Solution:
In each case the latitude is obtained from L = (90 - φ)
So: L1 = (90 - φ1) and L2 = (90 - φ2)
Then the difference in latitude is just the difference in zenith distances or:
L2 - L1 = φ2 - φ1 = 45 deg - 35 deg = 10 deg
2) Consider the system Epsilon Ursae Majoris which semi-major axis subtends an angle of 2½" and for which the parallax of the system is 0."127. Find the semi-major axis in astronomical units. (Hint: p" = 1/d)
Solution:
We have for the semi-major axis of a binary system whose plane is in the observer's visual plane:
a = (a" x d)
where: a" = 2 ½" and: p = 0."127
But p" = 1/d (from hint) so: d = 1/ p" = 1/ 0."127 = 7.87 pc
Then: a = (2 ½" x 7.87 pc) = 19.67 A.U.
3) What telescope magnification would be required to observe the planet Uranus as a disk 2 arcminutes in diameter?
Solution:
Uranus is 4" or 4 arcsec in diameter, unmagnified.
2 arcminutes = 2 x (60 arcsec) = 120 " since 60 arcsec = 1 arcmin
Then: M = 120"/ 4" = 30x
4) Find the Moon's angular diameter in radians.
Solution:
The Moon's angular diameter is 1/2 degree.
One radian (1 rd) can first be converted into arcsec as follows, given there are 3600 arcsec per degree.:
1 rd = 57.3 degrees = 57.3 deg/rad x (3600"/ deg)= 206 280 "
But 1/2 degree = 1800 arcsec = 1800"
Then:
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