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Orbital Mechanics

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The energy and angular momentum of a particle inside a central potential are given by:

E = ½*μ *(dr/dt)^2 + V(r) L = μ* r^2 *(dθ/dt)

V(r) = - (G*μ*M)/r + L^2/(2*μ*r^2) - (G*L^2*M)/(r^3*μ*c^2)

μ is the reduced mass and c is the speed of light. (Use units where c=1).

a) Solve these two equations for dr/dt and dθ/dt and show that:

dr/dt = +/- [( 2/μ)*(E - V(r))]^1/2 dθ/dt = L/(μ*r^2)

b) Write the equation of a) in a μ independent form by introducing e = E/μ and l = L/μ.

c) Use units such that G = -M = c = 1, and let e = -.02, l = 4.5. Let the initial position be given by r(t=0) = a, θ(t=0) = 0, a = -GM/(2e). Compute the initial velocities dr/dt(t=0) and dθ/dt(t=0) under the assumption both are positive.

d) Plot V(r)

Note: a = -GM/(2e)

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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