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<blockquote data-quote="The Sigil" data-source="post: 2536255" data-attributes="member: 2013"><p>Okay, this actually makes things pretty easy.</p><p></p><p>The angular size of an object ('a') as a function of distance to the object ('d') and the radius of the object ('r') is expressed by:</p><p></p><p>a = 2 tan^-1 (r/d)</p><p></p><p>However, since you have said that you want the angular sizes (i.e., how they appear to the earthbound observer) to be the same, we can get rid of all the nasty inverse tangent functions and come up with:</p><p></p><p>r1/d1 = r0/d0 or, rearranged slightly, d1 = d0*r1/r0</p><p></p><p>Where r1 is the fictitious planet's radius, r0 is the real planet's radius, d0 is the real distance to the planet, and d1 is the distance to the fictitious planet.</p><p></p><p>To make things as simple as possible, I'll assume all "planets" are exactly earth-sized; doubling the planet's size doubles the distance to the planet, and is a trivial exercise that can be done at home.</p><p></p><p>From here, it's a simple matter of taking average distance from earth to a planet (which can vary by quite a bit, but since you just wanted first-order approximations, we'll do that) and actual radius of a planet (expressed in "earth radii"). Since we're using "earth radii" as units of measurement, and we have said all fictional planets are exactly earth-sized, r1 = 1. Our formula for distance is then simply:</p><p></p><p>d1 = d0/r0</p><p></p><p>Here we go:</p><p></p><p>Sun - 1.37 million km (radius is 109 earths, average distance is 150 million km)</p><p>Moon - 1.41 million km (radius is .273 earths, average distance is 384 thousand km)</p><p>Jupiter - 69 million km (radius is 11.2 earths, average distance is 778 million km)</p><p>Saturn - 150 million km (radius is 9.45 earths, average distance is 1426 million km)</p><p>Venus - 158 million km (radius is .949 earths, average distance is 150 million km)</p><p>Mercury - 391 million km (radius is .383 earths, average distance is 150 million km)</p><p>Mars - 422 million km (radius is .533 earths, average distance is 225 million km)</p><p>Uranus - 715 million km (radius is 4.01 earths, average distance is 2870 million km)</p><p>Neptune - 1160 million km (radius is 3.88 earths, average distance is 4500 million km)</p><p></p><p>NOTES:</p><p></p><p>On a strict average over infinite time, every planet "outside" us is the same distance from us as it is from the Sun and every planet "inside" us is the same distance from us as the Sun is. I used semi-major orbital axes for all "average distance" measurements except earth-sun and earth-moon as these were more easily found on Wikipedia.</p><p></p><p>This number of roundings and <strong>huge oversimplifications</strong> (going from elliptical orbits to spherical orbits, assuming constant distance between the planets instead of varying distance due to different orbital speeds around the sun) means this system is <strong>GREATLY</strong> simplified... but is a "first order" approximation for you.</p><p></p><p>Luminosity equations (not shown) tell us that the Sun needs to be at the same temperature as it is now when reduced in size and distance to give the same amount of energy to the earth. (Chopping the size reduces the energy given off by the "reduction factor" squared, but bringing it closer reduces the energy needed by the "closer factor" squared; these wind up cancelling out)... so the Sun doesn't have to be any "hotter" than usual on the surface to provide the same energy to the earth.</p><p></p><p>Again, if you want to move planets farther out/closer in, you just enlarge/reduce them by a factor of the same amount; for instance, if you want the Sun to be twice as large as earth, move it twice as far away (to 2.74 million km out) to keep things the same. If you want the moon to be half of earth size, you can move it in to half its listed orbit (i.e., to about 700 thousand km out).</p></blockquote><p></p>
[QUOTE="The Sigil, post: 2536255, member: 2013"] Okay, this actually makes things pretty easy. The angular size of an object ('a') as a function of distance to the object ('d') and the radius of the object ('r') is expressed by: a = 2 tan^-1 (r/d) However, since you have said that you want the angular sizes (i.e., how they appear to the earthbound observer) to be the same, we can get rid of all the nasty inverse tangent functions and come up with: r1/d1 = r0/d0 or, rearranged slightly, d1 = d0*r1/r0 Where r1 is the fictitious planet's radius, r0 is the real planet's radius, d0 is the real distance to the planet, and d1 is the distance to the fictitious planet. To make things as simple as possible, I'll assume all "planets" are exactly earth-sized; doubling the planet's size doubles the distance to the planet, and is a trivial exercise that can be done at home. From here, it's a simple matter of taking average distance from earth to a planet (which can vary by quite a bit, but since you just wanted first-order approximations, we'll do that) and actual radius of a planet (expressed in "earth radii"). Since we're using "earth radii" as units of measurement, and we have said all fictional planets are exactly earth-sized, r1 = 1. Our formula for distance is then simply: d1 = d0/r0 Here we go: Sun - 1.37 million km (radius is 109 earths, average distance is 150 million km) Moon - 1.41 million km (radius is .273 earths, average distance is 384 thousand km) Jupiter - 69 million km (radius is 11.2 earths, average distance is 778 million km) Saturn - 150 million km (radius is 9.45 earths, average distance is 1426 million km) Venus - 158 million km (radius is .949 earths, average distance is 150 million km) Mercury - 391 million km (radius is .383 earths, average distance is 150 million km) Mars - 422 million km (radius is .533 earths, average distance is 225 million km) Uranus - 715 million km (radius is 4.01 earths, average distance is 2870 million km) Neptune - 1160 million km (radius is 3.88 earths, average distance is 4500 million km) NOTES: On a strict average over infinite time, every planet "outside" us is the same distance from us as it is from the Sun and every planet "inside" us is the same distance from us as the Sun is. I used semi-major orbital axes for all "average distance" measurements except earth-sun and earth-moon as these were more easily found on Wikipedia. This number of roundings and [b]huge oversimplifications[/b] (going from elliptical orbits to spherical orbits, assuming constant distance between the planets instead of varying distance due to different orbital speeds around the sun) means this system is [b]GREATLY[/b] simplified... but is a "first order" approximation for you. Luminosity equations (not shown) tell us that the Sun needs to be at the same temperature as it is now when reduced in size and distance to give the same amount of energy to the earth. (Chopping the size reduces the energy given off by the "reduction factor" squared, but bringing it closer reduces the energy needed by the "closer factor" squared; these wind up cancelling out)... so the Sun doesn't have to be any "hotter" than usual on the surface to provide the same energy to the earth. Again, if you want to move planets farther out/closer in, you just enlarge/reduce them by a factor of the same amount; for instance, if you want the Sun to be twice as large as earth, move it twice as far away (to 2.74 million km out) to keep things the same. If you want the moon to be half of earth size, you can move it in to half its listed orbit (i.e., to about 700 thousand km out). [/QUOTE]
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