Universe: Stars And Galaxies
Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 20, Problem 45Q
To determine

(a)

Themass of the core.

Expert Solution
Check Mark

Answer to Problem 45Q

The mass of the core in kilograms is 1.676×1030kg, and in solar mass, the mass is 0.84M.

Explanation of Solution

Given:

The diameter of the core is, d=20km.

The density of the core is, ρ=4×1017kg/m3.

Formula used:

The expression for the volume of the core is given by,

V=43πr3

Here, r is the radius of the core.

The expression for the density of the core is given by,

ρ=MV

Calculation:

The radius of the coreis calculated as,

r=d2=20km2=10km

The volume of the core is calculated as,

V=43πr3=43(π)(10km)3=43(π)(10km× 1000m 1km)3=4.19×1012m3

The mass of the core is calculated as,

M=ρV=(4× 10 17kg/ m 3)(4.19× 10 12m3)=(1.676× 10 30kg× 1 M 1.989× 10 30 kg)=0.84M

Conclusion:

The mass of the core in kilograms is 1.676×1030kg; and in solar mass, the mass is 0.84M.

To determine

(b)

Theforce of gravity on a 1kg object at the surface of the core and to compare it with the force if the object was kept on Earth.

Expert Solution
Check Mark

Answer to Problem 45Q

The force of gravity on the surface of the core is 1.12×1012N and is approximately 1011 times the force on the surface of the Earth.

Explanation of Solution

Given:

The radius of the core is, r=10km.

The mass of the object is, m=1kg.

Formula used:

The gravitational force between the two bodies is given by,

F=G(mMr2)

Here, M is the mass of the core.

Calculation:

The gravitational force between the two bodies is calculated as,

F=G( mM r 2 )=(6.67× 10 11N m 2/ kg 2)[( 1kg)( 1.676× 10 30 kg) ( 10km ) 2]=(6.67× 10 11N m 2/ kg 2)[( 1.676× 10 30 kg 2 ) ( 10km× 1000m 1km ) 2]=1.12×1012N

The ratio of the gravitational force on the surface of the core of the supernova and on the surface of the Earth is calculated as,

FcFe=1.12× 10 12N10N=1.12×1011N

Conclusion:

The force of gravity on the surface of the core is 1.12×1012N and is approximately 1011 times the force on the surface of the Earth.

To determine

(c)

The escape speed from the surface of the star’s core.

Expert Solution
Check Mark

Answer to Problem 45Q

The escape speed from the surface of the star’s core is 1.5×108m/s or 0.5c. Therefore, the escape speed is very large and has a value equal to half the speed of light. Thus, in order for the particles to escape the surface of the core, they have to attain a speed greater than half the speed of the light. This indicates that the explosion should be very large if the particles have to attain such a velocity.

Explanation of Solution

Given:

The radius of the core is, r=10km.

The mass of the object is, m=1kg.

Formula used:

The escape velocity is given by,

vc=2GMr

Calculation:

The escape velocity is calculated as,

vc= 2GMr= 2( 6.67× 10 11 N m 2 / kg 2 )( 1.676× 10 30 kg ) ( 10km )= 2.24× 10 20 N m 2 / kg ( 10km× 1000m 1km )=1.5×108m/s

Solve further,

vc=(1.5× 108m/s× 1c 3× 10 8 m/s )=0.5c

Conclusion:

The escape speed from the surface of the star’s core is 1.5×108m/s or 0.5c .Therefore, the escape speed is very large and has a value equal to half the speed of light. Thus, in order for the particles to escape the surface of the core, they have to attain a speed greater than half the speed of the light. This indicates that the explosion should be very large if the particles have to attain such a velocity.

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