First, calculate the distance to the Earth's shadow boundary at the zenith. The right triangle ABC with the right angle, abc, is formed from side A = segment from earth's center to surface, length = r; B = segment along earth's shadow boundary, length = r sin(theta); and segment C = segment ( hypotenuse) from earth's center to the zenith location of shadow. length = r + h where h = zenith altitude of satellite above surface of earth. Then you get:
2 2 2 2
r + r sin (theta) = ( r + h )
with the solution:
2 2 2 1/2
h = -r + 1/2( 4 r + 2 r sin (theta) )
from quadratic equation root.
So, at 90 minutes before sunrise, theta = 1.5h/24h x 360 = 22.5 degrees, and
for r = 6378 kilometers, h = -6378 + 6607 = 229 kilometers. So, if the
satellites orbit passed directly through your zenith position, if it is at a
distance of 229 kilometers above the earth, it would just be illuminated by
the sun 90 minutes before sunrise.
But, if it is located to the east of this zenith position, simple geometry shows that its elevation above the earth would have to be less than this so that it is still at the earth shadow boundary. At an elevation angle of 20 degrees, the geometry is a bit more complicated. As an approximation, I will guess that the altitude = zenith distance x sin( elevation angle) so you get something like 229 x sin(20) = 80 kilometers. But I know it is a bit more complicated than this. All you have to do is draw the figure and you will see it is a rather interesting and subtle problem.