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The gas cloud known as the Crab Nebula can be seen with

even a small telescope. It is the remnant of a supernova, a

cataclysmic explosion of a star. The explosion was seen

on the Earth on July 4, 1054 AD. The streamers in the

figure glow with the characteristic red color of heated

hydrogen gas. In the laboratory on earth, heated hydrogen

produces red light with frequency 4.568 × 1014 Hz; the red

light received from the streamers in the Crab Nebula

pointed at the Earth has a frequency of 4.586 × 1014 Hz.

a) Assume that the speed of the center of the nebula relative to the Earth is negligible.

Estimate the speed with which the outer edges of the Crab Nebula are expanding.

b) Assuming that the expansion has been constant since the supernova explosion, estimate the

average radius of the Crab Nebula in year 2006. (Immediately after the explosion, the size of the

Crab Nebula was neligible.) Give your answer in light years.

The angular size of the Crab Nebula, as seen from the earth, is about 4 arc minutes by 6 arc

minutes, for an average of 5 arc minutes. (1 arc minute = 1/60 of a degree)

c) Estimate the distance (in light years) to the Crab Nebula.

d) Estimate in what year BC the supernova explosion actually took place.

A light-year (ly) is the distance traveled by light in one year.

You may find it helpful to use the the first terms of the binomial expansion: (1+ε)^p ≅ 1 + pε

Hi, I'm not too sute how to approach this question but I think it has soething to do with the Doppler effect for light. If this is the case I would employ the following formula:

f_r = sqrt((c-v/c+v))(f_s)

where f_r is the frequency measured by the reciever and f_s the frequency of the source. What confuses me here is that both the frequencies are the same so I'm not sure how to set up the problem.

But, since they say to assume that the speed of the center of the nebula relative to the Earth is negligible, could I write the formula as follows?f_r = (1 + v/c)^(-1/2)

4.586 × 1014 Hz = (1 + v/(3.0 x 10^8))^-1/2

before I start solving for v, could you guys give me any hints on whether my setup is correct or not?

thank you !

even a small telescope. It is the remnant of a supernova, a

cataclysmic explosion of a star. The explosion was seen

on the Earth on July 4, 1054 AD. The streamers in the

figure glow with the characteristic red color of heated

hydrogen gas. In the laboratory on earth, heated hydrogen

produces red light with frequency 4.568 × 1014 Hz; the red

light received from the streamers in the Crab Nebula

pointed at the Earth has a frequency of 4.586 × 1014 Hz.

a) Assume that the speed of the center of the nebula relative to the Earth is negligible.

Estimate the speed with which the outer edges of the Crab Nebula are expanding.

b) Assuming that the expansion has been constant since the supernova explosion, estimate the

average radius of the Crab Nebula in year 2006. (Immediately after the explosion, the size of the

Crab Nebula was neligible.) Give your answer in light years.

The angular size of the Crab Nebula, as seen from the earth, is about 4 arc minutes by 6 arc

minutes, for an average of 5 arc minutes. (1 arc minute = 1/60 of a degree)

c) Estimate the distance (in light years) to the Crab Nebula.

d) Estimate in what year BC the supernova explosion actually took place.

A light-year (ly) is the distance traveled by light in one year.

You may find it helpful to use the the first terms of the binomial expansion: (1+ε)^p ≅ 1 + pε

Hi, I'm not too sute how to approach this question but I think it has soething to do with the Doppler effect for light. If this is the case I would employ the following formula:

f_r = sqrt((c-v/c+v))(f_s)

where f_r is the frequency measured by the reciever and f_s the frequency of the source. What confuses me here is that both the frequencies are the same so I'm not sure how to set up the problem.

But, since they say to assume that the speed of the center of the nebula relative to the Earth is negligible, could I write the formula as follows?f_r = (1 + v/c)^(-1/2)

4.586 × 1014 Hz = (1 + v/(3.0 x 10^8))^-1/2

before I start solving for v, could you guys give me any hints on whether my setup is correct or not?

thank you !

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