RHS, multiplied by $r$ on the outside. That way, all terms have the common factor $r$ in all the equations, so this factor cancels. This makes it clear that the contribution of parallax is always proportional to $R/r$ and vanishes when the star is at a great distance from the Earth.