Automated Calculations...

[Kristopher]

I think something like this has been posted here as a link before, but I'm looking for a website or program that automates the calculations for orbits, surface gravity, apparent brightness and size of stars, planets, and moons as seen from other planets and moons, etc.

 

Something that takes care of the kind of thing listed as this link.

 

 

For example, let's say I want to calculate the mass and density of a moon orbiting a gas giant in another system, which is 3/4 the diameter of Earth and has a surface gravity of 1.15 .

[Kristopher]

Re: Automated Calculations...

 

60 views and no suggestions?

 

That's depressing.

[Outsider]

Re: Automated Calculations...

 

Just doing the math is easier than it would seem at first.

 

At a given density, surface gravity varies directly with diameter. So a planet with half Earth's diameter will have half Earth's surface gravity. (Yes, you have only 1/8th the mass, but you're twice as close to it. Inverse Square Law means that that 1/8th mass pulls on you 4 times as hard since you're so much closer to it. Half diameter = half (one eighth * four) surface gravity)

 

In your example, this means that, were it as dense as Earth, your moon would have a surface gravity of .75 Earth normal. Since it has 1.15, that means it must be 1.15/.75 (1.5333) times as dense.

 

Getting the mass is then just a simple volume*density thing : 1.5333 * .75^3 = .647 Earth masses.

[Kristopher]

Re: Automated Calculations...

 

Just doing the math is easier than it would seem at first.

 

At a given density, surface gravity varies directly with diameter. So a planet with half Earth's diameter will have half Earth's surface gravity. (Yes, you have only 1/8th the mass, but you're twice as close to it. Inverse Square Law means that that 1/8th mass pulls on you 4 times as hard since you're so much closer to it. Half diameter = half (one eighth * four) surface gravity)

 

In your example, this means that, were it as dense as Earth, your moon would have a surface gravity of .75 Earth normal. Since it has 1.15, that means it must be 1.15/.75 (1.5333) times as dense.

 

Getting the mass is then just a simple volume*density thing : 1.5333 * .75^3 = .647 Earth masses.

 

So the moon in the example would about 8.46 g/cm^3, correct?

[Outsider]

Re: Automated Calculations...

 

So the moon in the example would about 8.46 g/cm^3' date=' correct?[/quote']

 

 

1.5333 times earth, which is 5.52, so yeah. It would have to be made of some pretty dense stuff. Iron is only 7.8. I'd expect this to be a pretty rich find for miners if 1.15 Gs is something they can deal with economically.

[Cancer]

Re: Automated Calculations...

 

What do you want, in terms of calculated quantities? Mass, radius, density, surface gravity are all tightly related to each other.

 

Orbital size and period are related also, but the mass of the thing being orbited is also needed.

 

Apparent brightness can be done, but now you need to specify your viewing point, and some things about the planet/moon being viewed that are unrelated to any of the preceding quantities.

[Kristopher]

Re: Automated Calculations...

 

What do you want, in terms of calculated quantities? Mass, radius, density, surface gravity are all tightly related to each other.

 

Orbital size and period are related also, but the mass of the thing being orbited is also needed.

 

Apparent brightness can be done, but now you need to specify your viewing point, and some things about the planet/moon being viewed that are unrelated to any of the preceding quantities.

 

Those sorts of things. Also, apparent size of one body in the sky of another, such as the size of a moon in the sky of the planet it orbits. Which I know is based on the distance and size of the moon, I just never remember how to calculate it.

[Outsider]

Re: Automated Calculations...

 

Apparent size is a direct linear relationship, atmospheric magnification notwithstanding.

 

If Luna were twice as close, it'd appear to have twice the diameter. If it had twice the diameter, it would appear to have twice the diameter too.

 

comparatively, its apparent size (in Lunar equivalents) would be :

Diameter in Lunar diameters/Distance from surface in Earth to Luna distances

 

 

The moon in your previous example was .75 Earth diameters. The Earth is about 3.7 times the diameter of the moon. So your moon would be about 2.8 Lunar diameters.

 

You never mentioned anything about its orbit other than that it orbited a gas giant. If (to extend the example) it orbited the gas giant at the same distance that Titan orbits Saturn (and the gas giant were the same mass as Saturn) then its distance from its primary would be roughtly 3.2 Earth-Luna distances. Meaning the apparent size from a floating pleasure barge in the upper atmosphere of the gas giant would be 2.8/3.2 times (about 90%) the apparent size of Luna from the surface of the earth.

 

 

One could easily set up a spreadsheet table to show apparent sizes.

 

 

Now, if you want orbital periods, and if your moon is tide locked and all that, THEN you're beyond my back of the hand comparison ability

[Cancer]

Re: Automated Calculations...

 

In case you need it, the Sun and Moon, on average, subtend 0.53 degrees (their angular diameter on the sky is a bit more than half a degree). The orbits of both Earth and Moon are elliptical so those vary by a few percent over the course of a year/month. The angular resolution of the naked eye is 1 to 2 arc minutes, that is, 1/60 to 1/30 of a degree.

 

Wikipedia lists angular diameters of other Solar System ojects.

 

FWIW, without an adaptive optics (AO) system of at least 1980's tech, with visual light and an Earthlike atmosphere, you won't be able to resolve anything with an imaging detector smaller than about half an arc second (and that's an extreme; 1 arc second is much more likely to be your limit). Without an atmosphere you get the diffraction limit of your instrument. With atmosphere and an AO system you can reach the diffraction limit.

[Kristopher]

Re: Automated Calculations...

 

Apparent size is a direct linear relationship, atmospheric magnification notwithstanding.

 

If Luna were twice as close, it'd appear to have twice the diameter. If it had twice the diameter, it would appear to have twice the diameter too.

 

comparatively, its apparent size (in Lunar equivalents) would be :

Diameter in Lunar diameters/Distance from surface in Earth to Luna distances

 

 

The moon in your previous example was .75 Earth diameters. The Earth is about 3.7 times the diameter of the moon. So your moon would be about 2.8 Lunar diameters.

 

You never mentioned anything about its orbit other than that it orbited a gas giant. If (to extend the example) it orbited the gas giant at the same distance that Titan orbits Saturn (and the gas giant were the same mass as Saturn) then its distance from its primary would be roughtly 3.2 Earth-Luna distances. Meaning the apparent size from a floating pleasure barge in the upper atmosphere of the gas giant would be 2.8/3.2 times (about 90%) the apparent size of Luna from the surface of the earth.

 

 

One could easily set up a spreadsheet table to show apparent sizes.

 

 

Now, if you want orbital periods, and if your moon is tide locked and all that, THEN you're beyond my back of the hand comparison ability

 

If we stick with the example situation, it would be orbiting a planet similar to Jupiter, at a distance that put it outside the worst of the radiation belts and allowed it to have its own magnetic field (that wasn't hopelessly entangled with the planet's).

 

How far out would it have to be to avoid being tide locked?