Lagrange point L1

acsinnz · Dec 17, 2008

Wikipedia seems a little ambiguous on the exact position of this point from the earths surface. Is it 1.5 x 10>3 km or 0.15 x 10>6 km or ???

UFmbutler · Dec 17, 2008

http://www.physics.montana.edu/faculty/cornish/lagrange.pdf Here is a derivation of them...it claims approximately 1.5x10^6 km.  Why do you ask?  Planning a trip? <div class="Discussion_UserSignature"> </div>

MeteorWayne · Dec 17, 2008

ACE orbits the L1 libration point which is a point of Earth-Sun gravitational equilibrium about 1.5 million km from Earth and 148.5 million km from the Sun. From its location at L1 ACE has a prime view of the solar wind, interplanetary magnetic field and higher energy particles accelerated by the Sun, as well as particles accelerated in the heliosphere and the galactic regions beyond. ----- ACE also provides near-real-time 24/7 continuous coverage of solar wind parameters and solar energetic particle intensities (space weather). When reporting space weather ACE provides an advance warning (about one hour) of geomagnetic storms that can overload power grids, disrupt communications on Earth, and present a hazard to astronauts. The spacecraft has enough propellant on board to maintain an orbit at L1 until ~2024. SOHO moves around the Sun in step with the Earth, by slowly orbiting around the First Lagrangian Point (L1), where the combined gravity of the Earth and Sun keep SOHO in an orbit locked to the Earth-Sun line. The L1 point is approximately 1.5 million kilometers away from Earth (about four times the distance of the Moon), in the direction of the Sun. There, SOHO enjoys an uninterrupted view of our daylight star. All previous solar observatories have orbited the Earth, from where their observations were periodically interrupted as our planet `eclipsed' the Sun. For reference, 1 AU (Earth's distance to the sun) is 150 x 10^6 km, the moon is 0.384 x 10^6 km from earth (average). <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

acsinnz · Dec 18, 2008

Thanks UF and Meteor The problem is that at the L1 point the attraction to the Earth should equal the attraction to the sun! Oliver & Boyd state that earth mass= 6.4 x 10>6 and sun mass=2.0 x 10>30 kg By using inverse square law this means that Attraction to earth = 6.4 x 10>6/1.5 x 1.5 x 10>18 = 2.844 x 10>-12 Attraction to sun = 2.0 x 10>30/1.48 x 1.49 x 10>22= 0.96 x 10>-8 = 9.6 x 10>-9 It does not seem to equate or is my maths wrong!

MeteorWayne · Dec 18, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Thanks UF and Meteor The problem is that at the L1 point the attraction to the Earth should equal the attraction to the sun! Oliver & Boyd state that earth mass= 6.4 x 10>6 and sun mass=2.0 x 10>30 kg By using inverse square law this means that Attraction to earth = 6.4 x 10>6/1.5 x 1.5 x 10>18 = 2.844 x 10>-12 Attraction to sun = 2.0 x 10>30/1.48 x 1.49 x 10>22= 0.96 x 10>-8 = 9.6 x 10>-9 It does not seem to equate or is my maths wrong! Posted by acsinnz</DIV> Your earth mass is wrong. According to NASA it is 5.9736 X 10^24 kg <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

acsinnz · Dec 19, 2008

Hi MeteorThe mass of the earth now corrected from 10^6 to 10 ^24Attraction to earth = 6.0 x 10>24/1.5 x 1.5 x 10>18 = 2.844 x 10>6 Attraction to sun = 2.0 x 10>30/1.48 x 1.49 x 10>22= 0.96 x 10>8 = 9.6 x 10>9but there is still a problem! maybe 0.15 million km from earth would be better? These 10^?? are real tricky

MeteorWayne · Dec 19, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Hi MeteorThe mass of the earth now corrected from 10^6 to 10 ^24Attraction to earth = 6.0 x 10>24/1.5 x 1.5 x 10>18 = 2.844 x 10>6 Attraction to sun = 2.0 x 10>30/1.48 x 1.49 x 10>22= 0.96 x 10>8 = 9.6 x 10>9but there is still a problem! maybe 0.15 million km from earth would be better? These 10^?? are real tricky   Posted by acsinnz</DIV> Indeed they are

What is the formula you are using for the gravitational attraction? <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

kelvinzero · Dec 20, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Thanks UF and Meteor The problem is that at the L1 point the attraction to the Earth should equal the attraction to the sun! Oliver & Boyd state that earth mass= 6.4 x 10>6 and sun mass=2.0 x 10>30 kg By using inverse square law this means that Attraction to earth = 6.4 x 10>6/1.5 x 1.5 x 10>18 = 2.844 x 10>-12 Attraction to sun = 2.0 x 10>30/1.48 x 1.49 x 10>22= 0.96 x 10>-8 = 9.6 x 10>-9 It does not seem to equate or is my maths wrong! Posted by acsinnz</DIV>Is it actually the point where the two gravities cancel? Maybe you have to factor in the v^2/r orbital force also? (that is just  a guess)

vogon13 · Dec 20, 2008

This L1 is for the barycenter of the earth/moon system.  Please add the mass of the moon.  (IIRC >1% e)    <div class="Discussion_UserSignature"> TPTB went to Dallas and all I got was Plucked !!So many people, so few recipes !!Let's clean up this stinkhole !! </div>

secondAttempt · Dec 20, 2008

The Lagrange 1 point is not where the acceleration of the sun and the acceleration of the Earth are equal but opposite.It is the point where the combined acceleration due to gravity from the sun the Earth are equal but opposite to the centrifugal force around the barycenter of the sun-Earth system.The sun and Earth pull in opposite directions, so acceleration from gravity is:GMsun/dsun^2 - GMearth/dearth^2, towards the sunThe distance from the L1 point to the sun-Earth barycenter is the Earth-sun distance minus the Earth-sun distance times the mass of the Earth, divided by the combined mass of the Earth and sun.The centrifugal acceleration is v^2/r, where v is your velocity, and r is the distance from L1 to the barycenter of the Earth-sun system. You can compute velocity if you know the period.The period must equal the period of Earth around the sun, so compute its period using 2*pi*sqrt(d^3/(G*(combined Earth / Sun mass)), where d is the Earth-sun distance. Square this and divide by the barycenter distance.Your gravitational accelerations from gravity and from centrifugal force should equal each other in magnitude.Note: This method does not compute the distance to the L1 point for you. It requires that already know this distance. This method only verifies your guess. To my knowledge, there is no analytic method of exactly computing the Lagrange 1 point. To find it, you must take your best guess, check it with the above method, refine your guess, check it again, and zero in on it this way. Here's an online calculator that will compute it for you using this numerical method:  http://www.orbitsimulator.com/formulas/LagrangePointFinder.html

acsinnz · Dec 29, 2008

Thanks 2nd attempt;  but I do not want to calculate the position I need to know approximately how near to the earths surface to the L1 point is.  Alternatively, some-one may know what the communication delay time in seconds is.  From this I can calculate  the distance by multiplying  by the speed of light. I believe I has told by someone that it around four or five times the distance  of the moon?

MeteorWayne · Dec 29, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Thanks 2nd attempt;  but I do not want to calculate the position I need to know approximately how near to the earths surface to the L1 point is.  Alternatively, some-one may know what the communication delay time in seconds is.  From this I can calculate  the distance by multiplying  by the speed of light. I believe I has told by someone that it around four or five times the distance  of the moon? Posted by acsinnz</DIV> SOHO moves around the Sun in step with the Earth, by slowly orbiting around the First Lagrangian Point (L1), where the combined gravity of the Earth and Sun keep SOHO in an orbit locked to the Earth-Sun line. The L1 point is approximately 1.5 million kilometers away from Earth <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

acsinnz · Dec 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>SOHO moves around the Sun in step with the Earth, by slowly orbiting around the First Lagrangian Point (L1), where the combined gravity of the Earth and Sun keep SOHO in an orbit locked to the Earth-Sun line. The L1 point is approximately 1.5 million kilometers away from Earth Posted by MeteorWayne</DIV>Thanks MeteorOn the SOHO website it says 149.2 km from earth BUT that they are not on the exact L1 point as there is far too much ?? interference from direct ? radiation and noise in the centre of the solar wind.  They therefore have chosen an orbit that rotates around the outside of the solar wind tunnel.  Do you know how far out away from the center of the tunnel that orbit is??  If more than say  6000 km then it will only measure the spillage around the edges of the main stream of incoming H+ ions!!

MeteorWayne · Dec 31, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Thanks MeteorOn the SOHO website it says 149.2 km from earth BUT that they are not on the exact L1 point as there is far too much ?? interference from direct ? radiation and noise in the centre of the solar wind.  They therefore have chosen an orbit that rotates around the outside of the solar wind tunnel.  Do you know how far out away from the center of the tunnel that orbit is??  If more than say  6000 km then it will only measure the spillage around the edges of the main stream of incoming H+ ions!! Posted by acsinnz</DIV> Sorry, but I can't understand what you are trying to ask/say.An object does not remain exactly at the L1 point due to various perterbations, in fact the several spacecraft there use small thrusting manueuvers to stay in that area.I have no idea what you mean by "the center of the tunnel of the orbit".I have no idea what you are trying to get at with the "spillage around the edges of the main stream of incoming H+ ions".Perhaps you should try asking one question per post, since when you have a lot it all comes out rather garbled.Wayne <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

acsinnz · Jan 3, 2009

Happy New Year Meteor  Sorry, I became distracted and wandered off in another subject completely.   You are correct try to finish Lagrange first. The null point L1 would be where a satellite weighing nothing would sit if only gravity is involvedAttraction to earth at distance d from L1= 6.62x 10^24/d x d   should equal Attraction to sun at L1  = 2.0 x 10^30/1.492 x 1.492 x 10^22= 0.90 x 10^8 Therefore d = sq.root of 6.62 x 10^24/ 0.9 x 10^8  =7.36 x 10^8 metres [about 736,000 km from earth] But due to the mass of the satellite we have to allow for the outward inertia force as it rotates the sun.   If the satellite were to weigh say 100 kg then the Lagrange point would need to move nearer to the sun to compensate for the sideways inertia [ball on a string effect]; so that it settles at around 1.5 million km from the earth.  I think second attempt may have the answer!

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