The fate of the Earth?
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From: Yousuf Khan on 14 Sep 2009 02:09 dow wrote: > On Sep 12, 5:08 pm, oriel36 <kelleher.ger...(a)gmail.com> wrote: > >> Repeat this as many times as you like,what looks good at close >> range,and in Kepler's era,things only went as far as Saturn,but his >> correlation between orbital periods and orbital radii does not work. > > Give me one example in which this equation does not work: > > T^2 = (2π)^2 a^3 / (G M) > > where G is the gravitational constant, T is the orbital period, a is > the semi-major axis of an elliptical orbit (which equals the radius if > the orbit is circular), and M is the mass of the primary, i.e. the > star or whatever that the object is going around. > > It works for all the planets, asteroids, etc., in our solar system, > using the mass of the sun for M, for all the satellites of Jupiter, > using Jupiter's mass for M, ditto for the satellites of the other > planets, and, yes, for Formalhaut b, using the mass of Formalhaut a as > the value of M. Actually, one thing that has always puzzled me is how one could use Kepler's laws to determine the mass of two objects whose mass is unknown? You see it all of the time, where scientists talk about determining the mass of one object by measuring it against another object, such as binary stars. Yousuf Khan
From: Sam Wormley on 14 Sep 2009 08:41 Yousuf Khan wrote: > dow wrote: >> On Sep 12, 5:08 pm, oriel36 <kelleher.ger...(a)gmail.com> wrote: >> >>> Repeat this as many times as you like,what looks good at close >>> range,and in Kepler's era,things only went as far as Saturn,but his >>> correlation between orbital periods and orbital radii does not work. >> >> Give me one example in which this equation does not work: >> >> T^2 = (2Ï)^2 a^3 / (G M) >> >> where G is the gravitational constant, T is the orbital period, a is >> the semi-major axis of an elliptical orbit (which equals the radius if >> the orbit is circular), and M is the mass of the primary, i.e. the >> star or whatever that the object is going around. >> >> It works for all the planets, asteroids, etc., in our solar system, >> using the mass of the sun for M, for all the satellites of Jupiter, >> using Jupiter's mass for M, ditto for the satellites of the other >> planets, and, yes, for Formalhaut b, using the mass of Formalhaut a as >> the value of M. > > > Actually, one thing that has always puzzled me is how one could use > Kepler's laws to determine the mass of two objects whose mass is > unknown? You see it all of the time, where scientists talk about > determining the mass of one object by measuring it against another > object, such as binary stars. > > Yousuf Khan T^2 = (2Ï)^2 a^3 / (G M+m) If one know the period and orbital distance and one of the masses... Background http://www2.sunysuffolk.edu/pappasm/AST101/Class7.pdf Some Detail http://www.physics.oregonstate.edu/~ketterj/COURSES/ph104spr/materials/lab/l7.pdf
From: dow on 14 Sep 2009 14:35 On Sep 14, 8:41Â am, Sam Wormley <sworml...(a)mchsi.com> wrote: > Yousuf Khan wrote: > > dow wrote: > >> On Sep 12, 5:08 pm, oriel36 <kelleher.ger...(a)gmail.com> wrote: > > >>> Repeat this as many times as you like,what looks good at close > >>> range,and in Kepler's era,things only went as far as Saturn,but his > >>> correlation between orbital periods and orbital radii does not work. > > >> Give me one example in which this equation does not work: > > >> Â T^2 = (2Ï)^2 a^3 / (G M) > > >> where G is the gravitational constant, T is the orbital period, a is > >> the semi-major axis of an elliptical orbit (which equals the radius if > >> the orbit is circular), and M is the mass of the primary, i.e. the > >> star or whatever that the object is going around. > > >> It works for all the planets, asteroids, etc., in our solar system, > >> using the mass of the sun for M, for all the satellites of Jupiter, > >> using Jupiter's mass for M, ditto for the satellites of the other > >> planets, and, yes, for Formalhaut b, using the mass of Formalhaut a as > >> the value of M. > > > Actually, one thing that has always puzzled me is how one could use > > Kepler's laws to determine the mass of two objects whose mass is > > unknown? You see it all of the time, where scientists talk about > > determining the mass of one object by measuring it against another > > object, such as binary stars. > > > Â Â Yousuf Khan > > Â Â T^2 = (2Ï)^2 a^3 / (G M+m) > > Â Â If one know the period and orbital distance and one of the masses... > > Â Â Background > Â Â Â http://www2.sunysuffolk.edu/pappasm/AST101/Class7.pdf > > Â Â Some Detail > Â Â Â http://www.physics.oregonstate.edu/~ketterj/COURSES/ph104spr/material...- Hide quoted text - > > - Show quoted text - You don't need to know one of the masses. All you need is the ratio of the masses, which can can be determined by observing the relative radii of their orbits. m1/m2 = r2/r1. dow
From: alien8er on 14 Sep 2009 16:05 On Sep 13, 4:53 am, oriel36 <kelleher.ger...(a)gmail.com> wrote: > Tell me how long it takes the Earth to turn once To "turn once" relative to what? Measured against the Sun, it varies through the year for several reasons(1); the mean value is currently 86,400 seconds and is defined as the "solar day". Measured against the Vernal equinox(2), it currently takes about 86164.09053083288 sidereal seconds. That's a so-called "sidereal day". Measured against the "fixed stars"(3), it currently takes 86164.098903691 sidereal seconds. Some call that period an "apparent sidereal day"; it's also called a "stellar day" by the International Earth Rotation and Reference Systems Service (IERS). (1) Mostly since the Earth's around the Sun isn't circular; the time it takes for the sun to return to its highest observed point changes as the Earth's orbital velocity changes. (2) The equinoxes don't hold still because the Earth orbits the sun. (3) The stars aren't really fixed either; that idea assumes there's something out there that they're attached to (the "crystal spheres"). The bottom line is that there's no fixed reference anywhere against which to measure the Earth's rotation period; it isn't even measurable against itself, though its absolute rotation _rate_ is measurable. Mark L. Fergerson
From: dow on 14 Sep 2009 16:36
> The bottom line is that there's no fixed reference anywhere against > which to measure the Earth's rotation period; it isn't even measurable > against itself, though its absolute rotation _rate_ is measurable. > > Mark L. Fergerson Ummm.. There is (or at least there is thought to be) an absolutely non- rotating frame of reference, independent of observations of stars, etc.. It's the frame in which there are no centrifugal or Coriolis forces. Theoretically, the absolute speed of rotation of he earth is its speed relative to this frame. However, there is no existing way in which this speed can be measured to more than a few digits of precision, so for practical purposes we have to use the "fixed" stars, even though we know they are not really fixed. dow |