Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 13, Problem 79CP

(a)

To determine

Find the total energy of the satellite-Earth system.

(a)

Expert Solution
Check Mark

Answer to Problem 79CP

The total energy of the satellite-Earth system is 3.67×107 J.

Explanation of Solution

Write the equation for total energy,

    E=12mvp2GMEmrp                                                                  (I)

Here, E is the total energy, m is the mass of satellite, vp is the speed satellite at perigee point, G is the gravitational constant, ME is the mass of Earth and rp is the distance between center of Earth to satellite.

Conclusion:

Substitute 1.60 kg for m, 8.23×103 m/s for vp, 6.67×1011 Nm2/kg2 for G, 5.98×1024 kg for ME and 7.02×106 m for rp in equation I.

    E=12(1.60 kg)(8.23×103 m/s)2(6.67×1011 Nm2/kg2)(5.98×1024 kg)(1.60 kg)7.02×106 m=3.67×107 J

Therefore, the total energy of the satellite-Earth system is 3.67×107 J.

(b)

To determine

Find the magnitude of the angular momentum of the satellite.

(b)

Expert Solution
Check Mark

Answer to Problem 79CP

The magnitude of the angular momentum of the satellite is 9.24×1010 kgm2/s.

Explanation of Solution

Write the equation for angular momentum.

    L=mvprpsinθ                                                         (II)

Conclusion:

Substitute 1.60 kg for m, 8.23×103 m/s for vp, 90.0° for θ and 7.02×106 m for rp in equation II.

    L=(1.60 kg)(8.23×103 m/s)(7.02×106 m)sin90.0°=9.24×1010 kgm2/s

Therefore, the magnitude of the angular momentum of the satellite is 9.24×1010 kgm2/s.

(c)

To determine

Find the satellite speed and distance from the center of Earth from the apogee.

(c)

Expert Solution
Check Mark

Answer to Problem 79CP

The satellite speed and distance from the center of Earth from the apogee are 5580 m/s and 1.04×107 m respectively.

Explanation of Solution

As the energy of the satellite-Earth system and angular momentum of the Earth are conserved then at apogee,

    12mva2GMmra=Emvarasin90.0°=L                                                         (III)

Conclusion:

Substitute the known values in equation III.

    12(1.60 kg)va2(6.67×1011 Nm2/kg2)(5.98×1024 kg)(1.60 kg)ra=3.67×107 J

    (1.60 kg)vara=9.24×1010 kgm2/s

Solving the above equations simultaneously and suppressing units.

    12(1.60)va2(6.67×1011)(5.98×1024)(1.60)(1.60)va9.24×1010=3.67×107

This further reduces to,

    0.800va211046va+3.6723×107=0

The above equation is the quadric equation then the solution is,

    va=11046±(11046)24(0.800)(3.6723×107)2(0.800)

The above equation gives two values such as, va=8230 m/s and va=5580 m/s. The smaller value refers to the velocity at the apogee and the larger value refers to the perigee.

Then the distance between satellite and Earth center is,

    ra=Lmva=9.24×1010 kgm2/s(1.60 kg)(5.58×103 m/s)=1.04×107 m

Therefore, satellite speed and distance from the center of Earth from the apogee are 5580 m/s and 1.04×107 m respectively.

(d)

To determine

Find the semi major axis of the satellite orbit.

(d)

Expert Solution
Check Mark

Answer to Problem 79CP

The semi major axis of the satellite orbit is 8.69×106 m.

Explanation of Solution

Write the equation for the major axis,

    2a=rp+ra                                                          (IV)

Here, a is the semi major axis

Conclusion:

Substitute 7.02×106 m for rp and 1.04×107 m for ra in the equation IV.

    a=12(7.02×106 m+1.04×107 m)=8.69×106 m

Therefore, the semi major axis of the satellite orbit is 8.69×106 m.

(e)

To determine

Find the period of satellite.

(e)

Expert Solution
Check Mark

Answer to Problem 79CP

The period of satellite is 134 min.

Explanation of Solution

Write the equation for the period,

    T=4π2a3GME                                                         (V)

Here, T is the period.

Conclusion:

Substitute 6.67×1011 Nm2/kg2 for G, 5.98×1024 kg for ME and 8.69×106 m for a in equation V.

    T=4π2(8.69×106 m)3(6.67×1011 Nm2/kg2)(5.98×1024 kg)=8060 s=134 min

Therefore, the period of satellite is 134 min.

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Chapter 13 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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