### Post by foxroe on May 13, 2017 7:46:30 GMT -6

In trying to wrap my head around the crazy orbital dynamics of the solar system of AS&SH (yeah, yeah, I know... it's just a game), I jotted down some notes and made some calculations that may be useful for some. Bear with me... it's been a while since my college astronomy and math courses. Hopefully someone who is really into this stuff can double check my calculations and assumptions. I'm just relying on basic Kepler laws here...

Hyperborean spin

================

Since Hyperborea is flat, in order for the ecliptic path of Helios to work how it does, the realm must be roughly parallel with the plane of the solar system. In order to account for total darkness/sunlight during 1/13 of the Hyperborean cycle, Hyperborea must be tilted on its central axis (the axis would pass through Vhuurmithadon, perpendicular to the plane of Hyperborea). This axis would "wobble" in synchronization with the Hyperborean cycle such that the face of the realm leans in toward Helios during the summer, and away from Helios in the winter. Since Helios never rises more than 25 degrees above the horizon, the axial tilt must also be about 25 degrees.

However, Hyperborea must not revolve around this axis. If it did, then there would be 6 years each of total darkness and sunlight, with Helios just appearing at different heights in the sky during the lighted years. In order for Helios to dip below the "horizon" for part of a daily revolution, Hyperborea must be spinning on another axis that is not quite parallel with the axis which dictates its tilt relative to the sun. Don't try to think about it too hard, you might hurt something (I'm pretty sure I did).

How far is Hyperborea from Helios?

==================================

Some math...

Orbital period (T) = 13 years = 410240376 sec

T = 2*pi*sqrt(a^3/u)

a = semimajor axis length for an elliptical orbit (or the radius for a circular orbit <- using this since the math is easier)

u = gravitational constant based on the combined mass of Helios and Hyperborea

u = G(m1+m2) = 6.67384 × 10^−11 [units=m^3/(kg*s^2)]

u = 6.67384x10^-11 *(1.99x10^30+insignificant mass) ---> I'm making the assumption that the Sun won't lose much mass...

u = 13.28x10^19

therefore, a = 827,045,361 km from sun

This puts Hyperborea's orbit just beyond Jupiter and about 600 million km from Saturn.

Where's Jupiter?

================

Ganymede (now one of Helios' planets) is Jupiter's largest moon. It is likely that during the evolution of the Sun into a red giant (Helios), Jupiter was consumed/destroyed, but Ganymede survived to become the innermost planet of Helios. Assuming Ganymede follows roughly the same orbit as Jupiter (778,400,000 km from the Sun), it is a mere 49 million km from Ganymede to Hyperborea, which is slightly further than the present distance between the Earth and Venus. Since Ganymede is about half as wide as Venus, it's about half as big/bright as Venus (when viewed from the Earth) when viewed from Hyperborea.

How "big" is Saturn?

====================

Based on the above calculations, Saturn would appear about twice as big/bright from Hyperborea as it does from Earth today.

How "big" is Helios?

====================

Scientists predict that the sun will expand to about Earth's orbit in its red giant phase - that's about 300,000,000 km wide, or well over 200 times its current size.

More math...

d = 2*arctan(2*r/2*D) = angular diameter (apparent width)

r = radius of object = 150,000,000 km

D = distance to the object = 827,045,361 km

d = 20.6

The sun's angular diameter as viewed from Hyperborea would be about 20 degrees; if you make both fists, extend your arms out fully, and put your fists together, this would be the apparent width of Helios as viewed from Hyperborea!

Hyperborean spin

================

Since Hyperborea is flat, in order for the ecliptic path of Helios to work how it does, the realm must be roughly parallel with the plane of the solar system. In order to account for total darkness/sunlight during 1/13 of the Hyperborean cycle, Hyperborea must be tilted on its central axis (the axis would pass through Vhuurmithadon, perpendicular to the plane of Hyperborea). This axis would "wobble" in synchronization with the Hyperborean cycle such that the face of the realm leans in toward Helios during the summer, and away from Helios in the winter. Since Helios never rises more than 25 degrees above the horizon, the axial tilt must also be about 25 degrees.

However, Hyperborea must not revolve around this axis. If it did, then there would be 6 years each of total darkness and sunlight, with Helios just appearing at different heights in the sky during the lighted years. In order for Helios to dip below the "horizon" for part of a daily revolution, Hyperborea must be spinning on another axis that is not quite parallel with the axis which dictates its tilt relative to the sun. Don't try to think about it too hard, you might hurt something (I'm pretty sure I did).

How far is Hyperborea from Helios?

==================================

Some math...

Orbital period (T) = 13 years = 410240376 sec

T = 2*pi*sqrt(a^3/u)

a = semimajor axis length for an elliptical orbit (or the radius for a circular orbit <- using this since the math is easier)

u = gravitational constant based on the combined mass of Helios and Hyperborea

u = G(m1+m2) = 6.67384 × 10^−11 [units=m^3/(kg*s^2)]

u = 6.67384x10^-11 *(1.99x10^30+insignificant mass) ---> I'm making the assumption that the Sun won't lose much mass...

u = 13.28x10^19

therefore, a = 827,045,361 km from sun

This puts Hyperborea's orbit just beyond Jupiter and about 600 million km from Saturn.

Where's Jupiter?

================

Ganymede (now one of Helios' planets) is Jupiter's largest moon. It is likely that during the evolution of the Sun into a red giant (Helios), Jupiter was consumed/destroyed, but Ganymede survived to become the innermost planet of Helios. Assuming Ganymede follows roughly the same orbit as Jupiter (778,400,000 km from the Sun), it is a mere 49 million km from Ganymede to Hyperborea, which is slightly further than the present distance between the Earth and Venus. Since Ganymede is about half as wide as Venus, it's about half as big/bright as Venus (when viewed from the Earth) when viewed from Hyperborea.

How "big" is Saturn?

====================

Based on the above calculations, Saturn would appear about twice as big/bright from Hyperborea as it does from Earth today.

How "big" is Helios?

====================

Scientists predict that the sun will expand to about Earth's orbit in its red giant phase - that's about 300,000,000 km wide, or well over 200 times its current size.

More math...

d = 2*arctan(2*r/2*D) = angular diameter (apparent width)

r = radius of object = 150,000,000 km

D = distance to the object = 827,045,361 km

d = 20.6

The sun's angular diameter as viewed from Hyperborea would be about 20 degrees; if you make both fists, extend your arms out fully, and put your fists together, this would be the apparent width of Helios as viewed from Hyperborea!