Universe
Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 20, Problem 62Q

(a)

To determine

The brightness of the star Betelgeuse in terms of a fraction of the Sun’s brightness. It is given that Betelgeuse is transformed from a red supergiant to a Type II supernova at the distance of 425 ly from Earth.

(a)

Expert Solution
Check Mark

Answer to Problem 62Q

Solution:

7.1×107b

Explanation of Solution

Given data:

The distance of the star from Earth is 425ly.

Formula used:

The expression for apparent magnitude of a supernova is,

m=M+5logd5

Here, m is the apparent magnitude, M is the absolute magnitude of Type II supernova and d is the distance of the star from Earth in parsecs.

The expression for ratio of brightness of two objects is,

b1b2=10(m1m2)2.5

Here, b1andb2 are the brightness and m1andm2 are the apparent magnitudes of the objects.

Explanation:

Convert the distance from light years to parsec as follows:

1ly=0.3pc

Therefore, the provided distance of the star from Earth in parsecs is,

d=425 ly(0.3 pc1ly)=130 pc

Write the formula for apparent magnitude of Type II supernova.

m=M+5logd5

The absolute magnitude for Type II supernova is 17. Substitute 17 for m and 130 for d.

m=17+5log(130)5=11.4

The apparent magnitude of the Sun is 26.8.

Write the expression for the brightness ratio of Betelgeuse and Sun.

bbS=10(mmS)2.5

Here, the subscript S refers to the corresponding quantities for the Sun and b is the brightness of Betelgeuse.

Substitute 26.8 for mS and 11.4 for m.

bBbS=1011.4(26.8)2.5=1015.42.5bB=7.1×107b

Conclusion:

So, the supernova is 7.1×107 times brighter than the Sun.

(b)

To determine

The comparison between the brightness of the supernova and that of Venus. It is given that it is transformed from a red supergiant to Type II supernova at the distance of 425 ly from Earth and the brightness of Venus is 109b.

(b)

Expert Solution
Check Mark

Answer to Problem 62Q

Solution:

The ratio of the brightness of the supernova to that of Venus is 710.

Explanation of Solution

Given data:

The brightness of Venus is 109b.

The distance of the star from Earth is 425ly.

Formula used:

The expression for apparent magnitude of a supernova is,

m=M+5logd5

Here, m is the apparent magnitude, M is the absolute magnitude of Type II supernova and d is the distance of the star from Earth in parsecs.

The expression for the ratio of brightness of two objects is,

b1b2=10(m1m2)2.5

Here, b1andb2 are the brightness and m1andm2 are the apparent magnitudes of the objects.

Explanation:

Refer to part (a). The brightness of the star with respect to that of the Sun is 7.1×107b.

In order to compare the brightness of the star with that of Venus, determine the ratio of their respective brightness (relative to the Sun), that is,

ratio=bbV

Substitute 7.1×107b for b and 109b for bV.

ratio=7.1×107b109b=710

Conclusion:

So, the supernova is 710 times brighter than Venus.

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