Write out both solutions and the steps taken to arrive at them. Box all final an
ID: 112202 • Letter: W
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Write out both solutions and the steps taken to arrive at them. Box all final answers. On September 6th, the Sun is at (RA 11 h 4 m, Dec. 6.0 degree). It crosses the meridian at 1: 26 PM as seen by Lincoln (latitude 41 degree). A. What is the Sun's altitude at 1: 26 PM? B. What is the sidereal time at 1: 26 PM? C What altitude and azimuth will the Sun be at when class ends at 3: 15 PM? Cassini measured Saturn's moon, Hyperion, as having a spin period of 5.1 days and an orbital period of 21.3 days around Saturn when it visited the moon in 2005. Using the math on sidereal and synodic periods, find how long it takes Saturn to make one full cycle in Hyperion's sky (rise, set, then rise again). This is a synodic period. Imagine we're sending a spacecraft to Jupiter. It is starting free of the Earth's gravity, but at 1.0 AU. It needs to go to Jupiter at 5.2 AU. Assume both planets are in circular orbits. A. We're going to construct a transfer orbit for our spacecraft. It will have aphelion at Jupiter's distance and perihelion at Earth's. What are the semi-major axes and eccentricity of this orbit? B. To get the spacecraft onto the transfer orbit, we need to change its speed from Earth's orbital speed to the right speed for this orbit. (They are in the same direction, so we can use scalar addition). How much is this difference? C. Once the spacecraft gets to Jupiter, we need to change orbits again to get it on an orbit like Jupiter's (so it can go into Jupiter orbit). How much does the spacecraft's speed need to change? D What is the time this trip would take? (Use Kepler's 3rd Law). E How does that compare to the New Horizons spacecraft, which made the Earth-Jupiter trip in 405 days?Explanation / Answer
1. A) Understanding that solar altitude is measured from the southern or northern point along the earth’s horizon; measurement begins at zero degrees and reaches a maximum of 90 degrees when the sun is directly overhead.
Calculate the altitude of the sun at solar noon for any starting latitude point with the following equation:
A = 90 - L +/- D
where A = altitude, L = latitude and D = declination. Enter the latitude of your starting location in degrees. Add D to L if the starting latitude is experiencing summer at the time of calculation; subtract D from L if the starting latitude is experiencing winter. So, in this case, according to the given date (September 6th) we use Add D experiencing summer. Values for D are equal to 23.5 on the June and December solstices and 0 on the March and September equinoxes.
A= 90- (41+23.5)
A= 90 - 64.5
A= 25.5
The altitude at 1:26 pm is 25.5°.
B) Fix the earth-sun line in [inertial] space. Let the earth rotate on its axis once a day, and let the celestial sphere rotate about the celestial poles once a year. As viewed from the north celestial pole, the earth will rotate counter clockwise, and the celestial sphere, clockwise. The rotation rate of the earth is one rotation every 24 solar hours (length of the solar day). The rotation rate of the celestial sphere is one rotation every tropical year (365.2422 days). The relative rotation rate of the earth and celestial sphere is
360o/day + 360o/(365.2422 days)
= 360o/day + 0.9856o/day
= 360.9856o/day
Thus, relative to the earth, the celestial sphere completes one rotation (sidereal day) in something less than a solar day. In fact, the sidereal day is just
(360/360.9856) x 24 hours = 23.9345 hours
= 23 hr 56 min 4.2 sec
i.e., approximately 3 min 56 sec shorter than the solar day. Another way to show the same thing is to form the ratio of the length of the sidereal day to that of the solar day:
(24 hr/solar day)/(23.9345 hr/sidereal day) = 1.0027 sidereal day/solar day
The tropical year is then
(365.2422 solar days) x 1.0027 = 366.2284 sidereal days
Now we may solve the problem of estimating the sidereal time on 1:26 pm, September 6th. From noon on March 21 to noon on September 6th is 168 solar days. From noon to 1:26 pm on September 6th is an additional 1.26/24 = 0.0525 solar days. Hence, the total elapsed time from the vernal equinox to 1:26 pm on September 6th is
168.0525 solar days
Now 168 solar days = 168 x 1.0027 = 168.453 sidereal days. Thus, solar noon on September 6th is
(0.453 sidereal days) x (23.9345 hr/sidereal day) = 10.842 hr
= 10 hr 56 min.
(i.e., a sidereal clock at noon solar time on September 6th reads 2 hr 21 min ahead of a solar clock).
Also, 0.0525 solar days = 0.0525 x 1.0027 = 0.0525 sidereal days (to within the accuracy of the calculation).
(0.0525 sidereal days) x (23.9345 hr/sidereal day) = 1.256 hr
= 1 hr 20 min (to within rounding accuracy)
The sidereal time at 1:26 pm, September 6th is, therefore
(10 hr 56 min) + (1 hr 20 min) = 11 hr 36 min
An alternative approach would be to convert 168.453 solar days into sidereal days in a single step:
(168.453 solar days) x 1.0027 = 168.908 sidereal days
If all clocks start at noon on the vernal equinox, then the sidereal clock, at 1:26 pm solar, September 6th, is reading 0.91 day past sidereal noon or
(0.908 sidereal day) x (23.9345 hr/sidereal day) = 21.732 hr
= 21 hr 50 min
Accuracy in the last digit is due to rounding error.
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