Solutions To Celestial Mechanics Problems

 Problems:

1) The Earth's aphelion distance is 1.01671 AU and its perihelion distance is 0.98329 AU. Given its eccentricity e = 0.016, then use this information and any other (from previous posts) to find:

a) the velocities at aphelion and perihelion

b) the energy constants C(A) and C(P) at each of these points, and h, the 'specific relative angular momentum'.

c) Hence or otherwise, use the vis viva equation to confirm the results you obtained in (a)

Solution:

The key here is to recognize "two birds" can be killed with one stone, that is,  obtaining the velocities for this section while also obtaining h for part b. This in turn depends on using specific algebraic properties to express h in terms of m, a and e. In so doing we get:

h = + [m a(1 - e²)]^½

where m= 1.33 x 10²⁰ Nm²/kg

(Note: for m, we already know G and m1= 1.99 x 10³⁰ kg (sun's mass) and m2 = 6.4 x 10²⁴ kg, Earth's mass)

Also: a = 1.496 x 10¹¹ m

Then h = 4.46 x 10¹⁵ N-m/kg = 4.46 x 10¹⁵ J/kg

The velocity at perihelion is then:

V_P = h/ a(1- e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 - 0.016)]

V_P = 3.03 x 10⁴ m/s

and the velocity at aphelion:

V_A = h/ a(1 + e) =  4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 + 0.016)]

V_A = 2.93 x 10⁴ m/s

b) We need the energy constants C(A) and C(P) at each of these points, and h, the 'specific relative angular momentum'

We already obtained h, in (a) so need only find the energy constants. We do so for each of the points, perihelion and aphelion.

Then:

C(P) =½ V_P ² - m/[a(1-e)]

C(P) = ½{3.03 x 10⁴ m/s}²

 - (1.33 x 10²⁰ Nm²/kg) / [1.496 x 10¹¹ m(1 - 0.016)]

C(P) = -4.45 x 10⁸ m²/s²

C(A) =½V_A² - m/[a(1+ e)]

C(A) =  

½{2.93 x 10⁴ m/s}²  -  (1.33 x 10²⁰ Nm²/kg) / [1.496 x 10¹¹ m(1 + 0.016)]

C(A) = -4.45 x 10⁸ m²/s²

Not surprisingly, we see they are the same (energy) constants.

c) To confirm the results obtained in (a):

Vis viva states:  V² = m (2/r - 1/a) or V = [m  (2/r - 1/a)]^½

To confirm the results in (a) we need the same velocities when:

r1 (perihelion radius vector) = 0.98329 AU = (0.98329)(1.496 x 10¹¹m)

Or:  r1 = 1.47 x 10¹¹ m

 And, r2( aphelion radius vector) = 1.01671 AU =(1.01671) (1.496 x 10¹¹m)

Or: r2 = 1.52 x  10¹¹ m

Then, call V1 the velocity at r1 (e.g. perihelion):

V1 = [m (2/r1 - 1/a)]^½  = 

[(1.33 x 10²⁰ Nm²/kg)[2/1.47 x 10¹¹ m -         1/1.496 x 10¹¹m]^½

V1 = 3.03 x 10⁴ m/s 

which is the same as V_P  obtained in (a) . Similarly:

V2 = [m (2/r2 - 1/a)]^½

V2 = [(1.33 x 10²⁰ Nm²/kg)[2/1.52 x 10¹¹ m - 1/1.496 x 10¹¹ m] ^½

V2 = 2.93 x 10⁴ m/s 

or the same as V_A obtained in (a). The vis viva equation therefore confirms the earlier results.

2)Explain the significance of the expression:

Z =  (W -  b)/  ( 1 -  Wb)

Solution:

The parameter b  is related to the planet's eccentricity (e) by:

(1  + b )/ (1 -   b) =   Ö {(1 + e) / (1 - e)}

The parameter Z is related to the true anomaly f by:

½   log Z =  i f/ 2 

The parameter W is related to both the true anomaly f  and Z as well as the eccentric anomaly E by:

ÖW  = ÖZ  / exp  (iE/2) = exp  (if/2)

 

In effect, the ratio (W -  b)/  ( 1 -  Wb) ties together the two critical orbital anomalies (E, f) with the planet's eccentricity e.

3)What  can you deduce regarding the relationship between: f,  E  and  w ?  What common factor unites them?

Ans.

The three angles act in concert define the orientation and the eccentricity e of the orbit.

The common factor uniting them is the eccentricity e .