Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 13, Problem 13.70AP

(a)

To determine

The initial speed of the satellite.

(a)

Expert Solution
Check Mark

Answer to Problem 13.70AP

The initial speed of the satellite is 7785.86m/s.

Explanation of Solution

The mass of the satellite is 100kg. The initial altitude of the satellite is 200km and the final altitude of the satellite is 100km.

Formula to calculate the initial speed of the satellite is,

    v1=GMERE+h1

Here, RE is the radius of the earth, h1 is the initial altitude of the satellite, G is the universal gravitational constant and ME is the mass of the Earth.

Substitute 5.972×1024kg for ME, 6.67×1011Nm2/kg2 for G, 200km for h1 and 6371000m for RE to find v1.

    v1=6.67×1011Nm2/kg2×5.972×1024kg6371000m+200km×1000m1km=7785.86m/s

Conclusion:

Therefore, the initial speed of the satellite is 7785.86m/s.

(b)

To determine

The final speed of the satellite.

(b)

Expert Solution
Check Mark

Answer to Problem 13.70AP

The final speed of the satellite is 7845.8m/s.

Explanation of Solution

Formula to calculate the final speed of the satellite is,

  v2=GMERE+h2

Here, h2 is the final altitude of the satellite.

Substitute 5.972×1024kg for ME, 6.67×1011Nm2/kg2 for G, 100km for h2 and 6371000m for RE to find v2.

v2=6.67×1011Nm2/kg2×5.972×1024kg6371000m+100km×1000m1km=7845.8m/s

Conclusion:

Therefore, the final speed of the satellite is 7845.8m/s.

(c)

To determine

The initial energy of the satellite-Earth system.

(c)

Expert Solution
Check Mark

Answer to Problem 13.70AP

The initial energy of the satellite-Earth system is 3.04×109J.

Explanation of Solution

Formula to calculate the initial energy of the satellite-Earth system is,

    E1=12mv12GMEmRE+h1

Here, m is the mass of the satellite.

Substitute 7845.8m/s for v1, 5.972×1024kg for ME, 6.67×1011Nm2/kg2 for G, 200km for h1 and 6371000m for RE to find E1.

  E1=12(100kg)(7785.86m/s)26.67×1011Nm2/kg2×5.972×1024kg×100kg6371000m+200km×1000m1km=3.03099×109J=3.04×109J

Conclusion:

Therefore, the initial energy of the satellite-Earth system is 3.04×109J.

(d)

To determine

The final energy of the satellite-Earth system.

(d)

Expert Solution
Check Mark

Answer to Problem 13.70AP

The final energy of the satellite-Earth system is 3.08×109J.

Explanation of Solution

Formula to calculate the final energy of the satellite-Earth system is,

    E2=12mv22GMEmRE+h2

Substitute 7845.8m/s for v2, 5.972×1024kg for ME, 6.67×1011Nm2/kg2 for G, 100km for h2 and 6371000m for RE to find E2.

    E2=12(100kg)(7845.8m/s)26.67×1011Nm2/kg2×5.972×1024kg×100kg6371000m+100km×1000m1km=3.0778×109J=3.08×109J

Conclusion:

Therefore, the final energy of the satellite-Earth system is 3.08×109J.

(e)

To determine

The mechanical energy of the system has decreased and estimates the amount of decrease mechanical energy of the system.

(e)

Expert Solution
Check Mark

Answer to Problem 13.70AP

The amount of decrease mechanical energy of the system is 4.69×107J.

Explanation of Solution

Formula to calculate the mechanical energy of the system is,

    T=E1E2

T is the amount of decreases mechanical energy of the system.

Substitute 3.0778×109J for E2 and 3.03099×109J for E1 to find T.

    T=(3.0309×109J)(3.0778×109J)=4.69×107J

Conclusion:

Therefore, the amount of decrease mechanical energy of the system is 4.69×107J and proved the mechanical energy of the system will be decreases.

(f)

To determine

What force makes the satellite’s speed increases.

(f)

Expert Solution
Check Mark

Answer to Problem 13.70AP

The component of the gravitational force pulls forward on the satellite and increases the speed of satellite.

Explanation of Solution

The only forces act on the satellite is the backward force of air resistance comparatively very small in magnitude to the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force, one component of the gravitational force pulls forward on the satellite to do positive work and makes speed increases.

Conclusion:

Therefore, component of the gravitational force pulls forward on the satellite and increases the speed of satellite.

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Chapter 13 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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