(a) The critical density of matter in the universe.

Question

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Answer 1

Question

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

To determine

(b)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

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Answer 2

Question

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

To determine

(b)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

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Answer 3

Question

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

To determine

(b)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

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Answer 4

Question

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

To determine

(b)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

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Answer 5

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

To determine

(b)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Answer 6

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Answer 7

Chapter 25, Problem 41Q

To determine

(a)

The critical density of matter in the universe.

Answer 8

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The Hubble constant is, H1=50 km/s⋅Mpc.

Formula used:

The critical density of matter in the universe is given by,

ρc=3H28πG

Calculation:

The gravitational constant is, G=6.67×10−11 m3/kg⋅s2

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 50 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 50 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 16.20× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=787.738× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=4.699×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 4.699×10−38 kg/m3.

Answer 13

To determine

(b)

The critical density of matter in the universe.

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Answer 14

To determine

(b)

The critical density of matter in the universe.

Answer 15

Expert Solution

Answer to Problem 41Q

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The Hubble constant is, H1=100 km/s⋅Mpc.

Calculation:

The critical density of matter in the universe is calculated as,

ρc=3H28πG=3 ( 100 kmMpc/ s⋅Mpc )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm )28π( 6.67× 10 −11 m 3 / kg⋅ s 2 )=3 ( 100 km/ s⋅Mpc )2 ( 1 Mpc 3.0856× 10 13 Mkm( 10 6 km 1 Mkm ) )2( 167.63 m 3 / kg⋅ s 2 )

Solve further,

ρc=3 ( 32.40× 10 −19 s −1 )2( 167.63 m 3 / kg⋅ s 2 )=1050.317× 10 −38 s −2( 167.63 m 3 / kg⋅ s 2 )=6.266×10−38 kg/m3

Conclusion:

The critical density of matter in the universe is 6.266×10−38 kg/m3.

Answer 20

Ch. 25 - Prob. 1Q Ch. 25 - Prob. 2Q Ch. 25 - Prob. 3Q Ch. 25 - Prob. 4Q Ch. 25 - Prob. 5Q Ch. 25 - Prob. 6Q Ch. 25 - Prob. 7Q Ch. 25 - Prob. 8Q Ch. 25 - Prob. 9Q Ch. 25 - Prob. 10Q

(a) The critical density of matter in the universe.

Concept explainers

(a)

The critical density of matter in the universe.

Answer to Problem 41Q

Explanation of Solution

(b)

The critical density of matter in the universe.

Answer to Problem 41Q

Explanation of Solution

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Chapter 25 Solutions

(a) The critical density of matter in the universe.

Concept explainers

(a) The critical density of matter in the universe.

Answer to Problem 41Q

Explanation of Solution

(b) The critical density of matter in the universe.

Answer to Problem 41Q

Explanation of Solution

Want to see more full solutions like this?

Chapter 25 Solutions

(a)

The critical density of matter in the universe.

(b)

The critical density of matter in the universe.