Wednesday, August 31, 2011

Solutions: Intermediate Astronomy Problems (4)

The problems at the end of the last instalment, and their solutions, in turn:

(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.

Solution

(a) We have:

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

By convention we assign '1' to the inner planet (Earth) and '2' to the outer (Saturn). We have a1 = 1 AU and for Saturn (from Kepler's third law):

T2 = (10,746.9/365.25) yr. = 29. 4yr.

a2 = {[T2]^2}^1/3 = [(29.4)^2]^1/3 = 9.50 AU

Whence:

(V1/V2) = (a2/a1)^½ = (9.50)^½ = 3.08

So, the Earth's orbital velocity should be about 3.1 x faster than Saturn's

(b) Saturn's mean orbital velocity is 9.664 km/s from an astrometric table.

From the sample problem, V(Earth) = 29.78 km/s

Then: (V1/V2) = (29.78 km/s)/ (9.664 km/s) = 3.08

(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.

Solution

(a) SP' = 1.53 AU is the hypoteneuse of the right triangle, SE'P'. SE' = 0.999 AU is one leg. Then, from Pythagoras' theorem:

P'E' = [(SP')^2 - (SE')^2]^½ = [1.53^2 - 0.999^2]^½

P'E' = 1.15 AU

(b) To obtain the angular velocity as observed from Earth we first need to obtain angle SP'E' since we must find φ, knowing that: (90 - S'P'E' = φ). From trigonometry:

cos (SP'E') = adj/ hyp = P'E'/ SP' = 1.15/ 1.53 = 0.757

SP'E' = arc cos (0.757) = 40.8 deg

then: φ = 90 - S'P'E' = 90 - 40.8 deg = 49.2 deg

The geocentric angular velocity at quadrature is then:

V_p sin (φ)/ E'P'

To get V_p, note:

(V1/V_p) = (a_p/a1)^½ = (1.53)^½ = 1.23

So: V_p = (1/1.23) (29.78 km/s) = 24.2 km/s

and:

V_p sin (φ)/ E'P' = [(24.2 km/s)sin (49.2)]/ 1.15 = 15.9 km/s

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

Mercury is interior to Earth so designate with '1':

T1 = 0.2408 yr.

a1 = {[T1]^2}^1/3 = [(0.2408)^2]^1/3 = 0.387 AU

V2/ V1 = (a1/a2)^½

So: V1 = (V2)/(a1/a2)^½ )

where: V2 is for Earth (29.78 km/s) and:

(a1/a2)^½ = (0.387/1)^½ = 0.622

V1 ('Merc') = V2/ 0.622 = (29.78 km/s)/0.622 = 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.

Solution

By convention we now assign '1' to the inner planet (Earth) and '2' to the outer (Uranus). We have a1 = 1 AU and for Uranus (from Kepler's third law):

T2 = 83.747 yrs.

By Kepler's 3rd law:

a2 = {[T2]^2}^1/3 = [(83.747)^2]^1/3 = 19.13

Whence:

(V1/V2) = (a2/a1)^½ = (19.13)^½ = 4.37

or: V2 ('Uranus') = V1/ (4.37)

where V1 = 29.78 km/s (Earth) so:

V2 = 29.78 km/s/ (4.37) = 6.8 km/s *

Explanation:

A cross check of other references (e.g. Abell, 'Exploration of the Universe', Appendix 8) discloses that this is the correct value! The 5.477 km/s from the Astrometric & Geodetic data sheet - is incorrect! It actually applies to Neptune, not Uranus.

We can check this: The semi-major axis of Neptune is 30.06 AU (from problem #5, 'Tackling Intermediate Astronomy 2').

By Kepler's 3rd law:

a2 = 30.06 AU

Whence:

(V1/V2) = (a2/a1)^½ = (30.06)^½ = 5.48

or: V2 ('Neptune') = V1/ (5.48) = (29.78 km/s)/ 5.48 = 5.43 km/s

It pays to double check! (Including the Examiner or the textbook!)