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5 million kilometers past the orbit of the Earth. The second telescope is a 2 m telescope that is planned to be placed on the far side of the moon (average distance to the moon is 380,000 km from the Earth).
(a). Which telescope will have the greater light gathering power (10 m or 2 m)?
The 2m telescope will have greater light gathering power.
(b). Explain your answer to part (a).
The L2 Langrangian point by definition would be in line with the Earth and the Sun, with the Earth always between the Sun and the telescope, limiting the amount of light that reaches the telescope and thus its ability to collect light.
If the 10m telescope were placed on the Earth on top of Mauna Kea in Hawaii at 14000 feet instead of at L2, which telescope NOW has the greater light gathering power, by what factor
(c). The 10 m telescope has a ____ (greater/smaller) light gathering power
Greater
(d). By a factor of ____ times.
8
(e). Compare your answers to parts (c) and (d) to your answer to part (a). Explain any differences.
A 2m telescope has an aperture area of 9.87, while a 10m telescope has an area of 78.54 -- about eight times...
The difference in distances would be negligible in terms of light gathering in light of the significant size differences.
There was a great debate concerning the value of the Hubble constant. (NOTE: The presently accepted value of the Hubble constant is 72 km/s/Mpc as listed in your formula sheet.) One party believed the Hubble constant (Ho) was closer to 50 km/s/Mpc and the other party believed the Hubble constant was closer to 100 km/s/Mpc. Assuming you measure a recessional velocity of a galaxy of 10,000 km/s, what is the distance to that galaxy given the two different Hubble constants
(a). Distance to galaxy using 100 km/s/Mpc = ____ Mpc
The equation v=Hd, where v is velocity, H is the constant, and d is distance, describes the relationship between observed velocity and galaxy distance. Rearranging to solve for distance yields d=v/H, or in this case d=10,000km/s/100km/s/Mpc, or d=100Mpc.
(b). Distance to galaxy using 50 km/s/Mpc = ____ Mpc
The same equation with 50km/s/Mpc used as the constant is d=10,000km/s/50km/s/Mpc or d=200Mpc.
(c). How do
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