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IOSR Journal of Applied Physics (IOSR-JAP) e-ISSN: 2278-4861.Volume 6, Issue 5 Ver. II (Sep.-Oct. 2014), PP 01-14 www.iosrjournals.org The General Formula To Find The Straight Line Distance Of Solar Planets From Sun With The Help Of Number Theory And Classical Physics Arjun. K VIth Semester,B.Sc Physics (Student), PTM Government College Perinthalmanna, University of Calicut, Kerala Abstract: Here a particular method is made to generate A Single Formula to find the the straight line distance of first eight planets from sun Keywords: Polynomials, Mean distance, Eccentricities, Keplerโs Law, Orbital Velocity, Escape Velocity, Bodeโs Law I. Introduction In our Solar system there are eight planets in orbit around sun. They can be classified into two different families based on the distance from sun. The outer family, major planets like Jupiter (whose radius is one tenth of that of sun, is the most massive among other planets). The outer family have widely spaced orbits; Consists mainly gaseous material. The four planets in our inner planet category contain the first four planets including our mother earth. All these planets have radius less than one tenth of the Jupiter. In 1772 Johann Bode noticed that the planetary orbit up to Saturn were close to fairly simple mathematical progression. Which became known as Bodeโs Law. The law is in the form ๐๐ = 0.3 + 0.4 ๐ 2๐ . Where the leading term is the radius of Mercuryโs orbit in Astronomical Unit. And for the remaining planets n=0,1,2,3... For Venus, Earth โฆetc. The value we get when n=3 is assign to Asteroids. Here I am trying to generate an equation to find the straight line distance of first eight planets from sun, via combining Classical Physics with Number Theory. The weapon we are taking from number theory is Mainly Polynomial Difference Theorem. And In Physics we are using Orbital Mechanics. Combining both these theories we can generate a single formula as said above. Here I am going to introduce a new method, Using this method we can find the n th term and sum of n terms of any different kinds of polynomial sequences. It is worth noting that using this method we can invent The General Formula to find the mean distance and eccentricity of planets as polynomial functions. Here we are ignoring attraction of each planets one by another. And we are also ignoring the rotation effect of planets around sun. II. The general method to find the nth term and sum of n terms of any polynomial sequences using Polynomial Difference Theorem Polynomial Difference Theorem Suppose the nth term of a sequence is a polynomial in n of degree m i.e.p(n) = a1nm + a2nmโ1 + a3nmโ2 + a4nmโ3 + ...... + am+1 Then its mth difference will be equal and (m + 1)th difference will be zero. 2.1 Case1:- (When m=1) Consider a polynomial sequence of power 1, i.e. p(n)=an+b First term= p(1)=a+b Second term=P(2)=2a+b Third term=p(3)=3a+b First Difference Second Difference First Term (1) a+b a 0 Second Term P(2) 2a+b a Third Term P(3) 3a+b 2.2 Case2:- (When m=2) Consider a polynomial sequence of power 2, i.e. p(n) = an2 + bn + c First term= p(1)=a+b+c Second term=P(2)=4a+2b+c 1 | Page The General formula to find the straight line distance of solar planets from sun with the help of Third term=p(3)=9a+3b+c Fourth term=p(4)=16a+4b+c First Difference Second Difference Third Difference First Term (1) Second Term P(2) Third Term P(3) a+b+c 3a+b 2a 0 4a+2b+c 5a+b 2a 9a+3b+c 7a+b Fourth Term P(4) 16a + 4b + c 2.3 Case3:- (When m=3) Consider a polynomial sequence of power 3, i.e. p(n) = an3 + bn2 + cn + d First term= p(1)=a+b+c+d Second term=P(2)=8a+4b+2c+d Third term=p(3)=27a+9b+3c+d Fourth term=p(4)=64a+16b+4c+d Fifth term=p(5)=125a+25b+5c+d First Term P(1) a+b+c+d 7a+3b+c 12a+2b 6a 0 D1 D2 D3 D4 Second Term P(2) 8a+4b+2c+d 19a+5b+c 18a+2b 6a Third Term P(3) 27a+9b+3c+d 37a+7b+c 24a+2b Fourth Term P(4) 64a+16b+4c+d 61a+9b+c Fifth Term P(6) 125a+25b+5c+d From the previous experience, the basic knowledge of the polynomial difference theorem can be put in another way. III. Inverse Polynomial Differnce theorem ! If the mth difference of a polynomial sequence is equal then its nth term will be a polynomial in n of degree m. First Difference is the difference between two consecutive terms. Second difference is the difference between two neighboring first differences. So on 3.1 Problem 1: Find the nth term of the triangular number sequence1,3,6,10,15,21,28,36,45,55... Solution First term=t1 = 1, t2 = 3, t3 = 6 , t4 = 10 , t5 = 15 Using polynomial difference theorem, we can find the nth term, D1 D2 D3 FirstTermt1P(1) 1 2 1 0 SecondTermt2P(2) 3 3 1 0 ThirdTermt3P(3) 6 4 1 FourthTermt4P(4) 10 5 FifthTermt5 15 Here Second difference is Equal, which means we can represent its nth term as a polynomial of degree 2, i.e. the general representation tn = p(n) =an2 + bn + c t1 = p(1) = a + b + c = 1 t2 = p(2) = 4a + 2b + c = 3 t3 = p(3) = 9a + 3b + c = 6 Solving this we get value of a=12; b= 12; c=0; ๐(๐ +1) Nth term=p(n) = 2 www.iosrjournals.org 2 | Page The General formula to find the straight line distance of solar planets from sun with the help of 3.2 Problem 2: the nth term of the number sequence solution Taking tn = ๐๐ t1 1 16 65 110 84 24 D1 D2 D3 D4 D5 t2 17 81 175 194 108 24 t3 98 256 369 302 132 ๐๐ t4 354 625 671 434 t5 979 1296 1105 t6 2275 2401 t7 4676 Let p(n) be the nth term , Since 5th difference is equal, the degree of nth term is 5 i.e.tn = p(n) = an5 + bn4 + cn3 + dn2 + en + f t1=p(1) = a+b+c+d+e+f =1 t2=p(2) = 32a+16b+8c+4d+2e+f =17 t3=p(3) = 243a+81b+27c+9d+3e+f =98 t4=p(4) = 1024a+256b+64c+16d+4e+f=354 t5=p(5) = 3125a+625b+125c+25d+5e+f=979 t6=p(6)=7776a+1296b+216c+36d+6e+f=2275 p(2)-p(1)=3a+b=2โ-(1) p(3)-p(2)=5a+b=3โ-(2) (2)-(1)=2a=1 a=12 Solving this we get ๐ ๐ = ๐ ๐=1 ๐ ๐+1 2๐ +1 (3๐ 2 +3๐โ1) 30 ๐ ๐ + 1 2๐ + 1 (3๐2 + 3๐ โ 1) ๐4 = 30 Using this we can find ๏ท Equation of mean distance From Sun(In 107 m) b n = ๏ท โ144464081 7 894094629 6 56448957031 5 11973677 4 5563041 3 ๐ + ๐ โ ๐ + ๐ โ ๐ 1000000 200000 1000000 32 4 5955539 4752953 + 2864028 ๐2 โ ๐+ 2 4 Equation of Orbital Period (In Earth days) T n = ๏ท โ6194601 7 376894501 6 467468457 5 3810907959 4 27956908203 3 ๐ + ๐ โ ๐ + ๐ โ ๐ 500000 1000000 100000 125000 250000 14593949 2 15086455 5979771 + ๐ โ ๐+ 64 64 64 Equation of orbital Eccentricity e n = ๏ท ๏ท 151 87493 6 1127483 5 1515287 4 28233521 3 3565267 2 ๐7 โ ๐ + ๐ โ ๐ + ๐ โ ๐ 393750 7200000 7200000 1440000 7200000 450000 10873869 1061 + ๐โ 1400000 400 Equation of Mean Orbital Velocity of planets in (km/s) 5969 7 53069 6 133961 5 879793 4 16029061 3 15941023 2 450101 ๐ โ ๐ + ๐ โ ๐ + ๐ โ ๐ + ๐ 252000 72000 14400 14400 72000 36000 1050 10809 โ 100 Equation of Gravity at the Equator( In g) www.iosrjournals.org 3 | Page The General formula to find the straight line distance of solar planets from sun with the help of โ 997 7 32633 6 90763 5 605341 4 11306057 3 11686051 2 710761 ๐ + ๐ โ ๐ + ๐ โ ๐ + ๐ โ ๐ 63000 72000 14400 14400 72000 36000 2100 3376 + 25 ๏ท Equation of Escape Velocity(In km/s) 10391 7 10883 6 2087581 5 1862227 4 104555069 3 14428591 2 37666903 โ ๐ + ๐ โ ๐ + ๐ โ ๐ + ๐ โ ๐ 36000 1200 18000 2400 36000 2400 6000 250573 + 100 Comparison of Bodoโs Equation V/s Our Equation In 1772 Johann Bode were close to a fairy simple mathematical progression, which became known as Bodeโs law. This law is illustrated in given below table Rn=0.4+0.3x2n Where the leading term is the distance from sun to mercury Planets Actual Distance (In AU) Bodes Distance Error in Bode(%) By above formula calculation Error % Mercury 0.4 0.4 0 0.4 (n=1) 0 Venus 0.7 0.7 (n=0) 0 0.7 (n=2) 0 Earth 1 1 (n=1) 0 1 (n=3) 0 Mars 1.5 1.6 (n=2) 6.67 1.5 (n=4) 0 Jupiter 5.2 5.2 (n=4) 0 5.2 (n=5) 0 Saturn 9.5 10 (n=5) 5.27 9.5 (n=6) 0 Uranus 19.6 19.6 (n=6) 0 19.6 (n=7) 0 Neptune 30 38.8(n=7) 29.3 30 (n=8) 0 IV. Equations of Motion in an Inertial frame (Two bodies problem) โ๐บ๐ 1 ๐ 2 Consider two point masses in free space, the mutual attraction between them will be shows attraction) (Universal law of gravitation) m ๐ +m ๐ The position of center of mass relative to an inertial frame ABC is ๐ ๐ = 1m ๐ +m 2 ๐ Each of the body is pulled by the other 1 ๐2 (-ve sign 2 ๐ญ๐ ๐๐ ๐ = Force exerted on 1 by 2. ๐ญ๐ ๐๐ ๐ = Force exerted on 2 by 1 ah www.iosrjournals.org 4 | Page The General formula to find the straight line distance of solar planets from sun with the help of Absolute velocity of G (Centre of mass ) is ๐ฝ๐บ = ๐๐ฎ = Absolute Acceleration of G is d๐๐ dt d 2๐ ๐ dt 2 = = m1 d ๐๐ d๐ +m 2 ๐ dt dt m 1 +m 2 d 2 ๐๐ d 2๐ m 1 2 +m 2 2๐ dt dt m 1 +m 2 ๐ซ Now ๐ be a vector pointing from m1 towards m2. I.e. ๐ = r Where r is the position vector of m2 relative to m1 . r = R2-R1. Thus Gm 1 m 2 ๐ญ๐ ๐๐ ๐ = opposite) . ๐ r2 Similarly ๐ญ๐ ๐๐ ๐ = โ Gm 1 m 2 r2 ๐ According to Newtonโs second law ๐ญ๐ ๐๐ ๐ = m1 d 2 ๐๐ d t2 d 2 ๐๐ = Gm 1 m 2 Gm 2 r2 ๐ ๐ญ๐ ๐๐ ๐ = m2 and d 2 ๐๐ Therefore dt 2 = r 2 ๐ and dt 2 = โ Thus G moves with a constant velocity. Gm 1 r2 ๐ (-ve sign indicates that the direction is d 2 ๐๐ dt 2 =โ Gm 1 m 2 r2 ๐ Substituting this on ๐๐ฎ we get ๐๐ฎ = 0. 4.1 Relative motion Equations We have d 2 ๐๐ dt 2 d 2 ๐๐ dt 2 = Gm 2 ๐ r2 Gm 1 =โ r2 ๐ d 2๐ซ d2 Therefore acceleration of m2 w.r.t m1 is 2 = 2 ๐ ๐ โ ๐ ๐ = โ dt dt Gravitational parameter ๐ = G(m1 + m2 ) d 2๐ซ G(m 1 +m 2 ) r2 ๐ ๐ Hence dt 2 = โ r 2 ๐ This equation will give information about the motion of m2 relative to m1. 4.2 Its time for some Orbit Formulas The Angular momentum of m2 relative to m1. ๐๐ ๐ณ๐๐ = ๐ ๐ ๐2 ๐๐ก ๐๐ is the velocity of m2 with respect to m1. ๐๐ก Now ๐ = ๐ณ๐๐ ๐2 =๐๐ ๐๐ ๐๐ก ๐๐ ๐๐ ๐๐ ๐2 ๐ ๐ ๐ = ๐ +๐๐ 2 = 0+๐๐ โ 2 ๐= ๐+๐๐ โ 3๐ซ = 0 ๐๐ก ๐๐ก ๐๐ก ๐๐ก r r ๐๐ ๐๐ก = 0 ๏จ l= a constant . ๐๐ ๐๐ ๐ ๐ ๏จ is a constant. Which means that at any given time r ( position vector) & (velocity ๐๐ก ๐๐ก vector) is in the same plane. The Cross product ๐ ๐ ๐ฅ ๐๐ ๐๐ก is perpendicular to that plane. ๐ = l๏จ Unit vector normal to the plane. www.iosrjournals.org 5 | Page The General formula to find the straight line distance of solar planets from sun with the help of Since it is a constant unit vector , m2 around m1 is in a single plane. ๐๐ Now let us resolve ๐๐ก into two components ๐๐ = V2 ๐ฎ2 + V1 ๐ฎ1 ๐๐ก ๐ฝ๐ is along the outward radial from m1 and ๐1 Is perpendicular to it. ๐๐ Then ๐ = ๐ ๐ ๐๐ก = ๐๐ฎ2 ๐ V2 ๐ฎ2 + V1 ๐ฎ1 ๐๐ ๐r Identity proof ๐ . ๐๐ก = r. ๐๐ก I.e. l = rV1 . We have ๐ซ. ๐ซ = r 2 ๐ d๐ซ ๐ซ. ๐ซ = 2๐ซ ๐๐ก dt ๐ dr r. r = 2r ๐๐ก dt 1 1 ๐๐ ๐ซ x ๐ฅ = 3 (๐ซ x ๐ x ) 3 r r ๐๐ก But A x (B xC) = B(A.C)-C(A.B) 1 d๐ซ d๐ซ ๐ซ ๐ซ. โ ๐ซ. ๐ซ 3 r dt dt 1 dr d๐ซ 2 = 3 ๐ซ r โ r r dt dt 1 dr d๐ซ = 2 ๐ซ โr r dt dt www.iosrjournals.org 6 | Page The General formula to find the straight line distance of solar planets from sun with the help of Let r covers an area โ๐ด in a time interval โ๐ก . 1 Therefore โA = bh 2 โ๐ด 1 ๐ = ๐๐1 = โ๐ก 2 2 l2 4.3 Proof for r = ฮผ(1+eCos ฮธ ) ๐๐ฅ ๐๐ ๐๐ฅ ๐. x๐ฅ = ๐๐ก ๐๐ก ๐๐ฅ ๐2 ๐ d ๐๐ x๐ฅ= x๐ฅ ๐๐ก 2 dt ๐๐ก But ๐2 ๐ ๐ x ๐ฅ = โ ๐ซx๐ฅ ๐๐ก 2 r3 ๐ ๐ซ ๐๐ก ๐ = r ๐๐ ๐r โ๐ ๐๐ก ๐๐ก ๐2 ๐ ๐ซ ๐ 1 = โ r3 ๐ซ x ๐ฅ l ๐๐ ๐๐ก ๐๐ Then๐1 = r ๐๐ก ๐๐ But ๐ = ๐๐1 = ๐ 2 ๐๐ก ๐ Thus ๐1 = ๐ ๐2 = I.e. ๐๐ ๐ซ x ๐ฅ โ ๐ = ๐ด ๐๐๐๐ ๐ก๐๐๐ก = ๐ช ๐๐ก ๐ Taking dot products on both sides by l. ๐๐ ๐ซ ๐. x๐ฅ โ ๐ = ๐ช. ๐ ๐๐ก ๐ Thus ๐ช. ๐ = ๐ I.e. C lies in the orbital plane. ๐๐ ๐ซ x๐ฅโ๐ =๐ช ๐๐ก ๐ Therefore ๐๐ ฮผ Proof For ๐2 = eSin(ฮธ) Angular velocity be d ๐๐ ๐ซ x๐ฅโ๐ =0 dt ๐๐ก ๐ ๐ซ ๐๐ง ๐๐ง =0 ๐๐ก ๐๐ง Since l is perpendicular to r l.r=0 From the textbox it is clear that ๐๐ก Thus ๐2 ๐ ๐ ๐ ๐ซ x๐ฅ=โ 3๐ซx๐ฅ=๐ 2 ๐๐ก r ๐๐ก ๐ d ๐๐ ๐ ๐ซ x๐ฅ = ๐ dt ๐๐ก ๐๐ก ๐ ๐๐ ๐๐ฆ ๐๐ฆ ๐๐ก ๐๐ฆ x๐ฅ ๐ซ ๐ l2 ฮผ l2 = ฮผ = ๐r ๐ l2 = ๐๐ก ๐๐ก ฮผ 1 + eCos ฮธ โeSinฮธ dฮธ 2 (1 + eCos ฮธ) dt โeSinฮธ l 2 (1 + eCos ฮธ) r 2 But ๐ = ฮผ x๐ฅ = ๐ช + ๐ ๐ ๏จ ๐๐ก๐ = ๐ + ๐ ๐๐ก ๐ ๐ = = ๐ท๐๐๐๐๐ ๐๐๐๐๐๐ ๐ ๐๐๐๐๐๐ก๐๐๐๐๐ก๐ฆ ๐ฃ๐๐๐ก๐๐ ๐ Now equation becomes ๐๐ x๐ฅ ๐ซ ๐๐ก = +๐ ๐ ๐ Taking dot product with r we get ๐๐ x๐ฅ ๐ซ ๐ซ. ๐๐ก = ๐ซ. + ๐ ๐ ๐ l2 1+eCos ฮธ Substituting this on above eqn we get ฮผ ๐2 = โ eSinฮธ l Taking the magnitude only ฮผ ๐2 = eSinฮธ l www.iosrjournals.org 7 | Page The General formula to find the straight line distance of solar planets from sun with the help of ๐๐ ๐ซ x ๐๐ก ๐ฅ. ๐ = r + ๐ซ. ๐ r + reCos ฮธ = l2 ฮผ or l2 ฮผ(1 + eCos ฮธ ) 4.4 Apse line and Periapsis r= We have ๐ = l2 ฮผ 1+eCos ฮธ ฮผ ๐ l ๐ ; ๐2 = eSinฮธ ; ๐1 = 1 + eCos ฮธ The point of closest approach lies on apse line and is called Periapsis. I.e. When ฮธ = 0 ๏จ ๐ = ฮผ ๐ l2 m2 comes close to m1 now. 1+e eSin ฮธ The flight path angle is ๐๐๐๐พ = ๐2 = 1+eCos ฮธ๏จThe orbit equation is symmetric about its axis. 1 4.5 Latus Rectum The latus rectum is the chord perpendicular to the apse line and passing through the center of attraction. By symmetry, the center of attraction divides the latus rectum into two equal parts, each of length p, Where ๐ = We have ๐= l2 ฮผ l2 ฮผ 1 + eCos ฮธ Elliptical Orbits( 0<e<1) ๐๐ = ฮผ ๐๐ = l2 1โe l2 ฮผ 1+e ๏จApoapsis ๏จPeriapsis From figure it is clear that 2๐ = ๐๐ + ๐๐ ๐=ฮผ l2 1โe 2 . ๐๐ = a(1 + e) ๐๐ = a(1 โ e) 4.6 Time Period According to Keplerโs law ๐๐ด = ๐ด ๐๐๐๐ ๐ก๐๐๐ก ๐๐ก In case of a complete rotation ๐๐ด = ๐๐๐; ๐๐ก = ๐ ๐๐๐ ๐ Therefore ๐ = 2 ๐= 2๐๐๐ 2๐๐ 2๐๐2 2๐ l2 = ๐ 1 โ ๐2 = 1 โ ๐2 = 1 โ ๐2 ๐ ๐ ๐ ๐ ฮผ 1 โ e2 www.iosrjournals.org 2 8 | Page The General formula to find the straight line distance of solar planets from sun with the help of 2๐ Solving we get ๐ = ๐ 2 ( On further solving = 2๐ ๐ ๐ 1โ๐ 2 )3 3 (๐)2 ; ๐ 2 = 4๐ 2 ๐ ๐3 4.7 Finding Mean distance Now we have ๐(ฮธ) = ฮผ l2 1+eCos ฮธ 2๐ 1 2๐ ๐ 1โe 2 Therefore ๐๐๐ฃ = 2๐ 0 ๐ ฮธ ๐ฮธ = 2๐ 0 1+eCos ฮธ ๐ 1โe 2 2๐ 1 ๐๐๐ฃ = 2๐ ๐ฮธ 0 1+eCos ฮธ 1 ๐ฮธ ฮธ ๐๐ข๐ก ๐ฅ = tan 2 ฮธ = 2tanโ1 x ๐= ๐ 2 2 = 1โx Cosฮธ = ๐ 1 + x2 1 + tan2 2 Since the function is an even function we can write it as following 1 โ tan2 ๐๐๐ฃ = 2 ๐๐๐ฃ ๐๐๐ฃ ๐๐๐ฃ ๐ 1 โ e2 =4 2๐ ๐ 1 โ e2 =4 2๐ ๐ 1 โ e2 =4 2๐ 1 โ e ๐๐๐ฃ = ๐๐๐ฃ ๐ ๐ 1 โ e2 2๐ 2๐ 1 + e ๐ 0 ๐ 1 1+e 1 โ x2 1 + x2 1 2 ๐x 1 + x2 1 + x2 + e 1 โ x2 1 ๐x 2 0 x 1โe +1+e ๐ 1 ๐x 1+e 0 x2 + 1โe ๐ 1 2 ๐x 0 1 + e x2 + 1โe 2๐ 1 + e = ๐ ๐2 l2 ฮผ 1 โ e2 l2 = ๐ 1 โ e2 ฮผ 1 1 ๐ฅ ๐๐ฅ = tanโ1 2 +๐ฅ ๐ ๐ ๐x 0 ๐ 1โe ๐ฅ 1โ๐ tanโ1 1+e 1+๐ โ 0 1 โ e2 ฯ ๐๐๐ฃ = ๐ 2 ๐๐๐ฃ = ๐ 1 โ e2 I.e. ๐๐๐ฃ = ๐ 1 โ e2 = ๐๐๐๐ ๐๐๐๐๐ ๐๐ฅ๐๐ = ๐ Or . ๐๐๐ฃ = ๐ 1 โ e2 = ra rp Where ๐๐ = a(1 + e) ; ๐๐ = a(1 โ e) 2๐ We know that ๐ = Therefore But ๐ = ๐๐ ๐๐ก ๐ ๐๐ ๐ 2 ๐๐ก V. Orbiting Path As A Time Function = ๐2 l2 ฮผ 1+eCos ฮธ www.iosrjournals.org 9 | Page The General formula to find the straight line distance of solar planets from sun with the help of ๐๐ ฮผ2 2 = 3 1 + eCos ฮธ ๐๐ก l t 2 ฮธ ฮผ 1 dt = 2 dฮธ 3 tp l 0 1 + eCos ฮธ Where t p time at periapsis passing. 5.1 Circular Orbit For circular orbits e=0. ฮผ2 t โ tp = ฮธ l3 If we take t p = 0 ฮผ2 t=ฮธ l3 On solving 3 t= r2 ฮธ ฮผ For a full revolution t = T, ฮธ = 2ฯ 3 T= 2ฯr 2 ฮผ 2ฯ Therefore ฮธ = T t 5.2 Elliptical Orbit (0<e<1) ฮธ 0 1 1 + eCos ฮธ ฮผ2 t= l3 1 1โ Rewrite it as ฮผ2 l3 t= 3 e2 2 1 3 1โe 2 2 Now let ME = 1 2 dฮธ = 1 โ e2 2ฯ T t= ฮผ2 l3 ฮผ2 l3 1โe ฮธ e 1 โ e2 Sinฮธ tan ] โ } 1+e 2 1 + eCosฮธ 1โe ฮธ e 1 โ e2 Sinฮธ tan ] โ } 1+e 2 1 + eCosฮธ { 2 tanโ1 [ { ME } or 3 2 { 2 tanโ1 [ 3 1 โ e2 2 t = ME 1 โ e2 3 2 t For the first Eight planets in our solar system e has a maximum value 0.20 in case of mercury. And e=0.0068 in case of venus (minimum) If we ignore mercury then e ranges between 0.0068 and 0.05 . Plotting X axis =ฦ and Y axis as corresponding ME. [Straightline is : y=x=theta] Mercury e=0.2056 Venus , e= 0.0068 www.iosrjournals.org 10 | Page The General formula to find the straight line distance of solar planets from sun with the help of Earth e=0.0167 Mars e=0.0934 Jupiter Saturn e=0.0560 e=0.0483 Uranus, e=0.0461 Neptune e=0.0097 AT A GLANCE www.iosrjournals.org 11 | Page The General formula to find the straight line distance of solar planets from sun with the help of 2ฯ I.e. we can approximate ME = T t N.B. We take t=0, at perihelion and starts counting from there. We know that ๐ = ฮผ l2 1+eCos ฮธ But l2 ฮผ = ๐ 1 โ e2 ๐ 1 โ e2 ๐= 1 + eCos ฮธ But b = ๐ 1 โ e2 Therefore b 1 โ e2 ๐= 2ฯ 1 + eCos T t VI. r(n) = Combining Classical Physics to Polynomial difference theorem b(n) 1 โ e(n) 2 2ฯ 1 + e(n) Cos T(n) t e(n)=Eccentricity of planet T(n) =Time period of planet Now let us recall our findings Equation of mean distance From Sun(In 107 m) ๏ท b n = โ144464081 7 894094629 6 56448957031 5 11973677 4 5563041 3 ๐ + ๐ โ ๐ + ๐ โ ๐ 1000000 200000 1000000 32 4 5955539 4752953 + 2864028 ๐2 โ ๐+ 2 4 ( n=1:Mercury, n=2:Venus, n=3:Earthโฆ..etc) Equation of Orbital Period (In Earth days) ๏ท T n = ๏ท โ6194601 7 376894501 6 467468457 5 3810907959 4 27956908203 3 ๐ + ๐ โ ๐ + ๐ โ ๐ 500000 1000000 100000 125000 250000 14593949 2 15086455 5979771 + ๐ โ ๐+ 64 64 64 Equation of orbital Eccentricity e n = 151 87493 6 1127483 5 1515287 4 28233521 3 3565267 2 ๐7 โ ๐ + ๐ โ ๐ + ๐ โ ๐ 393750 7200000 7200000 1440000 7200000 450000 10873869 1061 + ๐โ 1400000 400 I(n) = 50349 7 479215 6 865181 5 4268557 4 91522621 3 ๐ โ ๐ + ๐ โ ๐ + ๐ 3000000 900000 125000 90000 500000 195666519 2 417357269 40261711 โ ๐ + ๐โ 500000 1000000 250000 Gives the angle of inclination of each planet. ( n=1:Mercury, n=2:Venus, n=3:Earthโฆ..etc) N.B www.iosrjournals.org 12 | Page The General formula to find the straight line distance of solar planets from sun with the help of We have invented equation in such a way that when t=0 planet will be at perihelion and we will start counting from there onwards. VII. Letโs Make a reference point Let us try to make an equation of t based on some date as reference point. Here I am trying to make 1/12/1966 As a reference point. From planet Calendar which shows distance between each planet from Earth, After some mathematical treatments (Using simple Trigonometry) We can find the position/distance of planet from sun. In order to make a reference point function, we have applied ๐ก= T n b n ๐ถ๐๐ โ1 ( 2๐ 1โe n e n d 2 โd ) For each n=1,2,.. we will take corresponding planetโs distance d based on 1/12/1966 planetary calendar. Using that and Inverse polynomial Difference Theorem we can find a polynomial equation for this t for the specific date as โ32230339 7 469816681 6 5581737793 5 34875074219 4 ๐ก ๐ = ๐ + ๐ โ ๐ + ๐ 1000000 500000 500000 500000 15736047 3 15504011 2 15603855 3032237 โ ๐ + ๐ โ ๐+ 64 32 32 16 Now let us re write our equation as b(n) 1 โ e(n) 2 r(n) = 2ฯ 1 + e(n) Cos [t n + โt) T n โt =No of days from 1/12/1966 ( n=1:Mercury, n=2:Venus, n=3:Earthโฆ..etc) Let us take an Example, Suppose we want to find the distance between Earth and sun on 1 may 2014 Then โt = 01 โ 05 โ 2014 โ 01 โ 12 โ 1966 โ 17318.22 Days The distance between Earth and sun on May 1 =r 3 = 1โe 3 2 b 3 1+e 3 Cos 2ฯ T 3 = 1.016 AU t 3 +โt Let us take a look on other planets and to the value we got from planetary calendar 1 May 2014 n 1 2 3 4 5 6 7 8 Name of planet MERCURY VENUS EARTH MARS JUPITOR SATURN URANUS NEPTUNE Mean distance b n x 107 m Eccentricit ye n Time Period T n t(n) 5790.958283 10820.00392 14960.32528 22794.74907 77834.24688 142941.9512 287101.9634 450433.9838 0.2056 0.0068 0.0167 0.0934 0.0483 0.056 0.0461 0.0097 87.967214 224.695849 365.249328 686.925287 4332.53725 10759.08091 30684.1288 60188.37532 13.029 51.583032 54.09746 306.81414 1211.88 2753.92808 30739.04497 27365.2161 Value From Formula (In AU) 0.31 0.73 1.02 1.60 5.24 9.20 20.02 30.10 From Planetary calendar 0.30 0.73 1.01 1.61 5.23 9.92 20.04 30.03 Conclusion We have shown that, through this special type method we can find the โThe General formula to find the straight line distance of solar planets from sun at any instantโ. The General formula straight line distance of nth solar planets from sun At any instant www.iosrjournals.org 13 | Page The General formula to find the straight line distance of solar planets from sun with the help of r(n) = b(n) 1 โ e(n) 2 2ฯ 1 + e(n) Cos T(n) t We have invented equation in such a way that when t=0 planet will be at perihelion and we will start counting from there onwards. Where r(n)denotes the position of nth planet planet at t second. b n denotes the mean distance of nth planet from sun. e n denotes the orbital eccentricity of nth planet T n denotes equation of orbital period of nth planet. 50349 479215 865181 I(n)=3000000 ๐7 โ 900000 ๐6 + 125000 ๐5 โ 40261711 4268557 90000 ๐4 + 91522621 500000 ๐3 โ 195666519 500000 ๐2 + 417357269 1000000 ๐โ 250000 Gives the angle of inclination of each planet. ( n=1:Mercury, n=2:Venus, n=3:Earthโฆ..etc) But we can re-write the equation by making 1/12/1966 as a reference point as shown in the part VII Equation I.e. b(n) 1 โ e(n) 2 r(n) = 2ฯ 1 + e(n) Cos [t n + โt) T n โt =No of days from 1/12/1966 t n = Reference time function Acknowledgements I express my sincere thanks to K.Mohamed Haneefa, Professor of Physics, Malappuram Government College,Malappuram Kerala Rajesh.P.M , Asst.Professor of Physics, Government Engineering College, Thrissur Reeja Gopalakrishnan Nair, Asst.Professor of Physics, PTM Government College Perinthalmanna Sanal Raj, Asst.Professor of Physics , PTM Government College Perinthalmanna Sunil Kumar.P , HOD of Physics, PTM Government College Perinthalmanna For their valuable suggestions for the completion of this project and I especially thanks to my parents and classmates for all the support. References [1]. [2]. [3]. [4]. Planetary Science: The Science of planets around stars (2002) By George H A Cole, Michael M Woolfson Astrophysics of Solar System By KD Abhyankar (1999) Orbital Mechanics for Engineering Students (2005) By Howard Curtis The General formula to find the sum of first n Kth Dimensional S sided polygonal numbers and a simple way to find the n-th term of polynomial sequences By Arjun.K IOSR Journal of Mathematics (IOSR-JM) Volume 8 - Issue 3 10.9790/5728-0830110 [5]. [6]. [7]. http://www.nightskyatlas.com/planetscal.jsp http://www.windows2universe.org Classical Mechanics by Goldstein www.iosrjournals.org 14 | Page