Calculation of the orbit equation for the two-body problem

• start from energy    E =1/2 m/( r-dot^2 + (r theta-dot)^2) - k/r
• use l= m r ^2 theta-dot and get E = 1/2 m/( r-dot^2 + l^2/m^2r^2) - k/r
• rearrange to get r-dot = sqrt( 2E/m +2 k/(mr) -(1/r^2)(l/m)^2))
• rearrange to get dr/sqrt( 2E/m +2 k/(mr) -(1/r^2)(l/m)^2)) = dt
• write dt/dtheta dtheta = dtheta/theta-dot
• then cross-multiply by theta-dot and get
• dr theta-dot/sqrt( 2E/m +2 k/(mr) -(1/r^2)(l/m)^2)) = dtheta
• since theta-dot - l/mr^2 we have
• (dr/r^2( (l/m)/sqrt( 2E/m +2 k/(mr) -(1/r^2)(l/m)^2)) = dtheta
• let u = 1/r, so -du = dr/r^2 . Then we get
• -du(l/m)/sqrt( 2E/m +2 ku/(m) -u^2(l/m)^2)) = dtheta
• multiply on the left by m/l top and bottom and get
• -du/sqrt(2mE/l^2 +2mku/l^2 - u^2) = dtheta
• we complete the square in the denominator and get
• -du/sqrt(2mE/l^2 - (u-2mk/l^2 +(mk/l^2)^2) + (mk/l^2)^2) = dtheta.
• this is -du/sqrt(2mE/l^2+(mk/l^2)^2)-(u-mk/l^2)^2) = dtheta
• if we call a^2=2mE/l^2+(mk/l^2)^2) and b = u-mk/l^2 and p = l^2/(mk)
• we can rewrite as -du/sqrt(a^2-(u-1/p)^2) = dtheta
• a^2 = 2mE/l^2 + (1/p)^2.
• When we let u-1/p = a cos phi we get sqrt(a^2-(u-1/p)^2) = a sin phi
• since du = -a sin phi dphi the integral on the left reduces to
• du/sqrt(a^2-(u-1/p)^2) = dphi = dtheta
• This means phi=theta and 1/r-1/p = sqrt(2mE/l^2 + 1/p^2) cos theta
• multiplying by p on both sides gives p/r-1 = sqrt(p^2 2mE/l^2+1), or
• p/r = 1+ sqrt(1+2El^2/(mk^2) cos theta = 1 + e cos theta
• e = eccentricity = sqrt(1+2El^2/(mk^2)
• l = angular momentum, E = total energy, -k/r = U, p = l^2/(mk) = length