Solution to Celestial Mechanics Problems

Here we go back to show the solutions to the two earlier problems in celestial mechanics:

1)Problem: Estimate the energy constant, C, for Jupiter, and also its orbital angle of inclination, i, if a = 5.2 AU, and e = 0.048


We saw Jupiter’s mass m2 = 1.9 x 10^27 kg, and the Sun’s mass is 2 x 10^30 kg

And C = u/ 2a = G(m1 + m2)/ 2a , and h = +/- [(u a(1 - e^2))]^1/2

Then: C = 8.6 x 10^7 J/kg

To estimate the angle of inclination i, we make use of: C3/h = cos(i) and let:

C3 ~ C

Then:

cos(i) ~ C/h = +/- C/ [(m a(1 - e^2))]^1/2


i ~ arc cos(0.011) ~ 1 deg

This compares to the actual value of i= 3.1 deg



2) Problem: Show that for the Earth, the period P ~ 366 days. Use r = a = 1.49 x 10^11 m. Earth’s mean motion per day n = (T – to) is » 0.985647 deg/day.


There are various ways to approach this problem but I look at the most direct:

We have that, since: dA = (h/2) dt = (h/2) (T – t_o) then:

dA = dt = (T – t_o) = 0.985647 deg/day = 3.548 x 10^3 arcsec/day

where the units have been changed to arcsec/day from degrees/day by multiplying by 3600. (Note that the prescribed value for a is incorporated into the value for h).

The net area change in radians for dA (for one orbit – correlated to the period, or one full revolution in radians) is 2pi

To change this to arcsec:

(2pi radian) (57.3 deg/rad)(3600 arcsec/degree) = 1.296 x 10^6 arcsec

Then the period in days is:

P = [1.296 x 10^6 arcsec]/ [3.548 x 10^3 arcsec/day] = 365.3 days