hello-every-one-why-do-planets-of-the-solar-system-revolve-around-the-sun-in-an-eliptical-not-circular-orbit-

Question Number 97271 by M±th+et+s last updated on 07/Jun/20

h e l l o e v e r y o n e

w h y d o p l a n e t s o f t h e s o l a r s y s t e m

r e v o l v e a r o u n d t h e s u n i n a n e l i p t i c a l

n o t c i r c u l a r o r b i t

Commented by mr W last updated on 07/Jun/20

if we had a two−body system consisting of the sun and a planet and the only force acting between them is the gravitational force and both objects are seen as point masses, it can be proved that the orbit of the planet around the sun is an ellipse. the shape of the ellipse depents on the initial condition. in special condition, the orbit can be a circle which is also a special form of ellipse. but the real solar system is much more complicated than a two−body system, there are several planets which affect each other and both the sun and the planets are not point masses, therefore the real orbits of the planets are not perfect ellipses, only near ellipses. a circular orbit is too perfect, which can not be kept stable in nature, because any disturbance may let it leave this perfect form. nevertheless the orbit of Venus is almost a circle.

i f w e h a d a t w o - b o d y s y s t e m

c o n s i s t i n g o f t h e s u n a n d a p l a n e t

a n d t h e o n l y f o r c e a c t i n g b e t w e e n

t h e m i s t h e g r a v i t a t i o n a l f o r c e a n d

b o t h o b j e c t s a r e s e e n a s p o i n t m a s s e s,

i t c a n b e p r o v e d t h a t t h e o r b i t o f t h e

p l a n e t a r o u n d t h e s u n i s a n e l l i p s e .

t h e s h a p e o f t h e e l l i p s e d e p e n t s o n

t h e i n i t i a l c o n d i t i o n . i n s p e c i a l

c o n d i t i o n, t h e o r b i t c a n b e a c i r c l e

w h i c h i s a l s o a s p e c i a l f o r m o f e l l i p s e .

b u t t h e r e a l s o l a r s y s t e m i s m u c h m o r e

c o m p l i c a t e d t h a n a t w o - b o d y s y s t e m,

t h e r e a r e s e v e r a l p l a n e t s w h i c h

a f f e c t e a c h o t h e r a n d b o t h t h e s u n

a n d t h e p l a n e t s a r e n o t p o i n t m a s s e s,

t h e r e f o r e t h e r e a l o r b i t s o f t h e p l a n e t s

a r e n o t p e r f e c t e l l i p s e s, o n l y n e a r

e l l i p s e s . a c i r c u l a r o r b i t i s t o o p e r f e c t,

w h i c h c a n n o t b e k e p t s t a b l e i n n a t u r e,

b e c a u s e a n y d i s t u r b a n c e m a y l e t i t

l e a v e t h i s p e r f e c t f o r m . n e v e r t h e l e s s

t h e o r b i t o f V e n u s i s a l m o s t a c i r c l e .

Commented by M±th+et+s last updated on 08/Jun/20

t h a n k y o u s i r

Answered by Sourav mridha last updated on 07/Jun/20

okk let′s see the trajectory of body moving in central force field...but Here i am not using GTR..only the point of view of old classical concept of Newtonian Lagrangian Mechanics. let′s pick up the example of our solar system....where sun is at the focus and earth moves round sun in account of Gravitational force feild. so our Lagrangian constracted in plane polar coordinate like that L=(1/2)[(r^• )^2 +r^2 (𝛉^• )^2 ]−v(r) now using Euler−Lagrange eq^n (from Variational calculas) (∂L/∂r)−(d/dt)((∂L/∂r^• ))=0 and (∂L/∂𝛉)−(d/dt)((∂L/∂𝛉^• ))=0 from 1st eq^n we get r^(••) −r(𝛉^• )^2 =−(∂/∂r)[v(r)]=F(r)=(k/r^2 ) .................(i) F(r)=gravitetional force ,k=constant k=Gm_s m_e . and 2nd eq^n gives us r^2 𝛉^• =constant,multi:𝛍 both sides we get L=𝛍r^2 𝛉=constant ...........(ii) where 𝛍=((m_s m_e )/(m_s +m_e ))=reduce mass for central force− torque 𝛕 =r^→ ×f(r)=0 or (dL/dt)=constant where L=angular momentam. now let u=(1/r) now diff wrt to 𝛉 and using eq^n (ii) we get (du/d𝛉)=((−1)/r^2 ).((dr/dt)/(d𝛉/dt))=−(1/r^2 )(r^• /𝛉^• )=−(𝛍/L)r^• now (d^2 u/d𝛉^2 )=−(𝛍^2 /(L^2 u^2 )).r^(••) . so we get r^(••) =−((L^2 u^2 )/𝛍^2 ).(d^2 u/d𝛉^2 ) and r(𝛉^• )^2 =((u^3 L^2 )/𝛍^2 )... now putting this two results at eq^n (i) we get eq^(n ) of motion of earth around sun is (d^2 u/d𝛉^2 )+u=−((𝛍^2 k)/L^2 )(constant) by solving this easy 2nd ODE we get u(𝛉)=−((𝛍^2 k)/L^2 )+Acos𝛉+Bsin𝛉 or =−((𝛍^2 k)/L^2 )+𝛄cos𝚽 we considerate at first u=(1/r) so now r(𝚽)=(1/(−((𝛍^2 k)/L^2 )+γcos(𝚽))) =((−(L^2 /(𝛍^2 k)))/(1−((L^2 𝛄)/(μ^2 k))cos(𝚽))) compare this result with eq^n of a ellipse in plane polar coordinate r(𝛉)=(l/(1+_− ecos(𝛉))) in our case−− sami letus rectum of the ellipticak path of our planet is l=a(1−e^2 )=−(L^2 /(𝛍^2 k)) and escentricity e=((L^2 𝛄)/(𝛍^2 k)) this is why planets arw moving in a elliptical path... 1st proposed by Kepler in his 1st las of planetory motion..

\boldsymbol o k k \boldsymbol {l e t}^{'} \boldsymbol s \boldsymbol s e e \boldsymbol t h e \boldsymbol t r a j e c t o r y \boldsymbol o f

\boldsymbol b o d y \boldsymbol m o v i n g \boldsymbol i n \boldsymbol c e n t r a l \boldsymbol f o r c e

\boldsymbol f i e l d \dots \boldsymbol b u t \boldsymbol H ere \boldsymbol i \boldsymbol a m \boldsymbol n o t \boldsymbol u s i n g

\boldsymbol G T R . . only \boldsymbol t h e \boldsymbol p o i n t \boldsymbol o f \boldsymbol v i e w \boldsymbol o f

\boldsymbol o l d \boldsymbol c l a s s i c a l \boldsymbol c o n c e p t \boldsymbol o f \boldsymbol N e w t o n i a n

\boldsymbol L a g r a n g i a n \boldsymbol M e \boldsymbol c h a n i c s .

\boldsymbol {l e t}^{'} \boldsymbol s \boldsymbol p i c k \boldsymbol u p \boldsymbol t h e \boldsymbol e x a m p l e \boldsymbol o f

\boldsymbol o u r \boldsymbol s o l a r \boldsymbol s y s t e m \dots . \boldsymbol w h e r e \boldsymbol s u n \boldsymbol i s

\boldsymbol a t \boldsymbol t h e \boldsymbol f o c u s \boldsymbol a n d \boldsymbol e a r t h \boldsymbol m o v e s

\boldsymbol r o u n d \boldsymbol s u n \boldsymbol i n \boldsymbol a c c o u n t \boldsymbol o f \boldsymbol G r a v i t a t i o n a l \boldsymbol f o r c e

\boldsymbol f e i l d .

\boldsymbol s o \boldsymbol o u r \boldsymbol L a g r a n g i a n \boldsymbol c o n s t r a c t e d

\boldsymbol i n \boldsymbol p l a n e \boldsymbol p o l a r \boldsymbol c o o r d i n a t e \boldsymbol l i k e \boldsymbol t h a t

L = \frac{1}{2} [{(\overset{∙}{\boldsymbol r})}^{2} + \boldsymbol r^{2} {(\overset{∙}{\boldsymbol θ})}^{2}] - \boldsymbol v (\boldsymbol r)

\boldsymbol n o w \boldsymbol u s i n g \boldsymbol E u l e r - \boldsymbol L a g r a n g e \boldsymbol {e q}^{\boldsymbol n}

(\boldsymbol f r o m \boldsymbol V a \boldsymbol r i a t i o n a l \boldsymbol c a l c u l a s)

\frac{\partial L}{\partial \boldsymbol r} - \frac{\boldsymbol d}{\boldsymbol d t} (\frac{\partial L}{\partial \overset{∙}{\boldsymbol r}}) = 0 \boldsymbol a n d \frac{\partial L}{\partial \boldsymbol θ} - \frac{\boldsymbol d}{\boldsymbol d t} (\frac{\partial L}{\partial \overset{∙}{\boldsymbol θ}}) = 0

\boldsymbol f r o m 1 \boldsymbol s t \boldsymbol {e q}^{\boldsymbol n} \boldsymbol w e \boldsymbol g et

\overset{∙ ∙}{\boldsymbol r} - \boldsymbol r {(\overset{∙}{\boldsymbol θ})}^{2} = - \frac{\partial}{\partial \boldsymbol r} [\boldsymbol v (\boldsymbol r)] = \boldsymbol F (r) = \frac{\boldsymbol k}{\boldsymbol r^{2}}

\dots \dots \dots \dots \dots . . (\boldsymbol i)

\boldsymbol F (r) = \boldsymbol g r a v i t e t i o n a l \boldsymbol f o r c e, \boldsymbol k = \boldsymbol c o n s t a n t

\boldsymbol k = \boldsymbol {G m}_{\boldsymbol s} \boldsymbol m_{\boldsymbol e} .

\boldsymbol a n d 2 \boldsymbol n d \boldsymbol {e q}^{\boldsymbol n} \boldsymbol g i v e s \boldsymbol u s

\boldsymbol r^{2} \overset{∙}{\boldsymbol θ} = \boldsymbol c o n s t a n t, \boldsymbol m u l t i : \boldsymbol μ \boldsymbol b o t h \boldsymbol s i d e s

\boldsymbol w e \boldsymbol g e t \boldsymbol L = \boldsymbol {μ r}^{2} \boldsymbol θ = \boldsymbol c o n s t a n t

\dots \dots \dots . . (\boldsymbol i i)

\boldsymbol w h e r e \boldsymbol μ = \frac{\boldsymbol m_{\boldsymbol s} \boldsymbol m_{\boldsymbol e}}{\boldsymbol m_{\boldsymbol s} + \boldsymbol m_{\boldsymbol e}} = \boldsymbol r e d u c e \boldsymbol m a s s

\boldsymbol f o r \boldsymbol c e n t r a l \boldsymbol f o r c e -

\boldsymbol t o r q u e \boldsymbol τ = \vec{\boldsymbol r} \times \boldsymbol f (\boldsymbol r) = 0 \boldsymbol o r \frac{\boldsymbol d L}{\boldsymbol d t} = \boldsymbol c o n s t a n t

\boldsymbol w h e r e \boldsymbol L = \boldsymbol a n g u l a r \boldsymbol m o m e n t a m .

\boldsymbol n o w \boldsymbol l e t \boldsymbol u = \frac{1}{\boldsymbol r} \boldsymbol n o w \boldsymbol d i f f \boldsymbol w r t \boldsymbol t o

\boldsymbol θ \boldsymbol a n d \boldsymbol u s i n g \boldsymbol {e q}^{\boldsymbol n} (\boldsymbol i i) \boldsymbol w e \boldsymbol g e t

\frac{\boldsymbol d u}{\boldsymbol d θ} = \frac{- 1}{\boldsymbol r^{2}} . \frac{\frac{\boldsymbol d r}{\boldsymbol d t}}{\frac{\boldsymbol d θ}{\boldsymbol d t}} = - \frac{1}{\boldsymbol r^{2}} \frac{\overset{∙}{\boldsymbol r}}{\overset{∙}{\boldsymbol θ}} = - \frac{\boldsymbol μ}{\boldsymbol L} \overset{∙}{\boldsymbol r}

\boldsymbol n o w \frac{\boldsymbol d^{2} \boldsymbol u}{\boldsymbol {d θ}^{2}} = - \frac{\boldsymbol μ^{2}}{\boldsymbol L^{2} \boldsymbol u^{2}} . \overset{∙ ∙}{\boldsymbol r} .

\boldsymbol s o \boldsymbol w e \boldsymbol g e t \overset{∙ ∙}{\boldsymbol r} = - \frac{\boldsymbol L^{2} \boldsymbol u^{2}}{\boldsymbol μ^{2}} . \frac{\boldsymbol d^{2} \boldsymbol u}{\boldsymbol {d θ}^{2}} \boldsymbol a n d

\boldsymbol r {(\overset{∙}{\boldsymbol θ})}^{2} = \frac{\boldsymbol u^{3} \boldsymbol L^{2}}{\boldsymbol μ^{2}} \dots

\boldsymbol n o w \boldsymbol p u t t i n g \boldsymbol t h i s \boldsymbol t w o \boldsymbol r e s u l t s \boldsymbol a t

\boldsymbol {e q}^{\boldsymbol n} (\boldsymbol i) \boldsymbol w e \boldsymbol g e t \boldsymbol {e q}^{\boldsymbol n} \boldsymbol o f \boldsymbol m o t i o n \boldsymbol o f

\boldsymbol e a r t h \boldsymbol a r o u n d \boldsymbol s u n \boldsymbol i s

\frac{\boldsymbol d^{2} \boldsymbol u}{\boldsymbol {d θ}^{2}} + \boldsymbol u = - \frac{\boldsymbol μ^{2} \boldsymbol k}{\boldsymbol L^{2}} (\boldsymbol c o n s t a n t)

\boldsymbol b y \boldsymbol s o l v i n g \boldsymbol t h i s \boldsymbol e a s y 2 \boldsymbol n d \boldsymbol O D E

\boldsymbol w e \boldsymbol g e t \boldsymbol u (\boldsymbol θ) = - \frac{\boldsymbol μ^{2} \boldsymbol k}{\boldsymbol L^{2}} + \boldsymbol A c o s θ + \boldsymbol B s i n θ

or = - \frac{\boldsymbol μ^{2} \boldsymbol k}{\boldsymbol L^{2}} + \boldsymbol γ c o s Φ

\boldsymbol w e \boldsymbol c o n s i d e r a t e \boldsymbol a t \boldsymbol f i r s t \boldsymbol u = \frac{1}{\boldsymbol r}

\boldsymbol s o \boldsymbol n o w \boldsymbol r (\boldsymbol Φ) = \frac{1}{- \frac{\boldsymbol μ^{2} \boldsymbol k}{\boldsymbol L^{2}} + γ \boldsymbol c o s (\boldsymbol Φ)}

= \frac{- \frac{\boldsymbol L^{2}}{\boldsymbol μ^{2} \boldsymbol k}}{1 - \frac{\boldsymbol L^{2} \boldsymbol γ}{μ^{2} \boldsymbol k} \boldsymbol c o s (\boldsymbol Φ)}

\boldsymbol c o m p a r e \boldsymbol t h i s \boldsymbol r e s u l t \boldsymbol w i t h

\boldsymbol {e q}^{\boldsymbol n} \boldsymbol o f \boldsymbol a \boldsymbol e l l i p s e \boldsymbol i n \boldsymbol p l a n e \boldsymbol p o l a r

\boldsymbol c o o r d i n a t e \boldsymbol r (\boldsymbol θ) = \frac{\boldsymbol l}{1 \underline{+} \boldsymbol e c o s (\boldsymbol θ)}

\boldsymbol i n \boldsymbol o u r \boldsymbol c a s e - -

\boldsymbol s a m i \boldsymbol l e t u s \boldsymbol r e c t u m \boldsymbol o f \boldsymbol t h e \boldsymbol e l l i p t i c a k

\boldsymbol p a t h \boldsymbol o f \boldsymbol o u r \boldsymbol p l a n e t \boldsymbol i s

\boldsymbol l = \boldsymbol a (1 - \boldsymbol e^{2}) = - \frac{\boldsymbol L^{2}}{\boldsymbol μ^{2} \boldsymbol k}

\boldsymbol a n d \boldsymbol e s c e n t r i c i t y \boldsymbol e = \frac{\boldsymbol L^{2} \boldsymbol γ}{\boldsymbol μ^{2} \boldsymbol k}

\boldsymbol t h i s \boldsymbol i s \boldsymbol w h y \boldsymbol p l a n e t s \boldsymbol a r w \boldsymbol m o v i n g

\boldsymbol i n \boldsymbol a \boldsymbol e l l i p t i c a l \boldsymbol p a t h \dots

1 \boldsymbol s t \boldsymbol p r o p o s e d \boldsymbol b y \boldsymbol K e \boldsymbol p l e r \boldsymbol i n \boldsymbol h i s

1 \boldsymbol s t \boldsymbol l a s \boldsymbol o f \boldsymbol p l a n e t o r y \boldsymbol m o t i o n . .

Commented by smridha last updated on 08/Jun/20

ohh sorry it should be (dL/dt)=0 so L=constant.

\boldsymbol o h h \boldsymbol s o r r y \boldsymbol i t \boldsymbol s h o u l d \boldsymbol b e \frac{\boldsymbol d L}{d \boldsymbol t} = 0

\boldsymbol s o \boldsymbol L = \boldsymbol c o n s t a n t .

Commented by M±th+et+s last updated on 08/Jun/20

g r e a t w o r k t h a n k s

g r e a t w o r k t h a n k s

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