กฎนี้รู้จักในอีกชื่อหนึ่งที่ว่ากฎพื้นที่เท่า ซึ่งเป็นผลสืบเนื่องโดยตรงจากกฎการอนุรักษ์โมเมนตัมเชิงมุม (law of conservation of angular momentum) โปรดดูการการอนุพัทธ์ดังภาพ
การคำนวณมี 4 ขั้นดังนี้
1. คำนวณ มุมกวาดเฉลี่ย (mean anomaly) M จากสูตร
2. คำนวณ มุมกวาดเยื้องศูนย์กลาง eccentric anomaly E โดยการแก้ สมการของเค็พเพลอร์:
ขั้นที่สี่คือการคำนวณระยะห่างศูนย์สุริยะ r จากมุมกวาดจริง ν ด้วยกฎข้อแรกของเค็พเพลอร์:
การอนุพัทธ์ (Derivation) กฎของนิวตัน
การอนุพัทธ์ของกฎเค็พเพลอร์ข้อที่ 2
where is the tangential unit vector, and
So the position vector
is differentiated twice to give the velocity vector and the acceleration vector
Note that for constant distance, , the planet is subject to the centripetal acceleration, , and for constant angular speed, , the planet is subject to the coriolis acceleration, .
Inserting the acceleration vector into Newton's laws, and dividing by m, gives the vector equation of motion
Equating component, we get the two ordinary differential equations of motion, one for the radial acceleration and one for the tangential acceleration:
and integrate:
where is a constant of integration, and exponentiate:
This says that the specific angular momentum is a constant of motion, even if both the distance and the angular speed vary.
The area swept out from time t1 to time t2,
depends only on the duration t2−t1. This is Kepler's second law.
การอนุพัทธ์ของกฎเค็พเพลอร์ข้อที่ 1
and get
and
Differentiate
twice:
Substitute into the radial equation of motion
and get
Divide by
These solutions are
where and are arbitrary constants of integration. So the result is
Choosing the axis of the coordinate system such that , and inserting , gives:
Hyman, Andrew. "A Simple Cartesian Treatment of Planetary Motion" 2011-08-07 ที่ เวย์แบ็กแมชชีน, European Journal of Physics, Vol. 14, pp. 145-147 (1993).
"Kepler's Second Law" by Jeff Bryant with Oleksandr Pavlyk, The Wolfram Demonstrations Project.
ดูเพิ่ม
Kepler problem
Circular motion
Gravity
Two-body problem
Free-fall time
แหล่งข้อมูลอื่น
Crowell, Benjamin, Conservation Laws, http://www.lightandmatter.com/area1book2.html, an online book that gives a proof of the first law without the use of calculus. (see section 5.2, p.112)
David McNamara and Gianfranco Vidali, Kepler's Second Law -JAVA Interactive Tutorial, http://www.phy.syr.edu/courses/java/mc_html/kepler.html 2006-09-10 ที่ เวย์แบ็กแมชชีน, an interactive JAVA applet that aids in the understanding of Kepler's Second Law.
University of Tennessee's Dept. Physics & Astronomy: Astronomy 161 page on Johannes Kepler: The Laws of Planetary Motion [1]
Equant compared to Kepler: interactive model [2] 2008-12-26 ที่ เวย์แบ็กแมชชีน
Kepler's Third Law:interactive model[3] 2008-12-26 ที่ เวย์แบ็กแมชชีน
มกราคม 21, 2023
กฎการเคล, อนท, ของดาวเคราะห, งก, ามภาษา, ในบทความน, ไว, ให, านและผ, วมแก, ไขบทความศ, กษาเพ, มเต, มโดยสะดวก, เน, องจากว, เด, ยภาษาไทยย, งไม, บทความด, งกล, าว, กระน, ควรร, บสร, างเป, นบทความโดยเร, วท, ดของเค, พเพลอร, งกฤษ, kepler, laws, planetary, motion, อกฎทาง. lingkkhamphasa inbthkhwamni miiwihphuxanaelaphurwmaekikhbthkhwamsuksaephimetimodysadwk enuxngcakwikiphiediyphasaithyyngimmibthkhwamdngklaw krann khwrribsrangepnbthkhwamodyerwthisudkdkarekhluxnthikhxngdawekhraahkhxngekhphephlxr xngkvs Kepler s laws of planetary motion khuxkdthangkhnitsastr 3 khxthiklawthungkarekhluxnthikhxngdawekhraahinrabbsuriya nkkhnitsastraelankdarasastrchaweyxrmnchux oyhnenis ekhphephlxr ph s 2114 2173 epnphukhnphbphaphaesdngkd 3 khxkhxngekhphephlxrthimiwngokhcrdawekhraah 2 wng 1 wngokhcrepnwngridwycudofks f1 aela f2 sahrbdawekhraahdwngaerkaela f1 aela f3 sahrbdawekhraahdwngthi 2 dwngxathityxyuthicud f1 2 swnaerenga 2 swn A1 aela A2 miphiwphunethaknaelaewlathidawekhraah 1 thbphunthi A1 ethakbewlathithbphunthi 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karxnuphthth Derivation kdkhxngniwtn 3 1 karxnuphththkhxngkdekhphephlxrkhxthi 2 3 2 karxnuphththkhxngkdekhphephlxrkhxthi 1 3 3 kdekhphephlxrkhxthi 3 4 xangxing 5 duephim 6 aehlngkhxmulxunkd 3 khxkhxngekhphephlxr aekikhwngokhcrkhxngdawekhraahthukdwngepnwngri odymidwngxathityepncudsunyklangcudhnung wngriekidcakkarmicudsunyklang 2 suny dngphaph dngnnekhphephlxrcungkhdkhankhwamechuxinaenwkhxngxrisotetil potelmiaelaokhepxrnikhsthiwawngokhcrepnwngklm inkhnathidawekhraahekhluxnipinwngokhcr esntrngthiechuxmrahwangdawekhraahkbdwngxathitykwadphunthietha kninrayaewlaethakn sunghmaykhwamwadawekhraahokhcrerwkwaemuxxyuikldwngxathityaelachalngemuxxyuhangdwngxathity dwykdkhxni ekhphephlxridlmthvsdidarasastrxrisotetilthiwadawekhraahekhluxnthidwykhwamerwkhngthi kalngsxngkhxngkhabkarokhcrkhxngdawekhraahepnsdswnodytrngkbkalngsamkhxngkungaeknexk khrunghnungkhxngkhwamyawwngri khxngwngokhcr sunghmaykhwamwa imephiyngaetwngokhcrthiihykwaethannthimirayaewlanankwa aetxtrakhwamerwkhxngdawekhraahthimiwngokhcrthiihykwannkokhcrchakwawngokhcrthielkkwaxikdwykdkhxngekhphephlxridaesdngiwkhanglang aelaepnkdthimacakkdkhxngniwtnthiichphikdkhwsunysuriya heliocentric polar coordinate r 8 displaystyle r theta xyangirktam kdkhxngekhphephlxryngsamarthekhiynxyangxunidodyichphikdkharthiesiyn Cartesian coordinates 1 raylaexiydthangkhnitsastr aekikhkdkhxthi 1 aekikh kdekhphephlxrkhxthi 1 kdkhxaerkklawwa wngokhcrkhxngdawekhraahthukdwngepnrupwngrithimidwngxathityepncudofkscudhnung khnitsastrkhxngwngriepndngnismkarkhux r p 1 ϵ cos 8 displaystyle r frac p 1 epsilon cdot cos theta odythi p khux kungeltserktm semi latus rectum aela e khux khwameyuxngsunyklang eccentricity sungmikhamakkwahruxethakbsuny aelanxykwahnungemux 8 0 dawekhraahcaxyuthicudikldwngxathitythisud r m i n p 1 ϵ displaystyle r mathrm min frac p 1 epsilon emux 8 90 r p aelaemux 8 180 dawekhrahcaxyuthicudikldwngxathitythisud r m a x p 1 ϵ displaystyle r mathrm max frac p 1 epsilon kungaeknexkkhxngwngri a khuxmchchimelkhkhnitkhxng rmin aela rmax a p 1 ϵ 2 displaystyle a frac p 1 epsilon 2 kungaeknothkhxngwngri b khuxmchchimerkhakhnitkhxng rmin aela rmax b p 1 ϵ 2 displaystyle b frac p sqrt 1 epsilon 2 nxkcakniyngepnmchchimerkhakhnitrahwangkungaeknexkkbkungeltserktm a b b p displaystyle frac a b frac b p kdkhxthi 2 aekikh phaphaesdngkdekhphephlxrkhxthi 2 kdkhxthi 2 esntrngthiechuxmrahwangdawekhraahkbdwngxathity kwadphunthietha kninrayaewlaethakn 2 kdniruckinxikchuxhnungthiwakdphunthietha sungepnphlsubenuxngodytrngcakkdkarxnurksomemntmechingmum law of conservation of angular momentum oprddukarkarxnuphththdngphaphkarkhanwnmi 4 khndngni 1 khanwn mumkwadechliy mean anomaly M caksutrM 2 p t P displaystyle M frac 2 pi t P dd 2 khanwn mumkwadeyuxngsunyklang eccentric anomaly E odykaraek smkarkhxngekhphephlxr M E ϵ sin E displaystyle M E epsilon cdot sin E dd 3 khanwn mumkwadcring true anomaly 8 odyichsmkar tan 8 2 1 ϵ 1 ϵ tan E 2 displaystyle tan frac theta 2 sqrt frac 1 epsilon 1 epsilon cdot tan frac E 2 dd 4 khanwn rayahangsunysuriya heliocentric distance r cakkdkhxaerk r p 1 ϵ cos 8 displaystyle r frac p 1 epsilon cdot cos theta dd kdkhxthi 3 aekikh kdkhxthi 3 kalngsxngkhxngkhabkarokhcrkhxngdawekhraahepnsdswnodytrngkbkalngsamkhxngkungaeknexkkhxngwngokhcr dngnn imephiyngkhwamyawwngokhcrcaephimdwyrayathangaelw khwamerwkhxngkarokhcrcaldlngdwy karephimkhxngrayaewlakarokhcrcungepnmakkwakarepnsdswn P 2 a 3 displaystyle P 2 propto a 3 P displaystyle P khabkarokhcrkhxngdawekhraah a displaystyle a aeknkungexkkhxngwngokhcrdngnn P2 a 3 mikhaehmuxnknsahrbdawekhraahthukdwnginrabbsuriyarwmthngolk emuxhnwyhnungthukeluxk echn P thiwdepnpidarakhti sidereal year aela a inhnwydarasastr astronomical unit P2 a 3 mikha 1 sahrbdawekhraahthukdwnginrabbsuriya inhnwyexsix P 2 a 3 3 00 10 19 s 2 m 3 0 7 displaystyle frac P 2 a 3 3 00 times 10 19 frac s 2 m 3 pm 0 7 taaehnnginfngkchnkhxngewla aekikh pyhaekhphephlxrxnumankarokhcrwngriaelacud 4 cud s dwngxathity n ofkshnungkhxngwngri z cudikldwngxathitythisud c sunyklangkhxngwngri p dawekhraahaela a c z displaystyle a cz semimajor axis rayacaksunyklangthungcudikldwngxathitythisud nnkhuxkungaeknexk e c s a displaystyle varepsilon cs over a khwameyuxngsunyklang b a 1 e 2 displaystyle b a sqrt 1 varepsilon 2 kungaeknoth r s p displaystyle r sp rayacakdwngxathitythungdawekhraah n z s p displaystyle nu angle zsp taaehnngdawekhraahtamthiehncakdwngxathity nnkhux mumkwadcringpyhakhuxkarkhanwnphikdechingkhw r n khxngdawekhraahcakewlanbtngaetdawekhraahphancudikldwngxathitythisud t z s x a b z s p displaystyle zsx frac a b cdot zsp z c y z s x displaystyle zcy zsx aela M z c y displaystyle M angle zcy y cakthiehncaksunyklang nnkhuxmumkwadechliy z c y a 2 M 2 displaystyle zcy frac a 2 M 2 z s p b a z s x b a z c y b a a 2 M 2 a b M 2 displaystyle zsp frac b a cdot zsx frac b a cdot zcy frac b a cdot frac a 2 M 2 frac abM 2 M 2 p t T displaystyle M 2 pi t over T ody T khuxkhabkarokhcr z c y z s x z c x s c x displaystyle zcy zsx zcx scx a 2 M 2 a 2 E 2 a e a sin E 2 displaystyle frac a 2 M 2 frac a 2 E 2 frac a varepsilon cdot a sin E 2 Division by a 2 gives Kepler s equation M E e sin E displaystyle M E varepsilon cdot sin E E M e 1 8 e 3 sin M 1 2 e 2 sin 2 M 3 8 e 3 sin 3 M displaystyle E approx M left varepsilon frac 1 8 varepsilon 3 right sin M frac 1 2 varepsilon 2 sin 2M frac 3 8 varepsilon 3 sin 3M cdots a cos E a e r cos n displaystyle a cdot cos E a cdot varepsilon r cdot cos nu r a 1 e 2 1 e cos n displaystyle frac r a frac 1 varepsilon 2 1 varepsilon cdot cos nu to get cos E e 1 e 2 1 e cos n cos n e 1 e cos n 1 e 2 cos n 1 e cos n e cos n 1 e cos n displaystyle cos E varepsilon frac 1 varepsilon 2 1 varepsilon cdot cos nu cdot cos nu frac varepsilon cdot 1 varepsilon cdot cos nu 1 varepsilon 2 cdot cos nu 1 varepsilon cdot cos nu frac varepsilon cos nu 1 varepsilon cdot cos nu tan 2 x 2 1 cos x 1 cos x displaystyle tan 2 frac x 2 frac 1 cos x 1 cos x caid tan 2 E 2 1 cos E 1 cos E 1 e cos n 1 e cos n 1 e cos n 1 e cos n 1 e cos n e cos n 1 e cos n e cos n 1 e 1 e 1 cos n 1 cos n 1 e 1 e tan 2 n 2 displaystyle tan 2 frac E 2 frac 1 cos E 1 cos E frac 1 frac varepsilon cos nu 1 varepsilon cdot cos nu 1 frac varepsilon cos nu 1 varepsilon cdot cos nu frac 1 varepsilon cdot cos nu varepsilon cos nu 1 varepsilon cdot cos nu varepsilon cos nu frac 1 varepsilon 1 varepsilon cdot frac 1 cos nu 1 cos nu frac 1 varepsilon 1 varepsilon cdot tan 2 frac nu 2 khundwy 1 e 1 e aelaisrakthisxng caidphllphth tan n 2 1 e 1 e tan E 2 displaystyle tan frac nu 2 sqrt frac 1 varepsilon 1 varepsilon cdot tan frac E 2 inkhnthisamnieracaidkhwamechuxmoyngknrahwangewlakbtaaehnnginwngokhcrkhnthisikhuxkarkhanwnrayahangsunysuriya r cakmumkwadcring n dwykdkhxaerkkhxngekhphephlxr r a 1 e 2 1 e cos n displaystyle r a cdot frac 1 varepsilon 2 1 varepsilon cdot cos nu karxnuphthth Derivation kdkhxngniwtn aekikhkarxnuphththkhxngkdekhphephlxrkhxthi 2 aekikh m r M m r 2 r G displaystyle m cdot ddot mathbf r frac M cdot m r 2 cdot hat mathbf r cdot G r 8 8 displaystyle dot hat mathbf r dot theta hat boldsymbol theta where 8 displaystyle hat boldsymbol theta is the tangential unit vector and 8 8 r displaystyle dot hat boldsymbol theta dot theta hat mathbf r So the position vector r r r displaystyle mathbf r r hat mathbf r is differentiated twice to give the velocity vector and the acceleration vector r r r r r r r r 8 8 displaystyle dot mathbf r dot r hat mathbf r r dot hat mathbf r dot r hat mathbf r r dot theta hat boldsymbol theta r r r r r r 8 8 r 8 8 r 8 8 r r 8 2 r r 8 2 r 8 8 displaystyle ddot mathbf r ddot r hat mathbf r dot r dot hat mathbf r dot r dot theta hat boldsymbol theta r ddot theta hat boldsymbol theta r dot theta dot hat boldsymbol theta ddot r r dot theta 2 hat mathbf r r ddot theta 2 dot r dot theta hat boldsymbol theta Note that for constant distance r displaystyle r the planet is subject to the centripetal acceleration r 8 2 displaystyle r dot theta 2 and for constant angular speed 8 displaystyle dot theta the planet is subject to the coriolis acceleration 2 r 8 displaystyle 2 dot r dot theta Inserting the acceleration vector into Newton s laws and dividing by m gives the vector equation of motion r r 8 2 r r 8 2 r 8 8 G M r 2 r displaystyle ddot r r dot theta 2 hat mathbf r r ddot theta 2 dot r dot theta hat boldsymbol theta GMr 2 hat mathbf r Equating component we get the two ordinary differential equations of motion one for the radial acceleration and one for the tangential acceleration r r 8 2 G M r 2 displaystyle ddot r r dot theta 2 GMr 2 r 8 2 r 8 0 displaystyle r ddot theta 2 dot r dot theta 0 r 8 displaystyle r dot theta 8 8 2 r r 0 displaystyle frac ddot theta dot theta 2 frac dot r r 0 and integrate log 8 2 log r log ℓ displaystyle log dot theta 2 log r log ell where log ℓ displaystyle log ell is a constant of integration and exponentiate r 2 8 ℓ displaystyle r 2 dot theta ell This says that the specific angular momentum r 2 8 displaystyle r 2 dot theta is a constant of motion even if both the distance r displaystyle r and the angular speed 8 displaystyle dot theta vary The area swept out from time t1 to time t2 t 1 t 2 1 2 b a s e h e i g h t d t t 1 t 2 1 2 r r 8 d t 1 2 ℓ t 2 t 1 displaystyle int t 1 t 2 frac 1 2 cdot base cdot height cdot dt int t 1 t 2 frac 1 2 cdot r cdot r dot theta cdot dt frac 1 2 cdot ell cdot t 2 t 1 depends only on the duration t2 t1 This is Kepler s second law karxnuphththkhxngkdekhphephlxrkhxthi 1 aekikh p ℓ 2 G 1 M 1 displaystyle p ell 2 G 1 M 1 u p r 1 displaystyle u pr 1 and get G M r 2 ℓ 2 p 3 u 2 displaystyle GMr 2 ell 2 p 3 u 2 and 8 ℓ r 2 ℓ p 2 u 2 displaystyle dot theta ell r 2 ell p 2 u 2 X d X d 8 8 d X d 8 ℓ p 2 u 2 displaystyle dot X frac dX d theta cdot dot theta frac dX d theta cdot ell p 2 u 2 Differentiate r p u 1 displaystyle r pu 1 twice r d p u 1 d 8 ℓ p 2 u 2 p u 2 d u d 8 ℓ p 2 u 2 ℓ p 1 d u d 8 displaystyle dot r frac d pu 1 d theta cdot ell p 2 u 2 pu 2 frac du d theta cdot ell p 2 u 2 ell p 1 frac du d theta r d r d 8 ℓ p 2 u 2 d d 8 ℓ p 1 d u d 8 ℓ p 2 u 2 ℓ 2 p 3 u 2 d 2 u d 8 2 displaystyle ddot r frac d dot r d theta cdot ell p 2 u 2 frac d d theta ell p 1 frac du d theta cdot ell p 2 u 2 ell 2 p 3 u 2 frac d 2 u d theta 2 Substitute into the radial equation of motion r r 8 2 G M r 2 displaystyle ddot r r dot theta 2 GMr 2 and get ℓ 2 p 3 u 2 d 2 u d 8 2 p u 1 ℓ p 2 u 2 2 ℓ 2 p 3 u 2 displaystyle ell 2 p 3 u 2 frac d 2 u d theta 2 pu 1 ell p 2 u 2 2 ell 2 p 3 u 2 Divide by ℓ 2 p 3 u 2 displaystyle ell 2 p 3 u 2 d 2 u d 8 2 u 1 displaystyle frac d 2 u d theta 2 u 1 u 1 displaystyle u 1 d 2 u d 8 2 u 0 displaystyle frac d 2 u d theta 2 u 0 These solutions are u ϵ cos 8 A displaystyle u epsilon cdot cos theta A where ϵ displaystyle epsilon and A displaystyle A are arbitrary constants of integration So the result is u 1 ϵ cos 8 A displaystyle u 1 epsilon cdot cos theta A Choosing the axis of the coordinate system such that A 0 displaystyle A 0 and inserting u p r 1 displaystyle u pr 1 gives p r 1 1 ϵ cos 8 displaystyle pr 1 1 epsilon cdot cos theta If ϵ lt 1 displaystyle epsilon lt 1 this is Kepler s first law kdekhphephlxrkhxthi 3 aekikh T 2 4 p 2 G M r 3 displaystyle T 2 frac 4 pi 2 GM cdot r 3 where T planet s sidereal period r radius of the planet s circular orbit G the gravitational constant M mass of the sun T 2 4 p 2 G M m a 3 displaystyle T 2 frac 4 pi 2 G M m cdot a 3 ody T object s sidereal period a object s semimajor axis G the gravitational constant 6 67 10 11 N m kg M mass of one object m mass of the other object 1 2 1 ϵ a V A d t 1 2 1 ϵ a V B d t displaystyle begin matrix frac 1 2 end matrix cdot 1 epsilon a cdot V A dt begin matrix frac 1 2 end matrix cdot 1 epsilon a cdot V B dt 1 ϵ V A 1 ϵ V B displaystyle 1 epsilon cdot V A 1 epsilon cdot V B V A V B 1 ϵ 1 ϵ displaystyle V A V B cdot frac 1 epsilon 1 epsilon m V A 2 2 G m M 1 ϵ a m V B 2 2 G m M 1 ϵ a displaystyle frac mV A 2 2 frac GmM 1 epsilon a frac mV B 2 2 frac GmM 1 epsilon a V A 2 2 V B 2 2 G M 1 ϵ a G M 1 ϵ a displaystyle frac V A 2 2 frac V B 2 2 frac GM 1 epsilon a frac GM 1 epsilon a V A 2 V B 2 2 G M a 1 1 ϵ 1 1 ϵ displaystyle frac V A 2 V B 2 2 frac GM a cdot left frac 1 1 epsilon frac 1 1 epsilon right V B 1 ϵ 1 ϵ 2 V B 2 2 G M a 1 ϵ 1 ϵ 1 ϵ 1 ϵ displaystyle frac left V B cdot frac 1 epsilon 1 epsilon right 2 V B 2 2 frac GM a cdot left frac 1 epsilon 1 epsilon 1 epsilon 1 epsilon right V B 2 1 ϵ 1 ϵ 2 V B 2 2 G M a 2 ϵ 1 ϵ 1 ϵ displaystyle V B 2 cdot left frac 1 epsilon 1 epsilon right 2 V B 2 frac 2GM a cdot left frac 2 epsilon 1 epsilon 1 epsilon right V B 2 1 ϵ 2 1 ϵ 2 1 ϵ 2 4 G M ϵ a 1 ϵ 1 ϵ displaystyle V B 2 cdot left frac 1 epsilon 2 1 epsilon 2 1 epsilon 2 right frac 4GM epsilon a cdot 1 epsilon 1 epsilon V B 2 1 2 ϵ ϵ 2 1 2 ϵ ϵ 2 1 ϵ 2 4 G M ϵ a 1 ϵ 1 ϵ displaystyle V B 2 cdot left frac 1 2 epsilon epsilon 2 1 2 epsilon epsilon 2 1 epsilon 2 right frac 4GM epsilon a cdot 1 epsilon 1 epsilon V B 2 4 ϵ 4 G M ϵ 1 ϵ 2 a 1 ϵ 1 ϵ displaystyle V B 2 cdot 4 epsilon frac 4GM epsilon cdot 1 epsilon 2 a cdot 1 epsilon 1 epsilon V B G M 1 ϵ a 1 ϵ displaystyle V B sqrt frac GM cdot 1 epsilon a cdot 1 epsilon d A d t 1 2 1 ϵ a V B d t d t 1 2 1 ϵ a V B displaystyle frac dA dt frac frac 1 2 cdot 1 epsilon a cdot V B dt dt begin matrix frac 1 2 end matrix cdot 1 epsilon a cdot V B 1 2 1 ϵ a G M 1 ϵ a 1 ϵ 1 2 G M a 1 ϵ 1 ϵ displaystyle begin matrix frac 1 2 end matrix cdot 1 epsilon a cdot sqrt frac GM cdot 1 epsilon a cdot 1 epsilon begin matrix frac 1 2 end matrix cdot sqrt GMa cdot 1 epsilon 1 epsilon dd T d A d t p a 1 ϵ 2 a displaystyle T cdot frac dA dt pi a sqrt 1 epsilon 2 a T 1 2 G M a 1 ϵ 1 ϵ p 1 ϵ 2 a 2 displaystyle T cdot begin matrix frac 1 2 end matrix cdot sqrt GMa cdot 1 epsilon 1 epsilon pi sqrt 1 epsilon 2 a 2 T 2 p 1 ϵ 2 a 2 G M a 1 ϵ 1 ϵ 2 p a 2 G M a 2 p G M a 3 displaystyle T frac 2 pi sqrt 1 epsilon 2 a 2 sqrt GMa cdot 1 epsilon 1 epsilon frac 2 pi a 2 sqrt GMa frac 2 pi sqrt GM sqrt a 3 T 2 4 p 2 G M a 3 displaystyle T 2 frac 4 pi 2 GM a 3 T 2 4 p 2 G M m a 3 displaystyle T 2 frac 4 pi 2 G M m a 3 Q E D xangxing aekikh Hyman Andrew A Simple Cartesian Treatment of Planetary Motion ekbthawr 2011 08 07 thi ewyaebkaemchchin European Journal of Physics Vol 14 pp 145 147 1993 Kepler s Second Law by Jeff Bryant with Oleksandr Pavlyk The Wolfram Demonstrations Project duephim aekikhKepler problem Circular motion Gravity Two body problem Free fall timeaehlngkhxmulxun aekikhCrowell Benjamin Conservation Laws http www lightandmatter com area1book2 html an online book that gives a proof of the first law without the use of calculus see section 5 2 p 112 David McNamara and Gianfranco Vidali Kepler s Second Law JAVA Interactive Tutorial http www phy syr edu courses java mc html kepler html ekbthawr 2006 09 10 thi ewyaebkaemchchin an interactive JAVA applet that aids in the understanding of Kepler s Second Law University of Tennessee s Dept Physics amp Astronomy Astronomy 161 page on Johannes Kepler The Laws of Planetary Motion 1 Equant compared to Kepler interactive model 2 ekbthawr 2008 12 26 thi ewyaebkaemchchin Kepler s Third Law interactive model 3 ekbthawr 2008 12 26 thi ewyaebkaemchchin ekhathungcak https th wikipedia org w index php title kdkarekhluxnthikhxngdawekhraah amp oldid 10255525, wikipedia, วิกิ หนังสือ, หนังสือ, ห้องสมุด,