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