Question

(a) Keck-I has a 10m diameter mirror. HST (Hubble Space Telescope) has a 2.4m mirror.

What is the ratio of the light gathering power of Keck-I to the light-gathering power of HST?

[Compute the ratio (light gathering power of Keck)/(light gathering power of HST) and express as a decimal. Do not use scientific notation.]

(b) What is the ratio of the angular resolution limit of Keck to the angular resolution limit of HST? (Assume they are both observing at the same wavelength.)

[Compute the relevant ratio Keck/HST, express as a decimal, and enter below.]

(c) HST orbits at an altitude of 559 km above the Earth's surface. Assume its orbit is circular.

Compute its orbital period in hours.

(d) Compute its orbital speed (in km/s).

(e) If a satellite like Hubble was looking down at the Earth instead of up, what is the smallest physical size (in cm) on the Earth’s surface that it could resolve? [Hint: assume you are using the visible spectrum.]

Answer 1

a)

dkeck = diameter of keck-I = 10 m

dHST = diameter of HST = 2.4 m

(Light gathering power of Keck)/(Light gathering power of HST) = dkeck 2/dHST2 = 102/2.42 = 17.4

b)

dkeck = diameter of keck-I = 10 m

dHST = diameter of HST = 2.4 m

(Angular resolution limit of of Keck)/(Angular resolution limit of HST) = dkeck/dHST = 10/2.4 = 4.2

c)

h = height above the surface of earth = 559 km = 0.559 x 106 m

R = radius of earth = 6.4 x 106 m

r = radius of orbit = R + h = (6.4 x 106 ) + (0.559 x 106) = 6.959 x 106 m

T = Time period of the orbit

M = mass of earth = 5.98 x 1024 kg

Using Kepler's third law

T2 = 42r3/GM

T2 = 4(3.14)2(6.959 x 106)3/((6.67 x 10-11)(5.98 x 1024 ))

T = 5772.53 sec

T = 5772.53 sec (1h /3600 s)

T = 1.6 h

d)

v = orbital speed

orbital speed is given as

v = 2r/T

v = 2 (3.14)( 6.959 x 106)/5772.53

v = 7570.8 m/s