Chp 3

 

Chapter 3, Problem 3.

A 2-m telescope can collect a given amount of light in 1 hour. Under the same observing conditions, how much time would be required for a 6-m telescope to perform the same task?

Solution: About 7 minutes.
Mathematical Formula 15 tells us that the time to collect light is inversely proportional to the square of the diameter of the mirror:
.
For this problem the diameters are 2 meters for the small telescope and 6 meters for the large telescope. Thus the ratio is: ( 2 m / 6 m)² = (1/3)² = (1/9). Since it takes 1 hour for the small telescope to collect the light, it takes (1/9 hr)(60 min /hr) = 7 minutes for the large mirror to collect the light.

Chapter 3, Problem 4.

A certain space-based telescope can achieve (diffraction-limited) angular resolution of 0.05" for red light (of wavelength 700 nm). What would its resolution be (a) in the infrared, at 3.5 µm; and (b) in the ultravioletm at 140 nm?

Solution: (a) 0.25"; (b) 0.01".
The resolving power of a telescope can be found from the formula:


The formula includes both the wavelength of the light, and the diameter, D, of the telescope mirror. Here we need to solve for D first and then determine the resolution at other wavelengths.
D = (0.25) ( / ). In this case = 0.05 (in seconds of arc) and = 700 nm = 0.7 µm.
D = (0.25)(0.7 / 0.05) = 3.5 m.
Now solving the other parts for : (a) = (0.25)(3.5 / 3.5) = 0.25". (b) = 140 nm = 0.14 µm. = (0.25)(0.14 / 3.5) = 0.01".

CHP3 Problem 5 

Two identical stars are moving in a circular orbit around one another, with an orbital separation of 2 AU (Sec. 2.6). The system lies 200 light-years from Earth. If we happen to view the orbit face-on,how large a telescope would we need to resolve the stars, assuming diffraction-limited optics at a wavelength of 2 µm?

200 light years is 61.3 pc. From the definition of the parsec, 1 AU seen from 1 pc subtends an angle of 1" so 2 AU at 1 pc is 2". For a constant baseline the angle scales directly as the distance so the angular separation of 2 AU at 61.3 pc will be 1/61.3 times the angle at 1 pc (2") or 2/61.3 = 0.033". At a wavelength of 2 µm, solve for the mirror diameter = 0.25 (2 /0.033) = 15 m.

 

Chapter 3, Problem 8.

The Andromeda galaxy lies about 700,000 pc away. To what distancess do the angular resolutions of HST (0.05") and a radio interferometer (0.001") correspond at this distance?
Solution: HST: 0.17 pc; radio interferometer: 0.0034 pc = 700 AU.
This problem makes use of the small-angle formula:

We are solving here for d, the physical size of an object seen with the angular resolution given in the problem. Here D = 700,000 pc, so:
d = D ( / 206,265").
For the HST, = 0.05", so d = (700,000 pc)(0.05" / 206,265") = 0.17 pc.
For the radio interferometer, = 0.001", so d = (700,000 pc)(0.001" / 206,265") = (0.0034 pc)(3.09 X 10¹³ km / 1 pc)(1 AU/ 1.5 X 10⁸ km) = 700 AU.


Chapter 3 problem 10

angular resolution (arc sec) =		wavelength (µm)
	0.25	,
		telescope diameter (m)

 lambda=velocity per frequency
lambda=3*10^8 / 5*10^9
lambda=3/50

a-)0.25*10^(6)*3/50/5*10^6=0.003 arc sec

b-)0.25*1/50=0.005






CHP 4

problem 3

G=6.67*10^-11    mass earth=6*10^24

(M)m astroid=2*10^-4*6*10^-24=12*10^20 kg
r astroid= 0.073*6*10^6=438*10^3 meter

F=(GMm)/ (r**2) =0.03*10^3=30 N

F=ma=>  30=10*m  m=3 kg