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MrFun
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of Iowa
Nov 2000
time: 23:18
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I'm working on background for a fantasy novel I may write. Right now, I'm creating the star of the story's solar system called Lusteron, and the planet on which my characters will exist, which is called Hearthena.
Here are most of the specifications for Lusteron:
Lusteron (1) = Sol
Star Type: F6
Descriptive Type: yellow dwarf
Mass (1): 1.25
Luminosity (1): 2.15
Characteristics: higher level of flare activity, more ejections of charged particles, and many more spots
Here are most of the specifications for Hearthena:
Hearthena (1) = Earth
Planet Type: terrestrial
Distance from Lusteron (1): 1.55 AU (average distance in elliptical orbit)
Axis Tilt:
Diameter (1): 1.25 or 15,945.375 kilometers
Density: (5.52): 6.7
Mass (1): 1.15
Surface Gravity (1): 1.18 (150 pounds on Earth = 170 pounds on Hearthena)
Length of one completed orbit (1): 1.43 years or 16 months
Length of one rotation (1): 27 hours
My specifications are based on very rough estimates. As you can see, I have not yet decided on what degree of an axis tilt Hearthena will have. And from the specs above, you can see that Lusteron is a yellow star similar to our Sol, but it's somewhat larger and a little bit more than double the illumination. Hearthena is farther out from Lusteron than Earth is to Sol and this planet has a greater diameter than Earth.
For those who are knowledegable in physics and/or astrophysics, and if you're interested in helping out, I would greatly appreciate any insight you can provide after looking over the specifications.
I have a question. How many days will Hearthena have, given it takes 27 hours to rotate once, and 16 Earth months to orbit Lusteron once?
I can't think of any other questions that immediately come to mind right now, and right now, I need to leave to go somewhere. So when I come back, I might have one or two more questions.
And if you want to provide any other information on what kind of sun Lusteron would be, and what kind of planet Hearthena would be, please do so. And is there any other crucial aspects of this particular created star and planet system that I should keep in mind and never overlook or forget?
Thanks for anyone who is interested.
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reds4ever
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of the Spion Kop
Mar 2001
time: 05:18
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I dunno, I think it's good thing to have a "solid base" for a story, it doesn't take much effort and is worth it in the end. (IMHO).
MrFun doesn't have to use all the info on the star/planet in the actual story, it just helps if he knows in his mind what type of world he's writing about.
Last edited by reds4ever on 12-07-2005 at 05:32
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Omni Rex Draconis
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Sorry to trip you up so early in, but the numbers you presented don't quite add up.
If radius is 1.25 Earth, then volume is 1.95 Earth.
If density is 1.21 Earth, then mass is 1.95 x 1.21 = 2.36
The increased radius reduces gravity to 0.64 (inverse square),
Thus, gravity should by 2.36 x 0.64 = 1.51 Earth.
That is a little more than I would be comfortable with, both living on and writing about.
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Omni Rex Draconis
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Whoops, the orbital period is a bit off as well.
The formula for calculating orbits (assuming a star much bigger than the planet) is:
M x P^2 = D^3
Where M = the mass of the star (in Sol masses),
P = the period (in earth years), and
D = the distance (in AU).
According to your numbers then, the planet should only be 1.37 AU distant from the star, or have a year equal to 1.73 Earth years.
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Omni Rex Draconis
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Volume is 4/3 x pi x radius^3. You can leave out the first bit and just cube the radius to get "Earth volumes".
So 1.25^3 = 1.953125.
Gravity is reduced by the square of the distance. Thus, you can figure that gravity will increase in proportion to the radius.
For example, double the size is eight times the volume/mass, but one quarter grav due to distance. One quarter of eight is two, same as your size increase.
Density is just a straight multiplier. Twice as dense, twice the gravity.
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Omni Rex Draconis
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I noticed you are sticking to your 1.15 earth density. Got a bigger iron core in mind?
As for the solar distance, it works out pretty much earthlike. At 1.37 AU, your planet is receiving about 53% (inverse square law again) of what it would at Earth orbit. Your star has twice the luminosity, though, so it is all good (assuming that is what you wanted).
Remember that your star is hotter that Sol. The colour will be less red and more blue, and it will radiate more ultraviolet and other high energy particles. It would be a good idea to set the big iron core spinning for a larger magnetosphere; helps block out the ionized particles.
You can also change the numbers quite significantly once you get to the atmosphere stage. It can trap heat or reflect light, it can block UV or let it all through.
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Omni Rex Draconis
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You were worried about the physics, MrFun? These guys won't even let you get by with the names!
Or is this just a case of friends/ex-lovers having a bit o' fun with you?
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11-07-2005 18:43
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