Question

Hi there can you please answer question part (b) (ii) - thank you

Consider a test particle of mass m orbiting in a Schwarzschild black hole of mass M. If the particle orbits at a speed uc and at a distance r > ., where c is the speed of light and . = 2GM/? the Schwarzschild radius, we can use the usual Newtonian central force equations of motion to analyse the orbit. The effective potential, however, needs to be modified to include a correction term from general relativity: GMm U (r) - GME emr3 2m2 In the following, to make life easier, you can set G = c= 1. (a) For arbitrary values of l, m and M, this effective potential has three generic shapes. (i) Shape 1 supports one stable and one unstable circular orbit. (ii) Shape 2 supports one and only one unstable circular orbit. (iii) Shape 3 has no circular orbit. Find the conditions on t/(mm) that enable each of the above scenarios, and plot Ue(r) in each case. Where circular orbits exist (both stable and unstable), compute the corresponding orbit radius ro (b) Consider the case of a bound elliptical orbit. (1) Write down an equation of motion for u() (i.e., d’r/do ..), where u 1/1. (ii) Suppose M = 20. m = 1, and C = 100 (in some units). Let the particle be at the inner radius = 200 (again, in some appropriate units) at 0 = 0. Solve (numerically) the equation of motion you obtained in part (i) for the orbit r() from o=0 to o = 8. (iii) Plot r(ó). Does the particle get back to its starting point after 2x? If not, how much further in ø does it need to go until it reaches on again? /Note: what you have just computed is called the "perihelion shift (iv) Let 1 =r(6) coso and y=rlo) sino. Plot {r.y in a 2D plot parametric plot for 0 = 0 to 8

Answer 1

Answer:- Given that رأت (ط M=20 M = 1 cal 1 =100 G =1 du +0=0.002 + Gou? do² we will solve the equation perturbatively Let u=A+Bu putting this in the above equation and in powers of B Bº = 0 - GOA-A+a where a 0.002 So Ana B':0 - ū"+ū-120 Aű ů"+0.76 M =0 ūcos (0.910) So, the solution is 1 Y A+B cos (0.870) at d=0, ,V=200 200 = at B B 0.00 3 -13 a+B COS (0.870) where a 0.002,0 -0.003

at 0-0- we have at ¢ 2T we have Ys G+B cos (2110.87) so both are not same so perihelion shift is Δή = 2T1X 0.87 A0 = 1.74T