Earth's Orbital Velocity
Procedure
The plate scale (dispersion) of the spectrum of the the K giant star Arcturus can be found as shown in the figure below. Specifically, we must measure the distance m between the known reference lines of the spectrum of helium labelled "1" and "7" (see below).
The dispersion is then
(4,307.91 Å – 4,260.48 Å) ÷ m = 47.43 Å ÷ 123.3 mm = 0.3847 Å/mm .
Now, looking at Spectrum a—the spectrum of Arcturus observed when the distance between the earth and the giant star increases (a redshift), we see that the same lines in the star's spectrum appear in the reference spectra above and below the star's absorption spectrum. The lines in Spectrum a are all uniformly shifted to the right of the reference lines. In addition, the spectrum of Arcturus observed when the distance between the star and Earth is decreasing (Spectrum b) also shows the same absorption lines albeit shifted to the left of the reference lines. Spectrum b therefore shows a blueshift.
Velocity Determinations
To determine the amount of the redshift and blueshift,
we must choose a reference line and measure the position of the same
line in Spectrum a and b from that reference line (see the example for
line 5 shown in the figure below).
Be careful to pay attention to the sign of your measurement because measurements to the right of the reference line (as found in Spectrum a) are considered to be positive while measurements of Spectrum b (to the left of the reference line) are considered to be negative. It is also absolutely critical to measure the shift of the spectral lines to the nearest 0.1 mm. Using our example for line 5, we find that
Shift of line 5 in Spectrum a = +0.8 mm
Shift of line 5 in Spectrum b = –1.0 mm .
Therefore, the actual shifts—found using the dispersion—are
Δλa = +0.8 mm x 0.3847 Å/mm = +0.3078 Å
Δλb = –1.0 mm x 0.3847 Å/mm = –0.3847 Å .
Next, the Doppler velocity equation given on the back page of the lab handout
v = cΔλ/λo ,
can be used to calculate the redshifted and the blueshifted velocities of line 5:
VA = 300,000 km/s x (+0.3078Å) ÷ 4,294.13 Å = +21.50 km/s
VB = 300,000 km/s x (–0.3847 Å) ÷ 4,294.13 Å = –26.88 km/s .
Now, the orbital velocity of Earth is calculated using equation (1) to be
Vo = ½(VA – VB) = 0.5 x (+21.50 km/s – (–26.88 km/s)) = +24.19 km/s
while the radial (line-of-sight) velocity of Arcturus is found to be
Vs = ½(VA + VB) = 0.5 x (+21.50 km/s + (–26.88 km/s)) = –2.690 km/s .
Results
Arcturus does not lie exactly on the earth's orbital plane, however,
so we must make a minor correction to the orbital velocity of Earth to
account for Arcturus' declination. Given that the star is 30.8° above
the ecliptic, we must correct our calculation of the earth's orbital velocity
by a factor of cos 30.8°. Therefore, the corrected orbital velocity
of Earth is
Vc = +24.19 km/s ÷ cos 30.8° = +24.19 km.s ÷ 0.86 = 28.13 km/s .
In order to ensure the most accurate measurement of the orbital velocity, the entire set of calculations should be repeated for two other spectral line pairs and the three orbital velocities averaged.
Finally, the radius of the earth's orbit can be determined simply by using the average orbital velocity and knowing the orbital period (in other words, the number of seconds in one year). Using the relationship between velocity and the orbital period, we have
circular velocity = circumference of circle ÷ orbital period .
With our data,
Vc = 2π R÷ P
or, solving for the earth's orbital radius
R = Vc P ÷ 2π = 28.13 km/s x 31,560,000 s ÷ (2 x 3.14159) = 141,300,000 km
where P is simply the earth's orbital period expressed as the number of seconds in one year.
Given that the accepted value for the semimajor axis of Earth's orbit is 149,600,000 km, our calculation has an error of 5.6% .