Solutions to Intermediate Astronomy (6) - Part 1

We now look at the solutions to the previous problem set, focusing in on (1) - (3):

Problems:

1) Derive the vis viva equation using three or more of the equations given in this blog. (Hint: all incorporate u, and two incorporate e)


Solution:

To obtain the vis viva equation we proceed as follows:

(a) subtract C = ½V_P^2 - u/a(1 - e) from: ½V^2 - u/r = C

E.g.

½V^2 - u/r = C
- ½V_P^2 - u/a(1 – e) = C

= ½V^2 - ½V_P^2 - u/r - u/a(1 – e)

Then, substitute in: (V_P)^2 = u/a [{1 + e)/ (1 - e)] to eliminate V_P, viz.

½V^2 - [ u/a [{1 + e)/ (1 - e)] - u/r - u/a(1 – e)

Which, with appropriate algebraic simplification, yields:

V^2 = u (2/r - 1/a)

2) 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 blogs) 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:

a) 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 u, a and e. In so doing we get:

h = +/- [u a(1 - e^2)]^½

where u = 1.33 x 10^20 Nm^2/kg

(Note: for u, we already know G and m1= 1.99 x 10^30 kg (sun's mass) and m2 = 6.4 x 10^24 kg, Earth's mass)

Also: a = 1.496 x 10^11 m

Then h = 4.46 x 10^15 N.m/kg = 4.46 x 10^15 J/kg

The velocity at perihelion is then:

V_P = h/ a(1- e) = 4.46 x 10^15 J/kg / [1.496 x 10^11 m(1 - 0.016)]

V_P = 3.03 x 10^4 m/s

and the velocity at aphelion:

V_A = h/ a(1 + e) = 4.46 x 10^15 J/kg / [1.496 x 10^11 m(1 + 0.016)]

V_A = 2.93 x 10^4 m/s

b) 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) =½V_P^2 - u/[a(1-e)]

C(P) =½{3.03 x 10^4 m/s}^2 - (1.33 x 10^20 Nm^2/kg) / [1.496 x 10^11 m(1 - 0.016)]

C(P) = -4.45 x 10^8 m^2/s^2

C(A) =½V_A^2 - u/[a(1+ e)]

C(A) =½{2.93 x 10^4 m/s}^2 - (1.33 x 10^20 Nm^2/kg) / [1.496 x 10^11 m(1 + 0.016)]

C(A) = -4.45 x 10^8 m^2/s^2

And not suprisingly, we see they are the same (energy) constants.

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

Vis viva says:

V^2 = u (2/r - 1/a) or V = [u (2/r - 1/a)]^½

If it is to confirm the results in (a) then it should give the same velocities when:

r1 (perihelion radius vector) = 0.98329 AU = (0.98329)(1.496 x 10^11m)

r1 = 1.47 x 10^11 m

and r2( aphelion radius vector) = 1.01671 AU =(1.01671) (1.496 x 10^11m)

r2 = 1.52 x 10^11m

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

V1 = [u (2/r1 - 1/a)]^½

V1 = [(1.33 x 10^20 Nm^2/kg)[2/1.47 x 10^11 m - 1/1.496 x 10^11m]^½

V1 = 3.03 x 10^4 m/s which is the same as V_P obtained in (a)

Similarly:


V2 = [u (2/r2 - 1/a)]^½

V2 = [(1.33 x 10^20 Nm^2/kg)[2/1.52 x 10^11 m - 1/1.496 x 10^11m]^½

V2 = 2.93 x 10^4 m/s or the same as V_A, obtained in (a)

So, the vis viva equation confirms the results obtained in (a)


3) Attempt to obtain an improved value for Jupiter's mass (from what the first sample problem yields) using: T = 2π (a^3/u)^½ (Recall u = G(m1 + m2) and Jupiter's mass and the Sun's have already been given along with G, in the blog general information)

We have:

m1 = 1.99 x 10^30 kg

But the key in this problem is to use T = 2π (a^3/u)^½ to obtain u, and thence m2 for Jupiter as the sole unknown.

This means first arranging the equation to make u the subject, viz.

T^2 = 4π^2 ( a^3/u) (squaring both sides)

T^2 / 4π^2 = a^3/u

or:

u T^2 = 4π^2 a^3 (cross multiplying)

Tnen:

u = G(m1 + m2) = 4π^2 a^3 / T^2

Now, a = 5.2 AU = 7.77 x 10^11 m

And the period of Jupiter (from previous blogs) is T = 11.86 yrs. = 3.74 x 10^8s

Then:

4π^2 a^3 / T^2 = (4π^2)[7.77 x 10^11 m]^3 / [3.74 x 10^8s]^2

So:

G(m1 + m2) = 1.32 x 10^20 Nm^2/kg

whence (since we note m1 >> m2):

(m1 - m2) = 1.32 x 10^20 Nm^2/kg/ G

or:

(1.99 x 10^30 kg - m2) = 1.328 x 10^20 Nm^2/kg/ (6.7 x 10^-11 Nm^2/kg^2)


(1.99 x 10^30 kg - m2) = 1.981 x 10^30 kg

m2 = 1.99 x 10^30kg - 1.981 x 10^30 kg = 0.009 x 10^30 kg = 9 x 10^27 kg

compared with the value of 1.89 x 10^27 kg in tables.