Is there a physics major on here willing to help me with a few things?

[Impeesa]

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?

The Maxwell distribution depends on temperature. The higher in altitude you go, the lower the temperature, and thus the lower the average particle energy. The lower the average particle energy is, the lower the average velocity (of course). And as the average particle velocity gets lower, your chances of finding a particle that still has escape velocity become exponentially lower.

I could be completely wrong though, that's just the first explanation that comes to mind.

--Impeesa--

 

[Dr. Harry]

Why yes, I am a rocket scientist ...

I haven't scanned to see what has been posted; please forgive me if someone else has answered this already. The radius of Mars is 3,394 km, or 1.95 times the radius of the Moon. Angular size decreases linearly, so your planet would have to be ~751,000 km away from the Earth.

The period of the arrangement can be found using Kepler's Third Law, about 6.15 x 10^6 s, or about 71 days.

Kepler's Third Law links the period and the separation between the object's so if you want the "Mars-moon" to have a 28 day cycle (The Moon's sidereal cycle is 27.3 days, the solar month 29.5 days), the then Mars-moon would be at a distance slightly greater (1.05x) the Moon's current distance, and would be about 1.86 times the size that Luna is in our sky.

What determines the period of the tides is not (directly) the orbit of the Moon around the Earth, but the rotation of the Earth on its axis. On Earth, we have two tides every ~25 hours - twice a day adjusting for the movement of the Moon in the sky. The distance Mars-moon would have two tides a day (one tidal bulge on each side of the Earth) adjusted such that for eavery 71 days, there would be 70 tidal cycles. Every day, high tides would be about 20 minutes later.

Simply having a larger mass wouldn't automatically make the tides greater, what is important is the difference in the other body's gravitational pull from one side of the Earth to the other. For the Moon, this is a difference of 6.4%, a tad larger than what it would be if Mars were there. Mars at the farther orbit would have tides about half as great (3.3% differential).

Actually, you could do whatever the heck you wanted with the tides. The simple model on Earth can produce tides up to about ~1 m. The much larger tides in many places are due to the sloshing of the seas in their seabeds. A Ocean floor of the right shape, for the "tide wave" traveling around the Earth can produce MUCH greater tides if it is near resonance. (The same way a crystal wine glass can be oscillated into shaking itself a part when the sound waves of just the right note hit. The greatest tide differentials on Earth are, I believe, about 17 m.

The phobos object is entirely negligible.

Hope this was interesting.

Harry Leckenby

 

[Dr. Harry]

Harry again

I am a Professor of Physics doing astronomy at Texas A&M - Kingsville,so I do have some background in this.

BS Astronomy/Astrophysics Villanova
MS Space Science Florida Inst. of Technology
Ph.D. Physics Michigan Tech. Univ.

 

[Dr. Harry]

The center of mass would be at 1.14 Earth Radii, so the monthly wobble of the stars as the "Earth" and "Mars-moon" orbitted probably wouldn't be weird enough to see, but any astronomers would notice it.

 

[Dr. Harry]

Shadows

Well, the mars object is smaller than the earth but it appears as big as the sun in the sky, creating eclipses. Since the earth is the same distance from pseudo-mars as pseudo-mars is from the earth )) that means the earth would be bigger in the martian sky than pseudo-mars is in the earth's sky, which would imply that the shadow doesn't taper out, right? Of course I'll just fudge it but more than half of this discussion is just to satisfiy my cosmic curiousity.....

Your logic is correct. Theangular size of a distant object can be given as:

angle(") =(d/D)*206,265

where a is the angle in seconds of arc, d is the diameter of the object (in this case, Earth), D is the baseline distance (I used the 751 Kkm), and 206,265 is the number of arcseconds in one radian. Even at the farther distance, the Earth would have an angular size of 58' of arc, or almost a full degree. Since the Sun's angular size at the Earth's orbit is about 30', eclipses are possible. The size of the Earth's umbra at this distance would have a radius of 3,450 km -- just a bit more than Mars' radius of 3,394 km. Total eclipses are possible, but in a world like ours with the Moon's orbit at an angle (if about 5 degrees) to the plane of Earth's orbit, and the two points where the orbit crosses this plane (the line of nodes) precessing, a total eclipse would be very rare -- but, hey, it's your world.

 

[Kwalish Kid]

I just popped in to make sure that everyone is clear that high tides do occur 12 hours apart. The "sides" nearest to and furthest from the moon get the high tides.

Of course, ocean floor effects do have a huge influence on tidal size. This screwed up Newton and many others.

 

[tleilaxu]

wow... even better answers

 

[Nifft]

What's the closest that two Earth-sized planets could orbit each other "safely"? How big would each appear in the other's sky? Finally, how would such a close orbit affect the tides (in terms of size)?

Thanks, -- Nifft

 

[Umbran]

That "minimum safe distance" is called the "Roche Limit" for the system. It isn't exactly easy to calculate for "real" moons, especially when the two objects are roughly the same size, mass, and composition.

To a first approximation, though - if the planet and moon are of similar composition (meaning similar density and material strength), then the moon's center should not be any closer than about 2.4 planetary radii (R) from the planet's center. Closer than that, and it risks being pulled apart by tidal forces, and you'll get rings rather than a moon.

Now, here's where it gets sticky. If the two objects are of the same size (radius R),and there's oly 2.4R between their centers, this means that there's only 0.4 R between their surfaces.

Let's say these are planets like Earth. Radius 3960 miles. That means the surfaces at their closest are 1584 miles apart. Only half the width of the North American continent. At that distance, the "moon" would be huge. If you were standing right under it, it would seem to be... something approaching 136 degrees of the 180 degrees of sky.

Now, it's reasonable to say that the rough approximation will fail at such an extreme. However, Robert L Forward used some more sophisticated math to show that such a planetary system can actually be stable on geologic timescales. He uses them in the Rocheworld books.

The planets are tidally locked, and cease to be spherical. They are more like eggs, and they orbit each other with their points facing each other, always.

 

[bolen]

We do a lab in our freshman non-major course where the student show that saturn cant have a moon where the rings are. we assume that you have 2 1 kg chunks of rock/ice and use Newton's law of Gravity

F= G (m M)/r^2

to find the force between them we then find the difference between the pull of saturn and the two rocks assume that one is one Km from the other.

You find that the differental force by the pull of saturn is stronger then the pull of the two chunks together.

This lab is really an oversimplication of the problem. It ignores any rotational effect and thus angular momentum of the two orbiting rocks but the basic idea is right.

This might explain the Roche limit