Physics 1135 Astronomy: Practice Problems for Exam 2

Answers

1. You have travelled to the planet Tatooine, and one day observe the sun is directly overhead at your location. You are in radio contact with a colleague who is 1250 kilometers due north, who reports that the sun is 4 degrees away from the zenith at that location. Determine the circumference of Tatooine.
   (1250 *360)/4 = 112,500
2. The planet Dagobah has a radius of 5000 km. If you observe its moon from two points on the planet that are separated by 5000km, you will observe a parallax angle of 0.6 degrees. What is the distance to the moon in kilometers?
   (180 * 5000)/(3.14159 * 0.6) = 477,470
3. The moon of the planet Dagobah (see previous problem) has an angular diameter of 0.25 degrees. What is the diameter of the moon in kilometers? (477470 * 0.25 * 3.14159)/ 180 = 2080 acceleration of gravity at the surface of the planet.
   a =GM/r^2 The numerator of the fraction is:
   6.67 * 10^{-11} * 4 * 10^24= 2.66 * 10^14
   The denominator of the fraction is:
   (5 * 10^6)^2 = 2.5 *10^13
   The result is:
   (2.66 * 10^14)/(2.5 * 10^13) = 10.7
   The result has units of meters per second squared.
4. (b) Calculate how far an object will fall in 3 seconds on this planet, if air resistance is neglected.
   (10.7 * 3^2)/2 = 48.2
   The result is in meters.
   
   (c) Calculate the escape velocity at the surface of the planet.
   The result is in meters per second.

     
                /----------------------------------
               /  2 * 6.67 * 10^(-11) * 4 * 10^24
              /   ---------------------------------    = 10,330 meters per second
             /                5 * 10^6
           \/
   

5. What is the peak wavelength in the blackbody radiation for these objects: (a) a block of ice (T=273 K); (b) a person with body temperature 310 K; (c) a metal pan heated to the boiling point of water (T=373 K); (d) a metal object heated to 5000 K. Also calculate the frequency that corresponds to each of these wavelengths.
   • (a)
     (2.9 * 10^6)/273 = 10,600
     
   • (b)
     (2.9 * 10^6/310 = 9,350
     
   • (c)
     (2.9 * 10^6)/373 = 7,770
     
   • (d)
     (2.9 * 10^6)/5000 = 580
     In each case the result is in nanometers.
     To get the frequencies, use the formula:
     f=c/lambda
     (a) frequency:
     =2.83 * 10^13
     (b) frequency:
     =3.21 * 10^13
     (c) frequency:
     =3.86 *10^13
     (d) frequency:
     =5.17 * 10^14
     In each case the frequency is in hertz (cycles per second)
     
6. If you know the frequency of a photon, what is the formula that tells its energy?
   E = hf
   h is Planck's constant; f is frequency
7. If a matter/antimatter collision results in 10 kilograms of matter being converted into energy, what is the total amount of energy that will be produced (in joules)?
   E=mc^2
   10 * (3 * 10^8)^2 = 9 * 10^17 joules