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 42Q

(a)

To determine

The orbital period of a star, revolving around Sagittarius A* in years. It is given that the star has a circular orbit of radius 530au and an orbital speed of 2500km/s.

(a)

Expert Solution
Check Mark

Answer to Problem 42Q

Solution:

6.3years

Explanation of Solution

Given data:

Orbital velocity of the star is 2500km/s and the radius of its circular orbit is 530au.

Formula used:

The expression for the orbital period of a star:

P=2πrv

Here, P is the orbital period of a star, r is the radius of its orbit and v is its orbital speed.

Conversion from au to km,

1au =1.496×108km

Conversion from seconds to year,

1year =3.15×107s

Explanation:

Recall the expression for the period of an orbiting object.

P=2πrv

Substitute 2500km/s for v and 530au for r.

P=2π(530au)2500km/s=2π(530au(1.496×108km1au))2500km/s=1.99×108s=6.3 years

Conclusion:

Therefore, the orbital period of the star is 6.3 years.

(b)

To determine

The sum of the masses of Sagittarius A* and the star in terms of solar mass. It is given that the star orbits around Sagittarius A* in a circular orbit of radius 530au with an orbital speed of 2500km/s.

(b)

Expert Solution
Check Mark

Answer to Problem 42Q

Solution:

3.76×106M

Explanation of Solution

Given data:

Orbital velocity of the star is 2500km/s and the radius of its circular orbit is 530au.

Formula used:

Write the expression for Newton’s form of Kepler’s third law.

P2=4π2a3G(M+m)

Here, P is the orbital period of a star, r is the distance of the star from Sagittarius A*(radius of its orbit), v is the orbital speed of the star, M is the mass of Sagittarius A*, m is the mass of the star and G is the constant of gravitation, 6.673×1011 m3/kg-s2.

From the previous subpart, the orbital period of the star is 1.99×108s.

Conversion from au to m,

1au=1.496×1011m

Conversion from mass (in kg.) to solar mass,

1M=1.99×1030kg

Explanation:

In the previous subpart (b), the orbital period of the star was calculated as 1.99×108s.

Recall the expression for Newton’s form of Kepler’s third law.

P2=4π2a3G(M+m)

Rearrange the expression in terms of the masses of the star and Sagittarius A*,

(M+m)=4π2a3GP2

Substitute 1.99×108s for p, 530au for a and 6.673×1011 m3/kg-s2 for G,

M+m=4π2×(530au)3(6.673×1011m3/kg-s2)×(1.99×108s)2=4π2×(530au(1.496×1011m1au))3(6.673×1011 m3/kg-s2)×(1.99×108s)2=1.967×1043m32.642×106m3/kg=7.44×1036kg

The sum of masses in terms of solar mass of the star and Sagittarius A*.

M+m=7.44×1036kg(1M1.99×1030kg)=3.76×106M

Conclusion:

Hence, the sum of masses in terms of solar mass of Sagittarius A* and the star is 3.76×106M.

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