Universe - Text Only (Looseleaf)
Universe - Text Only (Looseleaf)
11th Edition
ISBN: 9781319115012
Author: Freedman
Publisher: MAC HIGHER
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Chapter 22, Problem 26Q

(a)

To determine

The orbital period of the cloud, if a gas cloud located in the spiral arm of a distant galaxy has an orbital velocity of 400 km/s while moving in a circular orbit and the distance from the center of the galaxy is 20,000 pc.

(a)

Expert Solution
Check Mark

Answer to Problem 26Q

Solution:

3.1×108yr

Explanation of Solution

Given data:

For a gas cloud located in the spiral arm of a distant galaxy, the orbital velocity is 400km/s and the distance from the center of the galaxy is 20,000pc.

Formula used:

The expression for time taken by an object in circular motion to complete an orbit is,

P=2πrv

Here, P is the orbital period, r is the distance from the center of the circular orbit and v is the orbital speed.

Conversion from km/s to m/s is performed as,

1km/s =1000m/s

Conversion from pc to m is performed as,

1pc =3.09×1016m

Conversion from 1 s to 1 year is performed as,

1year =365days1 day =24hours1 hour =60minute1minute=60second

Hence, 1year =3.16×107s

Explanation:

Write the expression for the period of the cloud’s orbit around the galactic center.

P=2πrv

Substitute 400km/s for v and 20,000pc for r.

P=2π(20,000pc)400km/s=2π(20,000pc(3.09×1016m1pc))(400km/s(1000m/s1m/s))=2π(6.18×1020m)(4×105m/s)=9.71×1015s

Further, solve as,

P=9.71×1015s(1yr3.15×107s)3.1×108 yr

Conclusion:

Therefore, the orbital period of the cloud is 3.1×108yr.

(b)

To determine

The mass of the galaxy contained within the cloud’s orbit, if a gas cloud located in the spiral arm of a distant galaxy has an orbital velocity of 400 km/s while moving in a circular orbit and the distance from the center of the galaxy is 20,000 pc.

(b)

Expert Solution
Check Mark

Answer to Problem 26Q

Solution:

7.4×1011 M

Explanation of Solution

Given data:

For a gas cloud located in the spiral arm of a distant galaxy, the orbital velocity is 400km/s and the distance from the center of the galaxy is 20,000pc.

Formula used:

The expression for mass of our galaxy within the cloud is,

M=rv2G

Here, r is the distance from the center of the galaxy when the cloud is moving in circular orbit, v is the orbital speed of the cloud and G is the constant of gravitation.

The value of G is 6.673×1011 m3/kgs2

Conversion from km/s to m/s is performed as,

1km/s =1000m/s

Conversion from pc to m is performed as,

1pc =3.09×1016m

Explanation:

Recall the expression for mass of our galaxy within the cloud.

M=rv2G

Substitute 400km/s for v, 20,000pc for r and 6.673×1011 m3/kgs2 for G.

M=(20,000pc(3.09×1016m1pc))(400km/s(1000m/s1km/s))26.673×1011 m3/kgs2=(6.18×1020m)(4×105m/s)26.673×1011 m3/kgs2=14.8×1041kg(1 M2×1030 kg)=7.4×1011 M

Conclusion:

Therefore, the mass of our galaxy within the cloud is 7.4×1011 M

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If astronomers find they have made a mistake and our Solar System is actually 7,350 pc rather than 8,300 pc from the center of the Milky Way Galaxy, but the orbital velocity of the Sun is still 225 km/s, what is the minimum mass (in solar masses, MSun) of the Galaxy within the orbit of the Sun?
You observe a star orbiting in the outer parts of a galaxy. The distance to this galaxy is known, and you are able to take a spectra of this star and determine its velocity. The star is 22 kpc from the galaxy center and moving in a circular orbit with speed 304 km/s. Compute the total mass of the galaxy internal to the star's orbit. You will get a large number; express it in scientific notation and in units of solar masses [e.g., 4.2e10]. [Hint: there is a Box in Chapter 22 of your textbook that will be of help. See also the course formula sheet.]
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