resolution

The resolution of a telescope, as we will learn, is the minimal angular size of features it can resolve. If we have a telescope with a resolution of 2'' (2 arc-seconds) what is the size, in kilometers (km), of the smallest crater we can make out on the moon, at a distance of 384,400 km? (The Moon's diameter is 3480 km, for comparison.) Round your answer to two significant figures.


ANS:

          a = 2"
     R = 384400km
   
     a/(360*3600)*(2pi*R)
     2/(360*3600)*(2pi*384400)    #I multiply 360 with 3600 for coversion to arcsec
     = 3.7272km

there is another way:

     (a/3600)*(pi/180)*R
     (2/3600)*(pi/180)*384400
     =3.7272km
  

A thought?

(with a serious face and the background music is as if you have found an invention, you wonder) 

"Sometimes I wonder?" (music stops) "HOW THE HECK AM I SITTING OVER THERE WHEN I AM STANDING OVER HERE!!!!!!!!!!!!!!!!!!!!!"

event horizon

I am going to introduce you guys to a new formula or concept, called the schwarzschild radius it is to find the event horizon of a black hole. It depends on the mass of the black hole. bigger mass bigger radius. 

Formula is:

               R = (2xGxM)/c^2
do not beleive in this formula while looking, think why, how and then believe in it.

related questions on there way.

Q.7 what is the escape velocity of the neutron star in question 6?

Ans:
       m = 3.96x10^30 Kg
       r = 10Km
       G = 6.67x10^-11 N/kg^2
     
       formula = (2 x G x m)/r^2
               = (2 x 6.67x10^-11 x 3.96x10^30) / 10^2
               = 5.28x10^18 km/s

Question 6

Q.6 Take a neutron star with 2Msun but a radius of 10Km! What is the accelaration of gravity at the surface of it? how much greater than g(9.8) is it? Finally how much will you weigh on it?(assume you somehow dont turn into a puddle of protoplasm)?

Ans:

          sun mass = 1.98x10^30
          m = 2 x 1.98x10^30 = 3.96x10^30 Kg
          radius = 10 Km
          G = 6.67x10^-11 N/Kg^2
          m2 = my weight = 40Kg (assuming)
          g = 9.8 m/s^2
          
          formula = (G x m)/r^2
                  = (6.67x10^-11 x 3.96x10^30) / 10^2
                  =  2.64132xx10^18 km/s^2
so:

          Gravity = 2.64x10^18 km/s^2
          g = 9.8 m/s^2
    
          times greater = gravity/g
                        = 2.64x10^18 / 9.8
                        = 2.69x10^17 times   g         
I would weigh:

          m2 = 40 Kg
          F = m x gravity 
            = 40 x 2.64x10^18
            = 1.055x10^20 Kg !!!!!!!!!!!!!! 

Look at this 360 panorama of Mars by Curiosity:

http://www.360cities.net/image/curiosity-mars

binry destruction

This is inter_activity which should be done in a blog. I will ask a problem, we will discuss, and then solve it.

This one interested me so here it is:

A. The star Sirius (the brightest star in the sky) has a white dwarf companion. Sirius has a mass of about
     2  Msun and it is still on the main sequence, while its companion is already a star corpse. Remember that                  
     a white dwarf cant have a mass greater  than 1.4 Msun. (Chandrasekhar limit http://en.wikipedia.org/wiki/Subrahmanyan_Chandrasekhar ). Assuming that they both formed on the same
     time have a group discuss how Sirius could have a white dwarf companion. HINT: Was the initial mass of                                               the white dwarf star more or lesser than Sirius?

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2a

2b

2c