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Is there a physics major on here willing to help me with a few things?
- Thread starter tleilaxu
- Start date
InvaderSquoosh
First Post
No problem, it's great to exercise old knowledge once in awhile. 
Fenlock
First Post
Regarding mars with an atmosphere, you might want to read kim stanley robinsons mars trilogy.
it is about terraforming mars and the scientific and politic implications - dont know how good his research was, but i found it an enjoyable read...
the last volume (green mars) has some nice description on living on mars with a breathable atmosphere (at lower altitutedes).
/fenlock
it is about terraforming mars and the scientific and politic implications - dont know how good his research was, but i found it an enjoyable read...
the last volume (green mars) has some nice description on living on mars with a breathable atmosphere (at lower altitutedes).
/fenlock
WayneLigon
Adventurer
tleilaxu said:
For unusual worlds. look to Larry Niven's work.
Mt. Lookitthat: One massive plateau about the size of California that rises high enough in the dense atmosphere that the air is breatheable there.
Jinx: heavy gravity world pulled into an egg shape by it's primary; the ends stick up out of the (spherical) atmosphere while the middle band is too dense to breathe.
The Smoke Ring: You don't even need a planet! An atmospheric band around a neutron star; the people live in zero-gravity.
Check out his Known Space stories for more.
WayneLigon
Adventurer
Some other things to think about: what would we see in a multi-moon scenario, like we see in so many fantasy novels? What if the moons are smaller than Luna? Larger? Would having a ring instead of a moon affect tides?
The info about the tilt of the moon was fascinating; I had no idea that would occur. What other effects can we get by varying the tilt? Longer full moons?
The info about the tilt of the moon was fascinating; I had no idea that would occur. What other effects can we get by varying the tilt? Longer full moons?
I may be wrong here but I kind of doubt Mars could hold an atmosphere very long. just look at the motion of air molecules and assume a Maxwell distribution of speeds. It seems to me that the atmostsphere would slowly go away (alot of the molecules would have escape velocity - just look at the tail of the distribution). What is wrong with my logic here?
oh for orbits it is kind of simple just use Kepler's law
Period^2 = semi major axis^3
The units work if you use period in earth years and semi-major axis in AU.
A Ring is formed because the tides are so strong they rip the mooon apart This is known as the Roche limit.
oh for orbits it is kind of simple just use Kepler's law
Period^2 = semi major axis^3
The units work if you use period in earth years and semi-major axis in AU.
A Ring is formed because the tides are so strong they rip the mooon apart This is known as the Roche limit.
Destil
Explorer
I find all of this facinating, being quite fond of moons for my game worlds and physics. nothing really to add, though.
Do you happen to know of anyplace online one could see such pictures? The visual I'm picturing in my mind is really quite interesting, though I'd like to accualy see it, as I may very well want to incorprate that into my next game moon...
The Sigil
Mr. 3000 (Words per post)
FOR A RECAP, GO TO THE BOTTOM OF THIS POST!
First of all, what you posit is a three-body problem, which means that for all intents and purposes, it is well-nigh unto empirically unsolveable. :-b It's why planets orbiting a dual-sun simply don't work well... the planetary orbit is somewhat unstable. The best we can do is "approximate" it but we know the approximation can't/won't be "spot on."
We can do a fairly close approximation by, as has been suggested, by assuming that the "sun" is fixed and the earth and "marsmoon" orbit around it, swinging about the center of mass.
Also, before I start, tide effects are due to the rotation of the earth and thereby occur every half-day. High tide is when the point on earth you are standing on is "in the line" that extends through both the center of the earth and the center of the moon. Low tide is when you are farthest away from that point (6 hours later, when you've moved from the "top" to the "side" of the earth).
From nasa.gov, we can note that Mars' mass is about 1/10th that of earth's, while its radius is about 1/2 that of earth's (this is not too far-fetched; the volume-linear ratio tells us that if mars and earth were equal density, we should see mars weighing 1/8th that of earth if it has 1/2 the radius). So depending on what you're referring to, "1/10th earth's size (mass)" and "1/2 earth's size (linear dimension)" are both correct.
The equatorial radius is 3397 km, so the diameter is (roughly) 6800 km.
The angular diameter of the moon is 30 minutes of arc (1/2 of a degree).
A (small) visual angle (less than 6 degrees) is described by the formula:
tan (angle) = size/distance
Since we are looking for distance, we swap things around and get:
distance = size / tan (angle) or in this case 6800 km/tan (1/2 degree) or 6800 km/0.00873 or (to two significant digits) 780,000 km.
Hence, the "marsmoon" must be 780,000 km away to appear the same "size" as our RL earth moon.
Due to Kepler's third law of planetary motion, we know that Distance^3 is proportional to Orbital Period^2; IOW, orbital period is proportional to distance^(3/2).
Let us normalize this equation to the moon; the moon averages 384,400 km from the earth and has a period of (roughly) 29 days.
Thus, 29 days*{(780,000 km)/(384,400 km)}^(3/2) = Orbital Period of Marsmoon
In other words, the Marsmoon would have a period of roughly 84 days... or slightly less than three months. That means it would be about 21 days between "phases" (full to half, half to new, new to half, half to full).
Varying the tilt does little, BTW, it just means that you'll have a slightly more eclectic orbit across the sky (instead of straight east-west you move the orbit towards north-south the grater the tilt).
Tides: This is a little tricky; gravity is proportional to mass over orbital distance^2. Therefore, we should see tides changed by a factor of
6.42x10^23 kg (mars):7.4x10^22 kg (moon)
divided by
{(780,000 km)/(384,400 km)}^2
or
8.68/(2.03)^2
or
2.1 times as much affect. Thus, tides would be about twice as high as they are now - or about 3.2 meters/10 feet on average (see below).
Well, I'll give you how big it would appear in the sky. From above:
tan (angle) = size/distance
or angle = tan^(-1) (size/distance)
tan^(-1) (6800 km/384,400 km) = tan^(-1) (0.01769) or 1.01 degrees.
Thus, if it were at a distance equal to that of our moon (and hence orbited at the same rate), the moon would appear to be twice as big across in the sky (RL moon has visual angle of 1/2 degree, this one hasangle of 1 degree) - it would appear to have a diameter twice as big, meaning the "area" it occupied in the sky would be 4 times as big.
Checking the effect on the tides:
The moon's "pull" on the tides can best be compared to our moon's simply by looking that the gravitational force it exerts; since gravity is proportional to Mass, we need only look at the factor of difference in masses of the moon and mars in this case (as they are at identical distances). 6.42x10^23 kg for mars, 7.4x10^22 kg for moon. The factor is 8.68... that means the tides would be almost ten times as powerful (yipes)!
Why is this "yipes"? Some sample tide heights. I assume this is from "low point to high point" rather than "average point to high point" - in which case the changes would be TWICE as big (up from the midpoint and then down from the midpoint in 6 hours). Average tide height on earth is about 1.6 meters (or about 4.5 feet). That means with a "Mars moon" this close, you're looking at average daily tides of 14 meters (46 feet)! That's about four stories! That means that your "coastal city" sits on a cliff and has to load and unload quickly at high tide before the water drops precipitously down the cliff! It also implies (to me) that you will have no beaches as we know them... only cliffs... and those cliffs will likeley erode rapidly.
The water is rising/falling at a rate of roughly 8 feet per hour (6 hours down, 6 hours up) or about 1 foot every 8 minutes. That's a HUGE effect. If I'm wrong and the "high tide" is off the "average", that means high tide is +46 feet and low tide is -46 feet; a difference of 90 feet every 6 hours. That's 1 foot every four minutes! That's a tide you can REALLY notice.
TO RECAP:
Mars at a distance that makes it look the same size as the moon to an earth-based observer:
Distance is 780,000 km (about twice that of moon)
"Month" is 84 days long (21 days from new to half or from half to full)
Tides are twice as high as they are on RL earth
Mars at moon distance:
Appears to be twice as big in diameter in the sky (area is 4x as big).
Tides are about TEN times as high(!) as on RL earth.
Hope that helps.
--The Sigil
First of all, what you posit is a three-body problem, which means that for all intents and purposes, it is well-nigh unto empirically unsolveable. :-b It's why planets orbiting a dual-sun simply don't work well... the planetary orbit is somewhat unstable. The best we can do is "approximate" it but we know the approximation can't/won't be "spot on."
We can do a fairly close approximation by, as has been suggested, by assuming that the "sun" is fixed and the earth and "marsmoon" orbit around it, swinging about the center of mass.
Also, before I start, tide effects are due to the rotation of the earth and thereby occur every half-day. High tide is when the point on earth you are standing on is "in the line" that extends through both the center of the earth and the center of the moon. Low tide is when you are farthest away from that point (6 hours later, when you've moved from the "top" to the "side" of the earth).
From nasa.gov, we can note that Mars' mass is about 1/10th that of earth's, while its radius is about 1/2 that of earth's (this is not too far-fetched; the volume-linear ratio tells us that if mars and earth were equal density, we should see mars weighing 1/8th that of earth if it has 1/2 the radius). So depending on what you're referring to, "1/10th earth's size (mass)" and "1/2 earth's size (linear dimension)" are both correct.
The equatorial radius is 3397 km, so the diameter is (roughly) 6800 km.
The angular diameter of the moon is 30 minutes of arc (1/2 of a degree).
A (small) visual angle (less than 6 degrees) is described by the formula:
tan (angle) = size/distance
Since we are looking for distance, we swap things around and get:
distance = size / tan (angle) or in this case 6800 km/tan (1/2 degree) or 6800 km/0.00873 or (to two significant digits) 780,000 km.
Hence, the "marsmoon" must be 780,000 km away to appear the same "size" as our RL earth moon.
Due to Kepler's third law of planetary motion, we know that Distance^3 is proportional to Orbital Period^2; IOW, orbital period is proportional to distance^(3/2).
Let us normalize this equation to the moon; the moon averages 384,400 km from the earth and has a period of (roughly) 29 days.
Thus, 29 days*{(780,000 km)/(384,400 km)}^(3/2) = Orbital Period of Marsmoon
In other words, the Marsmoon would have a period of roughly 84 days... or slightly less than three months. That means it would be about 21 days between "phases" (full to half, half to new, new to half, half to full).
Varying the tilt does little, BTW, it just means that you'll have a slightly more eclectic orbit across the sky (instead of straight east-west you move the orbit towards north-south the grater the tilt).
Tides: This is a little tricky; gravity is proportional to mass over orbital distance^2. Therefore, we should see tides changed by a factor of
6.42x10^23 kg (mars):7.4x10^22 kg (moon)
divided by
{(780,000 km)/(384,400 km)}^2
or
8.68/(2.03)^2
or
2.1 times as much affect. Thus, tides would be about twice as high as they are now - or about 3.2 meters/10 feet on average (see below).
Well, I'll give you how big it would appear in the sky. From above:
tan (angle) = size/distance
or angle = tan^(-1) (size/distance)
tan^(-1) (6800 km/384,400 km) = tan^(-1) (0.01769) or 1.01 degrees.
Thus, if it were at a distance equal to that of our moon (and hence orbited at the same rate), the moon would appear to be twice as big across in the sky (RL moon has visual angle of 1/2 degree, this one hasangle of 1 degree) - it would appear to have a diameter twice as big, meaning the "area" it occupied in the sky would be 4 times as big.
Checking the effect on the tides:
The moon's "pull" on the tides can best be compared to our moon's simply by looking that the gravitational force it exerts; since gravity is proportional to Mass, we need only look at the factor of difference in masses of the moon and mars in this case (as they are at identical distances). 6.42x10^23 kg for mars, 7.4x10^22 kg for moon. The factor is 8.68... that means the tides would be almost ten times as powerful (yipes)!
Why is this "yipes"? Some sample tide heights. I assume this is from "low point to high point" rather than "average point to high point" - in which case the changes would be TWICE as big (up from the midpoint and then down from the midpoint in 6 hours). Average tide height on earth is about 1.6 meters (or about 4.5 feet). That means with a "Mars moon" this close, you're looking at average daily tides of 14 meters (46 feet)! That's about four stories! That means that your "coastal city" sits on a cliff and has to load and unload quickly at high tide before the water drops precipitously down the cliff! It also implies (to me) that you will have no beaches as we know them... only cliffs... and those cliffs will likeley erode rapidly.
The water is rising/falling at a rate of roughly 8 feet per hour (6 hours down, 6 hours up) or about 1 foot every 8 minutes. That's a HUGE effect. If I'm wrong and the "high tide" is off the "average", that means high tide is +46 feet and low tide is -46 feet; a difference of 90 feet every 6 hours. That's 1 foot every four minutes! That's a tide you can REALLY notice.
Not meaningfully.
TO RECAP:
Mars at a distance that makes it look the same size as the moon to an earth-based observer:
Distance is 780,000 km (about twice that of moon)
"Month" is 84 days long (21 days from new to half or from half to full)
Tides are twice as high as they are on RL earth
Mars at moon distance:
Appears to be twice as big in diameter in the sky (area is 4x as big).
Tides are about TEN times as high(!) as on RL earth.
Hope that helps.
--The Sigil
Last edited:
The Sigil
Mr. 3000 (Words per post)
As I review my answer, I think I missed on the timing of the tides... high tide is when the side of the earth you're on is closest to the moon (the moon pulls the water up) and low tide comes 12 hours later (when the moon pulls the water "down", being on the other side of the earth).
But the height of the tides is still right.
Also, with the different answer... that's why I "showed my work" so that if I goofed, I can have somebody smack me upside the head and point it out. :b
--The Sigil
But the height of the tides is still right.
Also, with the different answer... that's why I "showed my work" so that if I goofed, I can have somebody smack me upside the head and point it out. :b
--The Sigil
