The Hubble space telescope can detect stars with apparent magnitudes of 29. How

Question

The Hubble space telescope can detect stars with apparent magnitudes of 29. How far away could e Sun be and still be detected with the telescope? Cepheid variable stars are very important distance indicators because they have large and well-known iminosities. How far away could a Cepheid variable with 20,000 times the luminosity of the Sun be, and still be detected with the Hubble Space Telescope? Express your answer in light years and compare it to the size of the Milky Way of approximately 100,000 light years

Explanation / Answer

The apparent magnitude of Sun which is to be made : m = 29

Absolute magnitude of Sun = M = 4.83

M = m - 2.5log(d/10)2

=> 4.83 = 29 - 2.5log(d/10)2

=> 9.668 = log(d/10)2

=> d = 682338.7 parsecs = 2224424.14 light years.

[where 1 parsec = 3.26 light years].

For the Cepheid variable star:

Absolute magnitude = - 3.6 and desired apparent magnitude = 29.

Substitute these values in the expression above and obtain distance d.

this comes out to be: d = 3.3113 x 107 parsecs = 1.0795 x 108 light years = 1079.5 times the size of the Milky Way.

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