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 39, Problem 65P

(a)

To determine

The radius of the orbit of GPS satellite.

(a)

Expert Solution
Check Mark

Answer to Problem 65P

The radius of the orbit of the GPS satellite is 2.66×107m_.

Explanation of Solution

Write the expression for the gravitational force acting on the satellite.

  Fg=GMEmr2                                                                                                         (I)

Here, Fg is the gravitational constant, G is the gravitational constant, ME is the mass of Earth, m is the mass of satellite, and r is the radius of the orbit.

Write the expression for the centripetal force acting on the satellite.

  Fc=mv2r                                                                                                                 (II)

Here, Fc is the centripetal force acting on the satellite, and v is the velocity of satellite.

Write the expression for the velocity of the satellite.

  v=2πrT                                                                                                                   (III)

Here, T is the time period of revolution of satellite around Earth.

Use expression (III) in (II).

  Fc=mr(2πrT)2                                                                                                       (IV)

Equate expressions (IV) and (I) and solve for r.

  GMEmr2=mr(2πrT)2GMET2=4π2r3r=(GMET24π2)13                                                                                              (V)

Conclusion:

Substitute 6.67×1011Nm2/kg2 for G, 5.98×1024kg for ME, and 11h58min for T in equation (V) to find r.

  r=((6.67×1011Nm2/kg2)(5.98×1024kg)(11h×3600s1hr+58min×60s1min)4π2)=2.66×107m

Therefore, the radius of the orbit of the GPS satellite is 2.66×107m_.

(b)

To determine

The speed of the satellite.

(b)

Expert Solution
Check Mark

Answer to Problem 65P

Speed of the satellite is 3.87×103m/s_.

Explanation of Solution

Equation (III) gives the orbital speed of the satellite.

  v=2πrT

Conclusion:

Substitute 2.66×107m for r, and 11h58min for T in the above equation to find the orbital speed.

  v=2π(2.66×107m)11h×3600s1h+58min×60s1min=3.87×103m/s

Therefore, speed of the satellite is 3.87×103m/s_.

(c)

To determine

The fractional change in the frequency due to time dilation.

(c)

Expert Solution
Check Mark

Answer to Problem 65P

The fractional change in the frequency due to time dilation is 8.34×1011_.

Explanation of Solution

Write the expression for the frequency.

  f=1T                                                                                                                     (VI)

Here, f is the frequency of satellite.

Differentiate expression.

  df=dTT2=fdTTdff=dTT                                                                                                      (VII)

The small fractional decrease in the frequency received from the satellite is equal to the fractional increase in period of oscillator due to the time dilation.

  dff=dTT=γΔtpΔtpΔtp=(γ1)                                                                                                   (VIII)

Write the expression for Lorentz factor.

  γ=11(vc)2                                                                                                         (IX)

Use expression (IX) in (VIII).

  dff=111(vc)2                                                                                                   (X)

Write the binomial expansion for 11(vc)2.

  11(vc)2=1+12(vc)2+...                                                                                     (XI)

Substitute expression (XI) in (X).

  dff=1(1+12(vc)2)=12(vc)2                                                                                             (XII)

Conclusion:

Substitute 3.87×103m/s for v, and 3.00×108m/s for c in equation (XII) to find dff.

  dff=12((3.87×103m/s3.00×108m/s)2)=8.34×1011

Therefore, the fractional change in the frequency due to time dilation is 8.34×1011_.

(d)

To determine

The fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position.

(d)

Expert Solution
Check Mark

Answer to Problem 65P

The fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position is 5.29×1010_.

Explanation of Solution

Write the expression for the change in gravitational potential energy.

  ΔUg=GMEm(1r21r1)                                                                                     (XIII)

Here, ΔUg is the change in gravitational potential energy, r2 is the final position, and r1 is the initial position of the satellite.

Write the given expression for the fractional change in frequency.

  Δff=ΔUgmc2                                                                                                           (XIV)

Conclusion:

Substitute 5.98×1024kg for ME, 6.67×1011Nm2/kg2 for G, 2.66×107m for r2, and 6.37×106m for r1 in equation (XIV) to find ΔUg.

  ΔUg=(6.67×1011Nm2/kg2)(5.98×1028kg)m[12.66×107m16.37×106m]=(4.76×107J/kg)m

Substitute (4.76×107J/kg)m for ΔUg, and 3.00×108m/s for c in equation (XIV) to find Δff.

  Δff=(4.76×107J/kg)mm(3.00×108m/s)2=5.29×1010

Therefore, the fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position is 5.29×1010_.

(e)

To determine

The overall fractional change in frequency due to both time dilation and gravitational blue shift.

(e)

Expert Solution
Check Mark

Answer to Problem 65P

The overall fractional change in frequency is 4.46×1010_.

Explanation of Solution

Write the expression for the overall fractional change in frequency.

  Δf=Δff+dff                                                                                                        (XV)

Here, Δf is the overall fractional change in frequency.

Conclusion:

Substitute 5.29×1010 for Δff, and 8.34×1011 for dff in equation (XV) to find Δf.

  Δf=8.34×1011+5.29×1010=4.46×1010

Therefore, the overall fractional change in frequency is 4.46×1010_.

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Chapter 39 Solutions

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

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