(a)
The radius of the orbit of GPS satellite.
(a)
Answer to Problem 65P
The radius of the orbit of the GPS satellite is
Explanation of Solution
Write the expression for the gravitational force acting on the satellite.
Here,
Write the expression for the centripetal force acting on the satellite.
Here,
Write the expression for the velocity of the satellite.
Here,
Use expression (III) in (II).
Equate expressions (IV) and (I) and solve for
Conclusion:
Substitute
Therefore, the radius of the orbit of the GPS satellite is
(b)
The speed of the satellite.
(b)
Answer to Problem 65P
Speed of the satellite is
Explanation of Solution
Equation (III) gives the orbital speed of the satellite.
Conclusion:
Substitute
Therefore, speed of the satellite is
(c)
The fractional change in the frequency due to time dilation.
(c)
Answer to Problem 65P
The fractional change in the frequency due to time dilation is
Explanation of Solution
Write the expression for the frequency.
Here,
Differentiate expression.
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.
Write the expression for Lorentz factor.
Use expression (IX) in (VIII).
Write the binomial expansion for
Substitute expression (XI) in (X).
Conclusion:
Substitute
Therefore, the fractional change in the frequency due to time dilation is
(d)
The fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position.
(d)
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
Explanation of Solution
Write the expression for the change in gravitational potential energy.
Here,
Write the given expression for the fractional change in frequency.
Conclusion:
Substitute
Substitute
Therefore, the fractional change in frequency due to the change in position of the satellite from Earth’s surface to its orbital position is
(e)
The overall fractional change in frequency due to both time dilation and gravitational blue shift.
(e)
Answer to Problem 65P
The overall fractional change in frequency is
Explanation of Solution
Write the expression for the overall fractional change in frequency.
Here,
Conclusion:
Substitute
Therefore, the overall fractional change in frequency is
Want to see more full solutions like this?
Chapter 39 Solutions
Physics: for Science.. With Modern. -Update (Looseleaf)
- An Earth satellite used in the Global Positioning System moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The satellite contains an oscillator producing the principal nonmilitary GPS signal. Its frequency is 1 575.42 MHz in the reference frame of the satellite. When it is received on the Earths surface, what is the fractional change in this frequency due to time dilation, as described by special relativity? (d) The gravitational blueshift of the frequency according to general relativity is a separate effect. The magnitude of that fractional change is given by ff=Ugmc2 where Ug/m is the change in gravitational potential energy per unit mass between the two points at which the signal is observed. Calculate this fractional change in frequency. (e) What is the overall fractional change in frequency? Superposed on both of these relativistic effects is a Doppler shift that is generally much larger. It can be a redshift or a blueshift, depending on the motion of a particular satellite relative to a GPS receiver (Fig. P1.39).arrow_forwardSuppose the primed and laboratory observers want to measure the length of a rod that rests on the ground horizontally in the space between the helicopter and the tower (Fig. 39.8B). To derive the length transformation L = L (Eq. 39.5), we had to assume that the positions of the two ends were determined simultaneously. What happens to the length transformation equation if both observers measure the end below the helicopter at one time t1 and the other end at a later time t2?arrow_forward(a) Find the value of for the following situation. An astronaut measures the length of his spaceship to be 100 m, while an observer measures it to be 25.0 m. (b) What is the of the spaceship relative to Earth?arrow_forward
- As measured by observers in a reference frame S, a particle having charge q moves with velocity v in a magnetic field B and an electric field E. The resulting force on the particle is then measured to be F = q(E + v × B). Another observer moves along with the charged particle and measures its charge to be q also but measures the electric field to be E′. If both observers are to measure the same force, F, show that E′ = E + v × B.arrow_forwardDescribe the following physical occurrences as events, that is, in the form (x, y, z, t): (a) A postman rings a doorbell of a house precisely at noon. (b) At the same lime as the doorbell is lung, a slice of bread pops out of a toaster that is located 10 1T1 from the door in the east direction from the door. (c) Tell seconds later, an airplane arrives at the airport, which is 10 km from the door in the east direction and 2 km to the south.arrow_forwardAn observer in a coasting spacecraft moves toward a mirror at speed v relative to the reference frame labeled by S in Figure P26.46. The mirror is stationary with respect to S. A light pulse emitted by the spacecraft travels toward the mirror and is reflected back to the spacecraft. The spacecraft is a distance d from the mirror (as measured by observers in S) at the moment the light pulse leaves the spacecraft. What is the total travel time of the pulse as measured by observers in (a) the S frame and (b) the spacecraft? Figure P26.46arrow_forward
- An observer in a coasting spacecraft moves toward a mirror at speed v relative to the reference frame labeled by S in Figure P26.46. The mirror is stationary with respect to S. A light pulse emitted by the spacecraft travels toward the mirror and is reflected back to the spacecraft. The spacecraft is a distance d from the mirror (as measured by observers in S) at the moment the light pulse leaves the spacecraft. What is the total travel time of the pulse as measured by observers in (a) the S frame and (b) the spacecraft? Figure P26.46arrow_forwardThe distance from earth to the nearest star is about 4 x 1016 meters. How long would it take a radio signal traveling at the speed of light, 3.0 x 108 m/s, to make a one-way trip to this star?arrow_forwardThe speed of light is c = 3.0 x 10^8 m/s. If the distance between the Earth and the moon is d = 2.39 x 10^5 mi (1.0 mi = 1.61 km). You send a laser from the Earth to the moon and back. How long does this take?arrow_forward
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning