Homework Help Pleaseeee!!

[ASTRONOMYCLASS]

Suppose you observe a star orbititing the galactic center at a speed of 1,000 km/s in a vircular orbit with a radius of 20 light-days ... what would your estimate be for the mass of the obect that the star is orbiting .. (mass of the black hole at teh center of our galaxy ) Express your answer in terms of the Sun's Mass ... I need to convert light days to meters and km/s to m/s

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Suppose you observe a star orbititing the galactic center at a speed of 1,000 km/s in a vircular orbit with a radius of 20 light-days ... what would your estimate be for the mass of the obect that the star is orbiting .. (mass of the black hole at teh center of our galaxy ) Express your answer in terms of the Sun's Mass ... I need to convert light days to meters and km/s to m/s <br /> Posted by ASTRONOMYCLASS</DIV></p><p>First, orbits are eliptical and not circular.&nbsp; Second, we would also need to know the mass of the star.&nbsp; Third, you would need to be familiar with Newton's laws of motion and Kepler's laws.&nbsp; Fourth, there are conversion calculators all over google.&nbsp; And, finally, fifth... don't expect homework answers.&nbsp; We would be glad to point you in the right direction and help you get your answer, but it will not be given freely unless you put some work into it first. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>First, orbits are eliptical and not circular.&nbsp; Second, we would also need to know the mass of the star.&nbsp; Third, you would need to be familiar with Newton's laws of motion and Kepler's laws.&nbsp; Fourth, there are conversion calculators all over google.&nbsp; And, finally, fifth... don't expect homework answers.&nbsp; We would be glad to point you in the right direction and help you get your answer, but it will not be given freely unless you put some work into it first. <br />Posted by derekmcd</DIV></p><p>A circle is an ellipse.&nbsp; It is just a special case in which the two foci coincide.&nbsp; Circular orbits are possible.&nbsp; Many elliptical orbits are in fact nearly circular.</p><p>If you make the very reasonable assumption that the mass if the star is much less than the mass of the black hole about which it is&nbsp; revolving, then you do not need to know the mass of the star to estimate the mass of the black hole.&nbsp; That is because you then treat the black hole as a point around which the star is revolving, and the gravitational force is then an inverse square relation to not only the center of the black hole but also the single focus of the circular orbit.&nbsp; So the radial acceleration from the kinematics of the orbit is equal to the acceleration that results from Newton's law of universal gravitation.</p><p>You are prudent in not giving away homework answers without some effort from the student.&nbsp;&nbsp; And I hope my discussion did not give away the candy store.&nbsp; I think I managed to write in terms that you will understand, but that still leave some work for the student.</p><p>BTW, welcome back.&nbsp; We need a sane voice or three around here.</p> <div class="Discussion_UserSignature"> </div>

 

[lildreamer]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>A circle is an ellipse.&nbsp; It is just a special case in which the two foci coincide.&nbsp; Circular orbits are possible.&nbsp; Many elliptical orbits are in fact nearly circular.If you make the very reasonable assumption that the mass if the star is much less than the mass of the black hole about which it is&nbsp; revolving, then you do not need to know the mass of the star to estimate the mass of the black hole.&nbsp; That is because you then treat the black hole as a point around which the star is revolving, and the gravitational force is then an inverse square relation to not only the center of the black hole but also the single focus of the circular orbit.&nbsp; So the radial acceleration from the kinematics of the orbit is equal to the acceleration that results from Newton's law of universal gravitation.You are prudent in not giving away homework answers without some effort from the student.&nbsp;&nbsp; And I hope my discussion did not give away the candy store.&nbsp; I think I managed to write in terms that you will understand, but that still leave some work for the student.BTW, welcome back.&nbsp; We need a sane voice or three around here. <br />Posted by DrRocket</DIV></p><p>Hi DrRocket&nbsp;</p><p>I was wondering by the wording of the OP question, if the object&nbsp;being referred to is in a Geostationary orbit - if so would resolving the answer be easier? </p> <div class="Discussion_UserSignature"> </div>

 

[MeteorWayne]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Hi DrRocket&nbsp;I was wondering by the wording of the OP question, if the object&nbsp;being referred to is in a Geostationary orbit - if so would resolving the answer be easier? <br />Posted by lildreamer</DIV><br /><br />An object in another galaxy cannot, by definition be in a geostationary orbit, since that word means staying directly above one point on the earth's surface <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[lildreamer]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>An object in another galaxy cannot, by definition be in a geostationary orbit, since that word means staying directly above one point on the earth's surface <br />Posted by MeteorWayne</DIV></p><p><font color="#ff0000">Suppose you observe a star orbititing the galactic center at a speed of 1,000 km/s in a vircular orbit with a radius of 20 light-days ... what would your estimate be for the mass of the obect that the star is orbiting .. (mass of the black hole at teh center of our galaxy ) Express your answer in terms of the Sun's Mass ... I need to convert light days to meters and km/s to m/s</font>&nbsp;&nbsp;</p><p>geostationary in reference to the Galactic centre&nbsp;&nbsp;- he did mention a black hole. So I was presuming he was talking about our galaxy. Geostationary does not necessarily means it has to be reference to Earth. Geostationary is relative to the object that it orbits.&nbsp; I don't mean to dis your explanation just offering up a different interpretation of the question. My question was if the object was Geostationary would the original OP be easier to answer?</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[adrenalynn]

<p>Well, yeah, by definition it DOES mean that...&nbsp; You can arbitrarily change the definition of accepted words, but'cha gotta let us know&nbsp;what it means to be &nbsp;"sitting on the table and eating off of the lamp" first. </p><p>&nbsp;Specifically, it's defined as an orbit directly above the equator of the Earth with a periodicity equal to that of the Earth, and an orbital eccentricity approaching zero.&nbsp; </p> <div class="Discussion_UserSignature"> <p>.</p><p><font size="3">bipartisan</font>  (<span style="color:blue" class="pointer"><span class="pron"><font face="Lucida Sans Unicode" size="2">bī-pär'tĭ-zən, -sən</font></span></span>) [Adj.]  Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.</p><p><strong><font color="#ff0000"><font color="#ff0000">IMPE</font><font color="#c0c0c0">ACH</font> <font color="#0000ff"><font color="#c0c0c0">O</font>BAMA</font>!</font></strong></p> </div>

 

[lildreamer]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Well, yeah, by definition it DOES mean that...&nbsp; You can arbitrarily change the definition of accepted words, but'cha gotta let us know&nbsp;what it means to be &nbsp;"sitting on the table and eating off of the lamp" first. &nbsp;Specifically, it's defined as an orbit directly above the equator of the Earth with a periodicity equal to that of the Earth, and an orbital eccentricity approaching zero.&nbsp; <br />Posted by adrenalynn</DIV><br /><br /><strong>GEOSTATIONARY ORBIT (GEO) &ndash; </strong>A special geosynchronous orbit which is circular and lying over the equator such that the satellite seems to remain stationary in the sky as seem from a location on the surface of Earth.</p><p><strong>GEOSYNCHRONOUS ORBIT (GSO) &ndash; </strong>An orbit around the Earth with an average distance from the center of Earth of about 26,000 miles, in which a satellite would have a period equal to the rotation period of the Earth.</p><p>Insert Saturn or Jupiter or any other planetary body instead of the term Earth <font color="#ff0000"><strong>and the concept remains the same</strong></font>. </p><p>I understand that the Latin term <font color="#000000"><strong>GEO means Earth</strong></font>but the concept remains the same - so what do you want to call other objects then demonstrating this principle - PlanetarySynchronous Orbit&nbsp; - PSO ????????????</p><p>either way I'm trying to ask whether or not if that particular&nbsp;object exhibited a GEO or GSO characteristic around the Black Hole would the question by the OP be easier to&nbsp;solve or not.....&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[MeteorWayne]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>GEOSTATIONARY ORBIT (GEO) &ndash; A special geosynchronous orbit which is circular and lying over the equator such that the satellite seems to remain stationary in the sky as seem from a location on the surface of Earth.GEOSYNCHRONOUS ORBIT (GSO) &ndash; An orbit around the Earth with an average distance from the center of Earth of about 26,000 miles, in which a satellite would have a period equal to the rotation period of the Earth.Insert Saturn or Jupiter or any other planetary body instead of the term Earth and the concept remains the same. I understand that the Latin term GEO means Earthbut the concept remains the same - so what do you want to call other objects then demonstrating this principle - PlanetarySynchronous Orbit&nbsp; - PSO ????????????either way I'm trying to ask whether or not if that particular&nbsp;object exhibited a GEO or GSO characteristic around the Black Hole would the question by the OP be easier to&nbsp;solve or not.....&nbsp; <br />Posted by lildreamer</DIV><br /><br />The question then becomes, how do you define a point of the suface of a black hole. How do you determine the rotation of the black hole. After all, it is a singularity; the only "surface" is the event horizon.</p><p>I realize you are trying to help, but...&nbsp;you are heading down a path that leads nowhere here </p><p>In any case, all that really matters is the mass of the black hole to solve the question.</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[lildreamer]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The question then becomes, how do you define a point of the suface of a black hole. How do you determine the rotation of the black hole. After all, it is a singularity; the only "surface" is the event horizon.I realize you are trying to help, but...&nbsp;you are heading down a path that leads nowhere here In any case, all that really matters is the mass of the black hole to solve the question. <br />Posted by MeteorWayne</DIV><br /><br />agreed ....going to have a Corona ...ciao.... <div class="Discussion_UserSignature"> </div>

 

[Mee_n_Mac]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Suppose you observe a star orbititing the galactic center at a speed of 1,000 km/s in a vircular orbit with a radius of 20 light-days ... what would your estimate be for the mass of the obect that the star is orbiting .. (mass of the black hole at teh center of our galaxy ) Express your answer in terms of the Sun's Mass ... I need to convert light days to meters and km/s to m/s <br />Posted by <strong>ASTRONOMYCLASS</strong></DIV><br /><br />We might help you but you've got to do the work.&nbsp; The last questions are the easy ones so start there.&nbsp; What's a light day ?&nbsp; It's how far light travels in one day.&nbsp; So you can look up the speed of light (in a vacuum) and then transform that to&nbsp;a convient system, say m/s.&nbsp; As for doing the transforms ... think just a bit.&nbsp; If I said you ran 14 km/fortnight, how far would you have run in 1 fortnight ?&nbsp; If I said 1 fortnight = 14 days, how far would have run in just 1 day ? <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Suppose you observe a star orbititing the galactic center at a speed of 1,000 km/s in a vircular orbit with a radius of 20 light-days ... what would your estimate be for the mass of the obect that the star is orbiting .. (mass of the black hole at teh center of our galaxy ) Express your answer in terms of the Sun's Mass ... I need to convert light days to meters and km/s to m/s&nbsp;&nbsp;geostationary in reference to the Galactic centre&nbsp;&nbsp;- he did mention a black hole. So I was presuming he was talking about our galaxy. Geostationary does not necessarily means it has to be reference to Earth. Geostationary is relative to the object that it orbits.&nbsp; I don't mean to dis your explanation just offering up a different interpretation of the question. My question was if the object was Geostationary would the original OP be easier to answer?&nbsp; <br />Posted by lildreamer</DIV></p><p>Wayne is correct, a geostationary orbit is an orbit that orbits the Earth and remains over a fixed location.&nbsp; Such orbits&nbsp; must be directly above the equator and they have an orbital period&nbsp;that matches the Earth's rotational period (24 hrs).&nbsp; Geostationary orbits are all at a fixed altitude, about 23000 miles if I recall correctly.&nbsp; The altitude is fixed by the mass of the Earth and the necessary orbital period, and of course the laws of orbital mechanics.&nbsp; Geostationary orbits are circular.&nbsp; The fact that an orbit is geostationary does not help in the calculations, since the orbital parameters are independent of the spinning of the Earth.</p><p>You cannot talk about an orbit that would be equivalent of geostationary about another body unless you know the rate at which that body rotates about its axis.&nbsp; Geostationary, or the equivalent, is not a particularly useful or interesting orbit about a black hole, even if it is spinning, since you can't see past the event&nbsp; horizon.&nbsp; </p><p>In this case, with some approximations one ought to be able to determine the mass of the black hole (or whatever is the large mass at the center of the orbit) from knowledge of the fact that the orbit is circular, the radius of the orbit and the linear speed.&nbsp; You don't need to know the rate of spin of the black hole, and it would not help if you did.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>

 

[adrenalynn]

<p>>> equivalent of geostationary about another body unless you know the rate at which that body rotates about its axis.</p><p>And the mass of the body as well as the mass of the satellite, correct?&nbsp; </p> <div class="Discussion_UserSignature"> <p>.</p><p><font size="3">bipartisan</font>  (<span style="color:blue" class="pointer"><span class="pron"><font face="Lucida Sans Unicode" size="2">bī-pär'tĭ-zən, -sən</font></span></span>) [Adj.]  Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.</p><p><strong><font color="#ff0000"><font color="#ff0000">IMPE</font><font color="#c0c0c0">ACH</font> <font color="#0000ff"><font color="#c0c0c0">O</font>BAMA</font>!</font></strong></p> </div>

 

[MeteorWayne]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>>> equivalent of geostationary about another body unless you know the rate at which that body rotates about its axis.And the mass of the body as well as the mass of the satellite, correct?&nbsp; <br />Posted by adrenalynn</DIV><br /><br />In most cases, the mass of the satellite (be it a planet around a star, a star around a galaxy center, or a moon around a planet) is insignificant to solving the equation. The rotation of the central mass has no effect on the calculations. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[adrenalynn]

<p>Yeah, ok, I'll buy that.&nbsp; <font size="3"><em>m</em>a</font><font size="1">c</font> = <font size="3"><em>m</em>g</font> </p><p>Newton says that's ok, so we'll go with him, he's a pretty bright guy, give or take that apple thang... </p><p>But at the Blackhole vs Planet level, I'm not sure I'd agree with you... Once we get over <em>M</em>&nbsp;<em>&mu;_err</em>/<em>&mu;</em>&asymp;10¹⁵&nbsp;kg, give or take, , our error is going to become pretty unacceptable, right?&nbsp; We'd be over the geocentric gravitational constant, no?&nbsp; <em>&mu;=</em>GM?</p> <div class="Discussion_UserSignature"> <p>.</p><p><font size="3">bipartisan</font>  (<span style="color:blue" class="pointer"><span class="pron"><font face="Lucida Sans Unicode" size="2">bī-pär'tĭ-zən, -sən</font></span></span>) [Adj.]  Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.</p><p><strong><font color="#ff0000"><font color="#ff0000">IMPE</font><font color="#c0c0c0">ACH</font> <font color="#0000ff"><font color="#c0c0c0">O</font>BAMA</font>!</font></strong></p> </div>

 

[MeteorWayne]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yeah, ok, I'll buy that.&nbsp; mac = mg Newton says that's ok, so we'll go with him, he's a pretty bright guy, give or take that apple thang... But at the Blackhole vs Planet level, I'm not sure I'd agree with you... Once we get over M&nbsp;&mu;err/&mu;&asymp;1015&nbsp;kg, give or take, , our error is going to become pretty unacceptable, right?&nbsp; We'd be over the geocentric gravitational constant, no?&nbsp; &mu;=GM? <br />Posted by adrenalynn</DIV><br /><br />All of that is irrelevant at the galactic scale for the object cited in the original post.</p><p>Sure we can nitpick about specific circumstances, but that wouldn't help with the homework assignment </p><p>BTW, it is REALLY good to see you back. SDC is a better place when you are around.</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[lildreamer]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Wayne is correct, a geostationary orbit is an orbit that orbits the Earth and remains over a fixed location.&nbsp; Such orbits&nbsp; must be directly above the equator and they have an orbital period&nbsp;that matches the Earth's rotational period (24 hrs).&nbsp; Geostationary orbits are all at a fixed altitude, about 23000 miles if I recall correctly.&nbsp; The altitude is fixed by the mass of the Earth and the necessary orbital period, and of course the laws of orbital mechanics.&nbsp; Geostationary orbits are circular.&nbsp; The fact that an orbit is geostationary does not help in the calculations, since the orbital parameters are independent of the spinning of the Earth.You cannot talk about an orbit that would be equivalent of geostationary about another body unless you know the rate at which that body rotates about its axis.&nbsp; Geostationary, or the equivalent, is not a particularly useful or interesting orbit about a black hole, even if it is spinning, since you can't see past the event&nbsp; horizon.&nbsp; In this case, with some approximations one ought to be able to determine the mass of the black hole (or whatever is the large mass at the center of the orbit) from knowledge of the fact that the orbit is circular, the radius of the orbit and the linear speed.&nbsp; You don't need to know the rate of spin of the black hole, and it would not help if you did.&nbsp; <br />Posted by DrRocket</DIV></p><p><font color="#ff0000">" You don't need to know the rate of spin of the black hole, and it would not help if you did."</font></p><p><font color="#000000">Thank You Dr.Rocket thats all I was asking for......</font></p><p><font color="#000000">Upon further digging I found the answer or the concept the OP is looking for...which several people here have already alluded to....</font></p><p><font color="#000000">By the way I stand corrected Geosynchronous orbit is specifically related to the Earth and only to the Earth...but the plain term "synchronous orbit" </font><font color="#000000">is used&nbsp; to explain&nbsp;an orbit in which an orbiting body (usually a satellite) has a period equal to the average rotational period of the body being orbited (usually a planet), and in the same direction of rotation as that body.</font></p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>A circle is an ellipse.&nbsp; It is just a special case in which the two foci coincide.&nbsp; Circular orbits are possible.&nbsp; Many elliptical orbits are in fact nearly circular.If you make the very reasonable assumption that the mass if the star is much less than the mass of the black hole about which it is&nbsp; revolving, then you do not need to know the mass of the star to estimate the mass of the black hole.&nbsp; That is because you then treat the black hole as a point around which the star is revolving, and the gravitational force is then an inverse square relation to not only the center of the black hole but also the single focus of the circular orbit.&nbsp; So the radial acceleration from the kinematics of the orbit is equal to the acceleration that results from Newton's law of universal gravitation.You are prudent in not giving away homework answers without some effort from the student.&nbsp;&nbsp; And I hope my discussion did not give away the candy store.&nbsp; I think I managed to write in terms that you will understand, but that still leave some work for the student.BTW, welcome back.&nbsp; We need a sane voice or three around here. <br /> Posted by DrRocket</DIV></p><p>I guess I'm just used to see orbital referred to as ellipses defined by their eccentricity.&nbsp; Low eccentricity can be approximated as circular, but I'm not aware of any orbits that are truely circular as in a perfect circle. &nbsp;</p><p>As for defining the mass of Sagittarius A* without knowing the mass of its satellites, a little common sense on my part should have recognized that.&nbsp; Speaking of which, here's an article that just came out today discussing this very thing:</p><p>&nbsp;</p><p>http://www.sciencedaily.com/releases/2008/12/081209221707.htm</p><h1 class="story"><font size="3">Unprecedented 16-year-long Study Tracks Stars Orbiting Milky Way Black Hole</font></h1><p><span class="date">ScienceDaily (Dec. 10, 2008)</span> &mdash;<em> By watching the motions of 28 stars orbiting the Milky Way's most central region with admirable patience and amazing precision, astronomers have been able to study the supermassive black hole lurking there. It is known as "Sagittarius A*" (pronounced "Sagittarius A star"). The new research marks the first time that the orbits of so many of these central stars have been calculated precisely and reveals information about the enigmatic formation of these stars &mdash; and about the black hole to which they are bound.</em></p><p><em>"The centre of the Galaxy is a unique laboratory where we can study the fundamental processes of strong gravity, stellar dynamics and star formation that are of great relevance to all other galactic nuclei, with a level of detail that will never be possible beyond our Galaxy," explains Reinhard Genzel, leader of the team from the Max-Planck-Institute for Extraterrestrial Physics in Garching near Munich.</em></p> <p><em>The interstellar dust that fills the Galaxy blocks our direct view of the Milky Way's central region in visible light. So astronomers used infrared wavelengths that can penetrate the dust to probe the region. While this is a technological challenge, it is well worth the effort. "The Galactic Centre harbours the closest supermassive black hole known. Hence, it is the best place to study black holes in detail," argues the study's first author, Stefan Gillessen.</em></p> <p><em>The team used the central stars as "test particles" by watching how they move around Sagittarius A*. Just as leaves caught in a wintry gust reveal a complex web of air currents, so does tracking the central stars show the nexus of forces at work at the Galactic Centre. These observations can then be used to infer important properties of the black hole itself, such as its mass and distance. The new study also showed that at least 95% of the mass sensed by the stars has to be in the black hole. There is thus little room left for other dark matter.</em></p> <p><em>"Undoubtedly the most spectacular aspect of our long term study is that it has delivered what is now considered to be the best empirical evidence that supermassive black holes do really exist. The stellar orbits in the Galactic Centre show that the central mass concentration of four million solar masses must be a black hole, beyond any reasonable doubt," says Genzel. The observations also allow astronomers to pinpoint our distance to the centre of the Galaxy with great precision, which is now measured to be 27 000 light-years.</em></p> <p><em>To build this unparalleled picture of the Milky Way's heart and calculate the orbits of the individual stars the team had to study the stars there for many years. These latest groundbreaking results therefore represent 16 years of dedicated work, which started with observations made in 1992 with the SHARP camera attached to ESO's 3.5-metre New Technology Telescope located at the La Silla observatory in Chile. More observations have subsequently been made since 2002 using two instruments mounted on ESO's 8.2 m Very Large Telescope (VLT). A total of roughly 50 nights of observing time with ESO telescopes, over the 16 years, has been used to complete this incredible set of observations.</em></p> <p><em>The new work improved the accuracy by which the astronomers can measure the positions of the stars by a factor of six compared to previous studies. The final precision is 300 microarcseconds, equivalent at seeing a one euro coin from a distance of roughly 10 000 km.</em></p><p>&nbsp;</p><p>More at link above.&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>>> equivalent of geostationary about another body unless you know the rate at which that body rotates about its axis.And the mass of the body as well as the mass of the satellite, correct?&nbsp; <br />Posted by adrenalynn</DIV></p><p>To determine the orbital parameters with absolute precision you need the mass of both bodies, but for most practical applications one of the bodies is much much larger than the other and it suffices to know only the mass of the larger body.&nbsp; That would let you calculate all the necessary orbital parameters for any satellite (of relatively small mass).</p><p>To determine a stationary orbit you need only know the rotational period in addition to the mass of the larger body.&nbsp; That is because all that you are doing is designating one particular orbit as stationary, based on the orbital period equalling the rotational period of the large body.<br /></p> <div class="Discussion_UserSignature"> </div>