Orbital Velocity of the Earth Using Stellar Spectra

Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
Experiment #4 — Orbital Velocity of the Earth Using Stellar Spectra
In this exercise you will find the orbital velocity of the Earth by measuring the Doppler shift in the spectrum of the star Arcturus. Among the very brightest of stars, shining with a soft orange light, Arcturus lights northern spring skies. While orbiting the Sun, the Earth moves towards the Arcturus half of its orbital period and it moves away from the star rest of the period as shown in figure. When the Earth is moving towards and away from the star, the observed light is respectively blue-shifted and red-shifted .
(a) Earth m Sun
(b) Earth moving towards
The photocopied figure (taken from SKY AND TELESCOPE) below displays two spectrograms of the star Arcturus taken six months apart. The top and the bottom spectrograms are of a reference light source at rest with respect to the telescope. Corresponding to each of the reference lines is a dark line in the star spectrum, displaced slightly to the right (red-shifted) in a, and to the left (blue-shifted) in b.
Select line 1 and measure the displacement of it in a and b to tength of millimeters. One way to do it is to use a straight edge to connect matching reference lines as shown in the figure on the lower right corner on the next page. The shift are shown by filled rectagles (red in a and blue in b) in the figure.Measure the shifts with a scale estimating to tength a tength of millimeter. A magnifying glass would be handy. Try to be as accurate as possible. The measurements give the shift ?? in mm for the line. Draw a table similar to one shown below and note down the measurements. Repeat the process for lines 1, 3, 5, and 7 and enter into the table.
???oving away
???????A
?rcturus
r
???????????????
line no.
??
??
original wavelength ?0 angstrom
shift ??
shift ??
vr ???c ?0
?
spectrum a mm
spectrum b mm
spectrum a angstrom
spectrum b angstrom
spectra a
spectra b
1
4260.48
Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
????????????????????????????Fig: Spectrum of Arcturus, when the Earth is at (a) and (b) positions.
The measurments you obtained are in millimeters in paper scale. These are not in the angstrom scale in which the the reference wavelengths are measured. You need to convert your measurements to the angstrom scales. By the way, 1 angstrom = 10-10 m. To do this, measure the distance between lines 1 and 7 in millimeters with the same accuracy mentioned in the previous paragraph. Divide the difference in actual wavelengths of line 1 (4260.48 angstroms) and line 7 (4307.91 angstroms) by the distance between lines 1 and 7.
???i.e., Conversion Factor (CF) ? ?4307.91 – 4260.48? angstroms /millimeter ???
It serves as the conversion factor to obtain the shift in wavelengths in angstroms. Multiply the numbers you have in the third column and fourth coulumn in mm of the above table by the CF and write down the results in fifth and sixth column of the table respectively. The new nubmers are the shifts in angstrom. Now we know the shift in spectrum due to the relative motion between source of light (Arcturus ) and the observer on the Earth. Now you can use the Doppler’s effect Eq.(6.3) from you text book to find the relative velocity between the Earth and the star.
a
???????????vr ? ?? ? ? ? b c ?0
where c = 300000 km/s is the speed of light and ?0 is the wavelength of
corresponding reference line. These values for line 2, 3, 4, 5 and 6 are respectively, 4271.16, 4271.76, 4282.41, 4294.13, and 4299.24.
????
Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
You can use above formula for both red-shift and blue-shift. Find the average vr from the second last column of the table. Let’s denote this number by vrr.
vrr = ……. km/s.
It represents the average relative velocity obtained from the red-shift in the spectrum a. This
velocity is related to speed of the star and the speed of the Earth by
vrr ?vStar ?vEarth.
Similarly, find the average vr from the last column of the table and denote it by
vrb = – ……. km/s.
Notice there is negative sign. We are adding this negative sign here for convenience. Actually it should be with the numbers in the last column of the table because the shift ?? is negative for the blue shift. Usually displacement in the right is taken as positive and in the left is taken as negative.
vrb ? vStar ? vEarth .
Our goal is to find vEarth , the orbital speed of the Earth. We can obtain it by eliminating vstar
from the above two equations. Please show the mathematical steps involved in the elimination process. I am writing the result below
vEarth ? vrr ? vrb . 2
Now plug in the values of vrr and vrb you found above and find vEarth . Remember to include –ve sign in vrb.
This speed is obtained by assuming that the Arcturus is located at ecliptic plane (In actuality, Arcturus lies at 31o latitude.). To get the true orbital value you need to divide above result by 0.86 because it is equal to cos31o. Compare your final result with the actual value of the orbital velocity of the Earth found in book or internet. Don’t forget to discuss the possible sources of error in the experiment while writing the conclusion section of your report.

Orbital Velocity of the Earth Using Stellar Spectra

Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
Experiment #4 — Orbital Velocity of the Earth Using Stellar Spectra
In this exercise you will find the orbital velocity of the Earth by measuring the Doppler shift in the spectrum of the star Arcturus. Among the very brightest of stars, shining with a soft orange light, Arcturus lights northern spring skies. While orbiting the Sun, the Earth moves towards the Arcturus half of its orbital period and it moves away from the star rest of the period as shown in figure. When the Earth is moving towards and away from the star, the observed light is respectively blue-shifted and red-shifted .
(a) Earth m Sun
(b) Earth moving towards
The photocopied figure (taken from SKY AND TELESCOPE) below displays two spectrograms of the star Arcturus taken six months apart. The top and the bottom spectrograms are of a reference light source at rest with respect to the telescope. Corresponding to each of the reference lines is a dark line in the star spectrum, displaced slightly to the right (red-shifted) in a, and to the left (blue-shifted) in b.
Select line 1 and measure the displacement of it in a and b to tength of millimeters. One way to do it is to use a straight edge to connect matching reference lines as shown in the figure on the lower right corner on the next page. The shift are shown by filled rectagles (red in a and blue in b) in the figure.Measure the shifts with a scale estimating to tength a tength of millimeter. A magnifying glass would be handy. Try to be as accurate as possible. The measurements give the shift ?? in mm for the line. Draw a table similar to one shown below and note down the measurements. Repeat the process for lines 1, 3, 5, and 7 and enter into the table.
???oving away
???????A
?rcturus
r
???????????????
line no.
??
??
original wavelength ?0 angstrom
shift ??
shift ??
vr ???c ?0
?
spectrum a mm
spectrum b mm
spectrum a angstrom
spectrum b angstrom
spectra a
spectra b
1
4260.48
Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
????????????????????????????Fig: Spectrum of Arcturus, when the Earth is at (a) and (b) positions.
The measurments you obtained are in millimeters in paper scale. These are not in the angstrom scale in which the the reference wavelengths are measured. You need to convert your measurements to the angstrom scales. By the way, 1 angstrom = 10-10 m. To do this, measure the distance between lines 1 and 7 in millimeters with the same accuracy mentioned in the previous paragraph. Divide the difference in actual wavelengths of line 1 (4260.48 angstroms) and line 7 (4307.91 angstroms) by the distance between lines 1 and 7.
???i.e., Conversion Factor (CF) ? ?4307.91 – 4260.48? angstroms /millimeter ???
It serves as the conversion factor to obtain the shift in wavelengths in angstroms. Multiply the numbers you have in the third column and fourth coulumn in mm of the above table by the CF and write down the results in fifth and sixth column of the table respectively. The new nubmers are the shifts in angstrom. Now we know the shift in spectrum due to the relative motion between source of light (Arcturus ) and the observer on the Earth. Now you can use the Doppler’s effect Eq.(6.3) from you text book to find the relative velocity between the Earth and the star.
a
???????????vr ? ?? ? ? ? b c ?0
where c = 300000 km/s is the speed of light and ?0 is the wavelength of
corresponding reference line. These values for line 2, 3, 4, 5 and 6 are respectively, 4271.16, 4271.76, 4282.41, 4294.13, and 4299.24.
????
Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
You can use above formula for both red-shift and blue-shift. Find the average vr from the second last column of the table. Let’s denote this number by vrr.
vrr = ……. km/s.
It represents the average relative velocity obtained from the red-shift in the spectrum a. This
velocity is related to speed of the star and the speed of the Earth by
vrr ?vStar ?vEarth.
Similarly, find the average vr from the last column of the table and denote it by
vrb = – ……. km/s.
Notice there is negative sign. We are adding this negative sign here for convenience. Actually it should be with the numbers in the last column of the table because the shift ?? is negative for the blue shift. Usually displacement in the right is taken as positive and in the left is taken as negative.
vrb ? vStar ? vEarth .
Our goal is to find vEarth , the orbital speed of the Earth. We can obtain it by eliminating vstar
from the above two equations. Please show the mathematical steps involved in the elimination process. I am writing the result below
vEarth ? vrr ? vrb . 2
Now plug in the values of vrr and vrb you found above and find vEarth . Remember to include –ve sign in vrb.
This speed is obtained by assuming that the Arcturus is located at ecliptic plane (In actuality, Arcturus lies at 31o latitude.). To get the true orbital value you need to divide above result by 0.86 because it is equal to cos31o. Compare your final result with the actual value of the orbital velocity of the Earth found in book or internet. Don’t forget to discuss the possible sources of error in the experiment while writing the conclusion section of your report.

Orbital Velocity of the Earth Using Stellar Spectra

Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
Experiment #4 — Orbital Velocity of the Earth Using Stellar Spectra
In this exercise you will find the orbital velocity of the Earth by measuring the Doppler shift in the spectrum of the star Arcturus. Among the very brightest of stars, shining with a soft orange light, Arcturus lights northern spring skies. While orbiting the Sun, the Earth moves towards the Arcturus half of its orbital period and it moves away from the star rest of the period as shown in figure. When the Earth is moving towards and away from the star, the observed light is respectively blue-shifted and red-shifted .
(a) Earth m Sun
(b) Earth moving towards
The photocopied figure (taken from SKY AND TELESCOPE) below displays two spectrograms of the star Arcturus taken six months apart. The top and the bottom spectrograms are of a reference light source at rest with respect to the telescope. Corresponding to each of the reference lines is a dark line in the star spectrum, displaced slightly to the right (red-shifted) in a, and to the left (blue-shifted) in b.
Select line 1 and measure the displacement of it in a and b to tength of millimeters. One way to do it is to use a straight edge to connect matching reference lines as shown in the figure on the lower right corner on the next page. The shift are shown by filled rectagles (red in a and blue in b) in the figure.Measure the shifts with a scale estimating to tength a tength of millimeter. A magnifying glass would be handy. Try to be as accurate as possible. The measurements give the shift ?? in mm for the line. Draw a table similar to one shown below and note down the measurements. Repeat the process for lines 1, 3, 5, and 7 and enter into the table.
???oving away
???????A
?rcturus
r
???????????????
line no.
??
??
original wavelength ?0 angstrom
shift ??
shift ??
vr ???c ?0
?
spectrum a mm
spectrum b mm
spectrum a angstrom
spectrum b angstrom
spectra a
spectra b
1
4260.48
Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
????????????????????????????Fig: Spectrum of Arcturus, when the Earth is at (a) and (b) positions.
The measurments you obtained are in millimeters in paper scale. These are not in the angstrom scale in which the the reference wavelengths are measured. You need to convert your measurements to the angstrom scales. By the way, 1 angstrom = 10-10 m. To do this, measure the distance between lines 1 and 7 in millimeters with the same accuracy mentioned in the previous paragraph. Divide the difference in actual wavelengths of line 1 (4260.48 angstroms) and line 7 (4307.91 angstroms) by the distance between lines 1 and 7.
???i.e., Conversion Factor (CF) ? ?4307.91 – 4260.48? angstroms /millimeter ???
It serves as the conversion factor to obtain the shift in wavelengths in angstroms. Multiply the numbers you have in the third column and fourth coulumn in mm of the above table by the CF and write down the results in fifth and sixth column of the table respectively. The new nubmers are the shifts in angstrom. Now we know the shift in spectrum due to the relative motion between source of light (Arcturus ) and the observer on the Earth. Now you can use the Doppler’s effect Eq.(6.3) from you text book to find the relative velocity between the Earth and the star.
a
???????????vr ? ?? ? ? ? b c ?0
where c = 300000 km/s is the speed of light and ?0 is the wavelength of
corresponding reference line. These values for line 2, 3, 4, 5 and 6 are respectively, 4271.16, 4271.76, 4282.41, 4294.13, and 4299.24.
????
Department of Physics, University of Colorado at Denver Physics 1052 (General Astronomy)
You can use above formula for both red-shift and blue-shift. Find the average vr from the second last column of the table. Let’s denote this number by vrr.
vrr = ……. km/s.
It represents the average relative velocity obtained from the red-shift in the spectrum a. This
velocity is related to speed of the star and the speed of the Earth by
vrr ?vStar ?vEarth.
Similarly, find the average vr from the last column of the table and denote it by
vrb = – ……. km/s.
Notice there is negative sign. We are adding this negative sign here for convenience. Actually it should be with the numbers in the last column of the table because the shift ?? is negative for the blue shift. Usually displacement in the right is taken as positive and in the left is taken as negative.
vrb ? vStar ? vEarth .
Our goal is to find vEarth , the orbital speed of the Earth. We can obtain it by eliminating vstar
from the above two equations. Please show the mathematical steps involved in the elimination process. I am writing the result below
vEarth ? vrr ? vrb . 2
Now plug in the values of vrr and vrb you found above and find vEarth . Remember to include –ve sign in vrb.
This speed is obtained by assuming that the Arcturus is located at ecliptic plane (In actuality, Arcturus lies at 31o latitude.). To get the true orbital value you need to divide above result by 0.86 because it is equal to cos31o. Compare your final result with the actual value of the orbital velocity of the Earth found in book or internet. Don’t forget to discuss the possible sources of error in the experiment while writing the conclusion section of your report.