Monday, August 29, 2011

Tackling Intermediate Astronomy Problems (4)

We now delve further into the intricacies of planetary motion, including using material presented in 'Tackling Simple Astronomy Problems', especially Kepler's 3rd or harmonic law. Recall from that law:

(T1/ T2)^2 = k(a1/ a2)^3

where T1, T2 are the periods, related to a1, a2 - the semi-major axes, as shown.

Now, it should be clear that once the sidereal period T of a planet is known, and also the semi-major axis a(or mean heliocentric distance) then the velocity of the planet in its orbit (assumed circular) can be computed, or:

V= 2π a/ T

Hence, for two planets, the ratio of their orbital velocities is:

V2/V1 = (a2/a1) (T1/T2)

where we intentionally allow the numbers 1 and 2 to refer to the inner and outer planets, respectively. As may deduced form Kepler's 3rd law:

T1 = [(a1)^3/k]^½ and T2 = [(a2)^3/k]^½

Substituting for T1 and T2 in the earlier form:

V2/ V1 = (a1/a2)^½

Now, in Fig. 1, the orbits for the Earth and a superior planet are shown, and the semi-majore axis are denoted by a and b, respectively. For any superior planet, b > a.

At opposition (the aligment SEP) the positions of Earth and planet are given as E and P, with velocity vectors V and V_p, tangential to their orbits. Now, from the expression for (V2/V1), V_p < V so the angular velocity of the planet as observed from Earth is:

(V_p - V)/ PE

and is in a direction opposite to the orbital motion, and hence is retrograde at opposition.

At the following quadrature, shown by configuration SE' P', the Earth's orbital velocity V is now along the line P'E' but the planet's velocity V_p has a component V_p sin(φ)) perpendicular to E'P'. The other component, V_p cos (φ) lies along the line P'E' and - like the Earth's velocity -doesn't contribute to the observed angular velocity of the planet.

The geocentric angular velocity at quadrature is then:

V_p sin (φ)/ E'P'

Example Problem:

a) Compare the orbital velocities of Venus and Earth, if the sidereal period for Venus is 224.69 d, and for Earth 365.25 d.

b) Verify this by using a Table of orbital velocities for the planets - given in km/s

We have:

V2/V1 = (a2/a1) (T1/T2)

By convention we assign '1' to the inner planet (Venus) and '2' to the outer (Earth). We have a2 = 1 AU and for Venus (from Kepler's third law):

T1 = (224.69/365.25) yr. = 0.6151 yr.

a1 = {[T1]^2}^1/3 = [(0.6151)^2]^1/3

a1 = 0.723 AU


V2/V1 = (0.7234)(1/0.6151)

V2/V1 = 1.175

(b) According to a Table of Orbital velocities in Astrometric & Geodetic Data:

V(Venus) = 35.02 km/s

V(Earth) = 29.78 km/s

Take the ratio (which must be 1.175)

V(Venus)/V(earth) = (35.02 km/s)/ (29.78 km/s) = 1.175

So, Venus' orbital velocity is 1.175 times Earth's

Other Problems:

(1)(a) Calculate the ratio of the Earth's tangential orbital velocity to Saturn's given Saturn's sidereal orbital period is 10,746.9 days.

(b) Validate this is approximately correct if Saturn's mean orbital velocity is 9.664 km/s from an astrometric table.

(2)At quadrature with the planet Mars, it is found (based on Fig. 1) that SP' = 1.53 AU, and SE' = 0.999 AU.

(a) From this deduce the distance P'E' and

(b) Hence, find the angular velocity of the planet as observed from Earth

(3) If Mercury's sidereal orbital period = 0.2408 yr. show that its tangential orbital velocity should be 47.87 km/s

(4) Based on the information in (3)and given that Uranus' sidereal period = 83.747 yrs., show that Uranus tangential orbital velocity is approximately 5.477 km/s.

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