EBK LOOSE-LEAF VERSION OF UNIVERSE
EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
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Chapter 20, Problem 30Q

(a)

To determine

The average density of a 1MΘ dwarf star whose diameter is same as that of

Earth.

(a)

Expert Solution
Check Mark

Answer to Problem 30Q

1.83×109kgm3

Explanation of Solution

Given data:

The mass of the star is 1MΘ and the radius of the star is 6378km.

Formula used:

The average density of the star can be calculated by using the following relation:

ρ¯=MV

Here, ρ¯ is the average density of the star, M is the mass of the star and V is the volume of the star.

Write the expression for the volume of a sphere.

V=43πR3

Here, V and R are the volume and radius of a spherical object, such as a star, respectively.

Explanation:

Since the mass of the star is given in solar mass, convert it into standard the mass unit, that is, kilograms.

1M=1.989×1030kg

Recall the expression for the volume of a sphere.

V=43πR3

Substitute 6.378×106 m for R.

V=43π(6.378×106 m)3=1.059×1021m3

Recall the expression for the average density of a star.

ρ¯=MV

Substitute 1.98×1030 kg for M and 1.059×1021 m3 for V.

ρ¯=1.98×1030 kg1.059×1021 m3=1.83×109kgm3

Conclusion:

The average density of the dwarf star, ρ¯=1.83×109kgm3.

(b)

To determine

Whether or not the star’s density is the same as the density of about an elephant per teaspoon, along with proper calculation.

(b)

Expert Solution
Check Mark

Answer to Problem 30Q

No, the star’s density is approximately twice that of about an elephant per teaspoon.

Explanation of Solution

Given data:

Let us consider a general elephant with mass 4000 kg and an ordinary teaspoon with volume Vts=5×106m3.

Formula used:

Write the expression of the density as -

ρ=mVts

Here, ρ is the density of about an elephant per teaspoon, m is the mass of an elephant and Vts is the volume of a teaspoon.

Explanation:

The density will be calculated as:

ρ=mVts

Substitute 4×103kg for m and 5×106m3 for Vts in the expression of density.

ρ=4×103kg5×106m3=8.1×108kg/m3

Compare ρ¯ (calculated in a-part) and ρ(calculated above).

ρ¯=2ρ

Conclusion:

The density of the dwarf star is approximately twice that of about an elephant per teaspoon. So, the provided statement is incorrect.

(c)

To determine

The escape speed required for a gas to eject from the star’s surface.

(c)

Expert Solution
Check Mark

Answer to Problem 30Q

6450km/s

Explanation of Solution

Given data:

The mass of the star is 1MΘ and the radius of the star is 6378km.

Formula used:

Write the expression for escape speed.

ve=2GMR

Here, ve is the escape speed of the gas, G is the gravitational constant and R is the radius of the star.

Explanation:

The escape speed is the minimum speed required to send an object out of the gravitational field of the body it is being sent from.

Recall the expression for escape speed.

ve=2GMR

Substitute 6.67×1011Nm2/kg2 for G, 1.98×1030 kg for M and 6.378×106 m for R in the expression of escape velocity.

ve=(2(6.67×1011Nm2/kg2)1.98×1030 kg6.378×106 m)12=6.45×106m/s(1 km/s1000 m/s)=6450km/s

Conclusion:

The escape speed for the gas is 6450km/s.

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Chapter 20 Solutions

EBK LOOSE-LEAF VERSION OF UNIVERSE

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