## Monday, November 7, 2016

### Solutions To Astronomy Angular Measure Problems

1) Two observers using cross staffs obtain zenith distances from their respective locations of   φ = 45 degrees, and  φ =  35 degrees. How far apart in latitude are their locations?

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: