Central Force Problem Consider two bodies of masses, say earth and moon, m E and m M moving under the influence of mutual gravitational force of potential V(r). Now Langangian of the system is where, µ = me.mm M and M = m E + M m Now, the generalized momenta are Select the case: L = µ(ṙ + r θ + r sin θ ϕ ) V (r) () P L ṙ = µṙ P θ = L θ = µr θ () P ϕ = L ϕ = µr sin θ ϕ θ(t = 0) = 90 θ(t = 0) = 0 (3) (always possible by orientation of the x, y, z coordinate system). The Euler-Lagrange Equation for θ is dp θ = µrṙ dt θ + µr θ = L θ = µr sin(θ) cos(θ) φ (4) Since all other terms are zero due to our choice, it must be true that also θ(t = 0) = 0 This can be expanded for higher derivatives, ultimately showing that θ must be constant at 90 degrees. This is of course due to the fact that both the magnitude and the direction of the angular momentum vector L is conserved, and the radius vector is always perpendicular to it. So if we choose the z-direction in the direction of L, the equations of motion for r(t) and ϕ(t), are restricted to the x-y plane. We have now reduced our analysis to that of a system with degrees of freedom, namely (r, ϕ). From now on, we assume that the force is pointing along the direction of the relative position r between the two objects. We can say that for such a central force the potential depends only on the distance r between the two objects. V (r) = V (r) = GMµ r (5)
Figure : Motion of two body system of reduced mass µ in the central force field Then the Langrangian for this system From Euler-Langrange equation (ELE) We get Then, the angular momentum l is constant L = µ(ṙ + r ϕ ) V (r) (6) d dt ( ) L L ϕ ϕ = 0 P ϕ = 0 = µr ϕ (7) P ϕ = L ϕ = µr ϕ = l = constant (8) Then Similarly, from ELE for r d dt ( ) L L ṙ 0 µ r µr ϕ + V (r) r P L r µ µr ϕ GMµ r = 0 (9) V (r) µ + P ϕ r µr 3 (0)
let a function h given by where V (r) = h = µ (ṙ + r ϕ ) + V (r) h = µ ṙ + P ϕ h = µ ṙ + + V (r) = E () µr l µr + V (r) = E h = µ ṙ + V (r) = E () l µr + V (r) is pseudo-potential and E is the total energy. From Eq. (0), ṙ = ± E P ϕ µ µr V (r) It doesn t alter the trajec- where the sign ± depends on r(t) is increasing or decreasing at time t. tory.taking + sign, we get r r dr = µ E P ϕ µr V (r) t t dt = t t (3) Kepler s Second Law From Eq. (7) The differential area swept out in time dt is ϕ = dϕ dt = l µr Figure : Area swept out by the radius vector r in time dt 3
da = r (r dϕ) = r ϕ dt A = da dt = r ϕ = l = constant (4) µ Thus, the particle sweeps away equal area in equal interval of time, which is Kepler s second law. Again ϕ dϕ = ϕ ϕ = l t ϕ µr dt = l t µr (t t ) (5) from Eq. () and Eq. (4), we get ϕ ϕ = l µ r r dr µ r E P ϕ µr V (r) ϕ ϕ = l r dr/r µ r E P ϕ µr V (r) (6) Again using Eq. (7), we can write dϕ = l µr dt d dt = l d µr dϕ ṙ = dr dt = l dr µr dϕ l dṙ µr dϕ = l dṙ µr dϕ = l ( ) d l dr µr dϕ µr dϕ (7) Again Let u Using Eq.(5) and Eq.(7), we get, taking + sign, ϕ ϕ = l µ dr dϕ = dr du du dϕ = du u dϕ = r du dϕ dr = du (8) r du (9) E l u µ V (u) Using Eq.(6) in Eq.(9), l u µ ( d u dφ l u 3 ) dv + u = 0 µ du d u dφ = u + µ dv l u (0) 4
Considering the power law function of r for the potential such that V (r) = k r n+ () V (u) = k u (n+) () Again, Eq. (8) becomes dϕ = du (3) µe l µk l u (n+) u E = µ ṙ + l µr + V (r) = µ ṙ + V (r) At r min, r max, and r 0, equilibrium position ṙ = 0 For equilibrium position r 0, V (r) r = 0 and E =E 0 (4) For a mass µ on a spring with spring constant k s, V = ks r, so k = k s / and n =. For Kepler s problem, n =, k = GMµ, and V (u) = ku: l µr0 3 + k r 0 = 0 and r 0 = l µk (5) E 0 = l µ ( ) µk l k µk l E 0 = µk l = V/ = T (6) [Note: Alternative way to find maximum and minimum values of r ( From Goldstein Text) For maximum and minimum values of r, E = l µr k r Er + kr l µ = 0 This equation is quadratic in r, so we will have two roots given by: k ± k + El µ E = k E ( ± = a( ± e) + El µk ) 5
with, We get From Eq. (9) a = k E and e = + El µk ] u + u = µk l u(ϕ) = Acos(ϕ ϕ 0) + µk l Acos(ϕ ϕ 0 ) + µk l = Acos(ϕ ϕ 0 ) + C therefore, C( + e cos(ϕ ϕ 0 )) (7) Without loss of generality, let us assume that ϕ 0 = 0 at t = 0, so the above Eq. (30) becomes C( + e cosϕ) (8) where C = µk l and e = A C Now, for r min is the eccentricity of the orbit of the particle. E = l µr min k r min = l µ C ( + e) kc( + e) E = µk l ( e ) e = + El µk (9) At equilibrium E 0 = µk l The nature of the orbit depends upon the magnitude of e according to the following scheme: e = 0, E = µk l : circle e =, E = 0 : parabola e >, E > 0 : hyperbola e <, E < 0 : ellipse 6
Figure 3: Trajectory of the body with varying eccentricity in the central force field For e <, case of ellipse, C( + e) = C( e) rmin = rmax The major half axis, a is defined by the relation a = r + r0 a = rmin + rmax a= + C + e e a= C( e ) k l = = E e µk 7 (30)
So, Choose ϕ 0 such that r(ϕ 0 ) = r min From Fig.(3), C( + e cosϕ) = a( e ) ( + e cosϕ) r(ϕ b )( cosϕ b ) = a e (3) or, or, a( e )cosϕ b + ecosϕ b = ae (e + e cosϕ b ) = ( e )cosϕ b or, cosϕ b = e r b (ϕ b )( cosϕ b ) = r b e = ae r b = a (3) Then, we get, b = r b a e = a e = e l µk = l a µk (33) The equation C( + e cosϕ) is actually an equation of an ellipse with shifted co-ordinates x and y (or x and y, original co-ordinate system) with, x o = ae, a = x (34) a + y b = (35) (x + x 0 ) a + y b = (36) C( e ), b = C (37) e This can be proven by using y = y = r(ϕ) sin(ϕ) and x = ae + r(ϕ) cos(ϕ) and plugging in. Kepler s Third Law Now area of ellipse A = πab The period of elliptical motion T is the ratio of the total area of the ellipse (A) to the areal velocity ( A) and is given as : Because T = πab l/µ = πµab l T = πa 3/ µ k = πa3/ GM T = 4π a 3 µ k (38) b = a e = b = a / l µk ( ) k. El E µk = k E. l µk (39) 8
The Eq.(38) shows that the square of the periods of the object in central force is proportional to the cube of the major half axis i. e T a 3, which is Kepler s third law. [Note: If a planetory object of mass m is in the motion under the potential of central force, we should replace the reduced mass µ by mass m of the planet] 9