Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 9, Problem 43P

(a)

To determine

The radius of the orbit.

(a)

Expert Solution
Check Mark

Answer to Problem 43P

The radius of the orbit is 2.66×107m_.

Explanation of Solution

The gravitational force acting on the satellite is equal centripetal force.

Write the expression for the force acting on the satellite.

  F=ma        (I)

Here, m is the mass of the satellite, a is the acceleration of the satellite.

Equate the gravitational force, and centripetal force.

  GMEmr2=mυ2r        (II)

Here, ME is the mass of Earth, r is the radius of Earth, υ is the speed of the satellite, G is the gravitational constant.

Substitute, 2πrT for υ in equation (II) and rewrite to obtain an expression for r.

  GMEmr2=mr(2πrT)2GMET2=4π2r3r=(GMET24π2)        (III)

Here, T is the time period.

Conclusion

Substitute, 6.67×1011Nm2/kg2 for G, 5.98×1024kg for ME, 43080s for T in equation (III) to find the value of r.

  r=(6.67×1011Nm2/kg2×5.98×1024kg×(43080s)24(3.14)2)=2.66×107m

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

(b)

To determine

The speed of the satellite.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

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

Explanation of Solution

Write the expression for speed in terms of period.

  υ=2πrT        (IV)

Conclusion:

Substitute, 2.66×107m for r, and 43080s for T in equation (IV) to find the speed of the satellite.

  υ=2π(2.66×107m)43080s=3.87×103m/s

Therefore, the 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 43P

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        (V)

Here, f is the frequency.

Take the derivative of equation (V) on both sides.

  df=dTT2        (VI)

Substitute, f for 1T in equation (VI).

  df=f(dTT)dff=dTT        (VII)

The fractional change in the frequency is equal to the fractional change in time period according to equation (VII).

Substitute, γΔtptp for dT, and Δtp for T in equation (VII).

  dff=γΔtptpΔtp=(γ1)        (VIII)

Substitute, 11(υ2/c2) for γ in equation (VIII).

  dff=(11(υ2/c2)1)1[1+12(υ2/c2)]=12(υ2/c2)        (IX)

Here, c is the speed of the light.

Conclusion:

Substitute, 3.87×103m/s for υ, and 3.00×108m/s for c in equation (IX) to obtain the fractional change in frequency.

  dff=12((3.87×103m/s)2/(3.00×108m/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.

(d)

Expert Solution
Check Mark

Answer to Problem 43P

The fractional change in frequency due to the change in position of the satellite is +5.29×1010_.

Explanation of Solution

Write the expression for the gravitational potential.

  ΔUg=GMEmr        (X)

The fractional change in frequency due to the change in position of the satellite is.

  Δff=ΔUgmc2        (XI)

Conclusion:

Substitute, 6.67×1011Nm2/kg2) for G, 5.98×1024kg for ME, and 12.66×107m16.37×106m for r in equation (XI).

  ΔUg=(6.67×1011Nm2/kg2)(5.98×1024kg)m12.66×107m16.37×106m=(4.76×107J/kg)m

Substitute, (4.76×107J/kg)m for ΔUg, and 3.00×108m/s for c in equation (XI) to find fractional change in frequency due to the change in position of the satellite.

  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 is +5.29×1010_.

(e)

To determine

The overall fractional change in the frequency.

(e)

Expert Solution
Check Mark

Answer to Problem 43P

The overall fractional change in the frequency is +4.45×1010_.

Explanation of Solution

The fractional change in frequency due to the change in position of the satellite is +5.29×1010, and the fractional change in the frequency due to time dilation is 8.34×1011.

Hence the overall change in the frequency is.

  8.34×1011++5.29×1010=+4.46×1010

Conclusion:

Therefore, the overall fractional change in the frequency is +4.45×1010_.

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Chapter 9 Solutions

Principles of Physics: A Calculus-Based Text

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