To do so, I'm following what's described in this document and implementing C++ code.
That is a very old document you are using. It's using the 1982 IAU precession model. There have been multiple updates since then. The latest set of changes are working toward eliminating the concept of Greenwich Sidereal Time. (In software terms, GMST and GAST are deprecated concepts. While still currently supported, support will be dropped in some future release of the Earth rotation model.)
Other than using an outdated version of a deprecated concept, what did you do wrong? The expression you are using to compute GMST is at the top of page A-26:
$$\Lambda = H_0 + \Delta H + \omega^{\ast}\,(t-\Delta t)$$
The $\Delta H$ term takes one from GMST to GAST. Dropping that term, the expression for GMST is
$$H = H_0 + \omega^{\ast}\,(t-\Delta t)\tag1$$
where
$$\begin{align}
H_0 = &\,24110.54841 + 8640184.812866 T_u + 0.093104 {T_u}^2 - 6.2\times10^{-6} {T_u}^3\tag 2 \\
&\,\text{is the Greenwich sidereal time at midnight, in seconds of time} \\
\omega^{\ast} = &\,7.2921158553\times10^{-5} + 4.3\times10^{-15} T_u\tag3 \\
&\,\text{is the Earth's sidereal rotation rate, in radians}/\text{second} \\
t\phantom{\Delta}\quad&\text{is UTC time of day, in seconds, and} \\
\Delta t\quad&\text{is the difference between UT1 and UTC.}
\end{align}$$
Note that there's a units mismatch in equation (1). Equation (2) yields a result in units of sidereal seconds while equation (3), when multiplied by time of day, yields a result in units of radians. The latter needs to be converted to sidereal seconds. To do this, simply multiple both terms in equation (3) by $\frac{86400}{2\pi}$:
$$\begin{align}
\omega^{\ast} = &\,1.00273790935 + 5.9\times10^{-11} T_u\tag4 \\
&\,\text{is the Earth's sidereal rotation rate, in sidereal seconds}/\text{UT second}
\end{align}$$
Applying this to 2019 Jan 01 08:00:00 UTC yields the following:
- The Julian date at midnight UTC 2019 Jan 01 is $2458484.5$, or $2451545+6939.5$. See footnote 1 for an explanation of this convention.
- Dividing the residual 6939.5 by 36525 yields $T_u$: $T_u = 6939.5 / 36525 = 0.1899931553730322$.
- Applying equation (2) yields the Greenwich Mean Sidereal Time at midnight: $H_0 = 1665686.527373333$ sidereal seconds.
- Applying equation (4) and multiplying by $t=8*3600$ seconds yields another 28878.85178960284 sidereal seconds to the above. What about $\Delta t$, aka dUT1? I've ignored it (i.e., treated it as if it was zero.) See footnote 2.
- 1665686.527373333 + 28878.85178960284 = 52965.37916293584 (modulo 86400). This is the mean sidereal time, in seconds. Converting to hours, minutes and seconds results in 14:42:45.4. The 0.4 seconds is bogus because I ignored dUT1.
Footnotes:
There's a problem with using Julian dates to express time when high precision is of concern. The problem is that adding 2.45 million and change to the time since noon on 1 Jan 2000 results in potential truncation problems. The next representable Julian date after midnight 1 Jan 2000 is 40 microseconds after midnight. I myself try to stay a couple of orders of magnitude away from the quantization level, which means problems for dates at the millisecond level.
Almost every scientific package that supports Julian dates either uses a pair of double precision numbers or uses some sort of modified Julian date, such as time since 12 Noon 16 Nov 1858 (omits the leading 2.4 million in Julian date), time since 12 noon 23 May 1968 (omits the leading 2.44 million), or time since J2000.
The term $\Delta t$ in equations (2) and (4) is better known as dUT1. This is the difference between UT1 (time using the mean rotating Earth as a clock) and UTC (time that runs at the same rate as an atomic clock, but with leap seconds occasionally inserted to keep UT1 and UTC in sync). Those leap seconds keep $|\text{UT1}-\text{UTC}| < 0.9\,\text{seconds}$. Simply ignore the $\Delta t$ term if that's the case -- but remember that fractions of a second are meaningless.
