Here is an approximate geometrical solution.
Suppose your position at height
$h$ above the Earth's surface is $O$, the
center of the Earth is $C$, a point of the Earth's surface that you
see at the horizon is $B$, and the radius of the
Earth where you are is $r$. Then the sine of $COB$ is $r / (r + h)$, so if $A$ is a point on the "normal" horizon (perpendicular to the zenith) directly
above $B$, this will also be the cosine of the
complementary angle $BOA$. Therefore, the horizon has a dip angle
of $d_1=\arccos (r/(r + h))$ below the "normal" horizon.
Projecting these points to the celestial sphere, if the center of the Sun sets below or rises above
the dipped horizon at $B$, the north celestial pole is at $P$,
and the direction towards north on the normal horizon is at $N$, then
$BAN$ and $ANP$ are right angles, $AB$ equals the dip angle ($d$ say),
$NP$ equals the latitude $L > 0$, and $PB$ equals 90° minus the Sun's declination ($90^\circ - \delta$, say.)
Solving using the spherical laws of sines and cosines gives
$$
\cos BPN = \frac{\sin d + \sin L \sin \delta}{\cos L \cos \delta}.
$$
To find the dip angle
$d$ for a sunrise or sunset at the Earth's surface, you can use that
the Sun's angular radius is approximately 16 arcminutes, depending on where the Earth is in its orbit, and
also apply a refractive correction of about 34 arcminutes
([1], §1.1, p. 2.) This gives approximately $d=d_0=50^\prime$. To find the dip angle for a sunrise or sunset high above the surface,
you could just take $d=d_0+d_1$ (but see below.) The difference in times
then will be about the time it takes the Earth to rotate by an angle of
$$
\alpha=
\arccos\frac{\sin d_0 + \sin L \sin \delta}{\cos L \cos \delta}
-
\arccos\frac{\sin (d_0 + d_1) + \sin L \sin \delta}{\cos L \cos \delta}.
$$
Plugging in values of $h=50 \ \rm km$, $r=6367.49 \ \rm km$ at $45^\circ N$,
$L=45^\circ$, $d_0=50^\prime$, and $\delta=\pm 23^\circ 26^\prime$ for the
solstices or $\delta=0$ for the equinox gives estimates of
$$
\begin{array}{cl}
\alpha=13^\circ 15^\prime & \hbox{for the summer solstice,}\\
\alpha=10^\circ 10^\prime & \hbox{for the equinox, and}\\
\alpha=11^\circ 36^\prime & \hbox{for the winter solstice.}
\end{array}
$$
Converting these to times gives approximately
$$
\begin{array}{cl}
53 \ \hbox{minutes} & \hbox{for the summer solstice,}\\
41 \ \hbox{minutes} & \hbox{for the equinox, and}\\
46 \ \hbox{minutes} & \hbox{for the winter solstice.}
\end{array}
$$
A major source of error in these estimates is the refractive correction.
According to [1], §1.1, p. 2, it's conventional to
use a correction of $34^\prime$, and this is what's often
used in published tables of sunrises and sunsets, but it's not always
accurate and can lead to errors of several minutes. The correction will surely also be different 50 km above the Earth's surface. [1] gives some more accurate refractive models, which might help to compute a better estimate. In its article on the sunrise equation, Wikipedia suggests that you take
$$
d_1=2.076 \sqrt{h},
\qquad \hbox{if $d_1$ is in arcminutes and $h$ is in meters.}
$$
Using this estimate would give
$$
\begin{array}{cl}
58 \ \hbox{minutes} & \hbox{for the summer solstice,}\\
44 \ \hbox{minutes} & \hbox{for the equinox, and}\\
50 \ \hbox{minutes} & \hbox{for the winter solstice,}
\end{array}
$$
but I doubt that this formula is intended to be accurate at a height of
50 km.
Addendum: Here are some graphs showing how the time varies
with the latitude and the Sun's declination. They are computed
using the formulae above and $d_1=\arccos (r/(r+h))$.
In the first graph, the declination where the time is smallest,
which, interestingly, is not zero (unless you are at the equator), is marked with an "x".
References [2], [3], [4], [5] and
[6] discuss the similar problem of finding when the shortest
twilight occurs.
References:
[1]: "Evaluating the Effectiveness of Current Atmospheric Refraction Models in Predicting Sunrise and Sunset Times", Teresa Wilson, Ph. D. thesis, Michigan Technological University, 2018; DOI 10.37099/mtu.dc.etdr/697.
[2]: "The shortest twilight", B. G. Marsden and R. F. Griffith, The Observatory, Feb. 2000, vol. 120, pp. 62-66.
[3], [4], [5], [6]: "The Twilight", Orrin E. Harmon, Popular Astronomy, Sep. 1896, vol. 4, pp. 148-154; Oct. 1896, vol. 4, pp. 211-214; Nov. 1896, vol. 4, pp. 252-258; and Sep. 1897, vol. 5, pp. 257-263.
