Upgrade your account to a Community Supporter account and remove most of the site ads.

Reply to thread

Message: <blockquote data-quote="The Sigil" data-source="post: 655167" data-attributes="member: 2013">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/distanceSince 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 MarsmoonIn 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)}^2or8.68/(2.03)^2or2.1 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/distanceor 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"? <a href="http://www.arbovirus.health.nsw.gov.au/areas/arbovirus/climate/tideheights.htm" target="_blank">Some sample tide heights.</a>  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 earthMars 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. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile    :)"  data-smilie="1"data-shortname=":)" />--The Sigil</blockquote>

Verification