Specific angular momentum (with respect to a planetary body) of a test mass that orbits a planet and accretion dosk Specific angular momentum of a test mass that orbits a planet and accretion dosk

(See attached file for full problem description with proper symbols)

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Q2a

The specific angular momentum (with respect to a planetary body) of a test mass that orbits a planet at a distance r with angular speed  is j = r2   (scalars). Show how this follows from the expression of the angular momentum of this test mass, J = r  p (vectors), where p is the linear momentum of the test mass. (Note: J, r, p, in this latter expression are vectors, and  denotes the vector product, while j,r and  in the former are scalars, and  denotes a simple multiplication)


I have answered this question as follows, please advise if I have made a mistake and what I have done wrong. Thank you.

a) Starting with the angular momentum of the test mass,

(1)

Where p = the linear momentum of the test mass =  

Equation 1 implies that the angular momentum of the test mass,

1) Is a vector
2) Its magnitude is J = rp sin θ
3) Its direction is perpendicular to the plane defined by the vectors r and p and in that sense is specified by the right hand rule.

So the angular momentum J of the test mass will be directed along the z – axis, and can be written,

J = (0, 0, Jz)

Where

(2)

Where = angular speed =  

It then follows from equation 2 that the specific angular momentum of the test mass can be written as,




b)   Derive an expression for the specific angular momentum jd of a test mass that                executes a circular Keplerian orbit around the planetary object, Assume that the radius is twice the radius of the planetary object RPO.

For this question I am less sure as it seems to be asking me to repeat the first question, please advise where I am going wrong. Below is the answer as I see it so far. Thank you.

b)   If the test mass is executing a circular Keplerian orbit then from Kepler’s law of orbits it has an eccentricity of zero at the inner edge of the accretion disc. As such the test mass’ radius from the white dwarf will be equal to its semi major axis, in this case r = 2RWD. So we can negate the effect of eccentricity from the calculations.

Therefore from equation 2

(2)

It can be seen that

Attached file(s):

Q2a.doc  View File