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# Question on an Orbital Mechanics Problem

The energy and angular momentum of a particle inside a central potential are given by:
E = ½*μ *(dr/dt) + V(r) L = μ* r^2 *(dθ/dt)^2

V(r) = (G*μ*M)/r + L^2/(2*μ*r^2)

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 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)

See attached file for full problem description.

#### Solution Preview

There are a few typos in your formula for the energy. Let's derive this formula first. I'll denote the mass of the particle by m and the mass of the object it is orbiting by M. The gravitational potential is:

V_{grav}(r) = -GMm/r (1)

Note the minus sign! Minus the gradient of the potential is the force. So, if you differentiate
V_{grav}(r) w.r.t. r and multiply by minus one you should get the force in the r-hat direction, which is indeed the case.

Now let's write down the formula for the kinetic energy:

E_{kin}=1/2 m v^2 (2)

In (2) you write the velocity in polar coordinates. A small displacement vector ds can be written as:

ds = dr r-hat + r d θ θ-hat

and thus:

v = ds/dt = dr/dt r-hat + r d θ/dt θ-hat (3)

squaring (i.e. taking the inner product of the vector with itself gives)

v^2 = (dr/dt )^2 + r^2 (d θ/dt)^2

and the kinetic energy (2) can be written as:

E_{kin} = 1/2 m (dr/dt )^2 + 1/2 m r^2 (d θ/dt)^2

The total energy is:

E = E_{kin} + V_{grav}(r) =

1/2 m (dr/dt )^2 + 1/2 m r^2 (d θ/dt)^2 - GMm/r (4)

The angular momentum is given by m r v_{t}, where v_{t} is the θ-hat component (a.k.a. the tangential component) of the velocity, see (3):

L = m r^2 d θ/dt (4)

If you solve this for d θ/dt you get:

d θ/dt = L/(m r^2) (5)

If you substitute (5) in (4) you get:

E = 1/2 m ...

#### Solution Summary

A detailed solution is given. Orbital mechanics problems are determined.

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