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23 Nov 2019

A communications satellite is in a synchronous orbit that is 3.6*10^7 m directly above the equator. The satellite is located midway between Quito, Equador, and Belem, Brazil, two cities almost on the equator that are separated by a distance of 3.5*10^6 m. Find the time it takes for a telephone call to go by way of satellite between the cities. Ignore the curvature of the Earth.

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Sixta Kovacek
Sixta KovacekLv2
7 Sep 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A communications satellite is in synchronous orbit that is2.7 107 mdirectly above the equator. The satellite islocated midway betweenQuito, Equador and Belem, Brazil, two citieson the equator that areseparated by a distance of 3.5 106 m.Find the time it takes for atelephone call to go by way ofsatellite between these two cities.Ignore the curvature of theearth.

(1) Using Equation delta.d = 2do+ n glass d glass - d1 and Figure 1-3 and 1-4 located in theExperimental
Procedure, determine the speed of light assuming that the distancebetween
the transmitter-side mirror and the face of the corner-cube do is1.7m and the
distance between the two mirrors d1 is 4 em. (Note: The corner-cubeis
designed so that light always travels 6.7 em within the corner-cube- i.e.
dglass)). nglass=1.5
(2) Calculate the time-of-flight you expect for light to propagatea distance of 20
m in (a) vacuum and (b) a medium with an index of refraction n =1.5.
Express your answer in nanoseconds (ns).
(3) In telephone calls using a satellite, you sometimes can hear adelay in the
response of the other person. Satellites are usually either inlow-Earth orbit
Uust above the atmosphere) or much higher in geosynchronous orbit.The
latter is at an altitude of 35,785 kilometers (22,236 miles), wherethe satellite
completes one orbit in exactly one day. Because the orbitalvelocity matches
the spin rate of the Earth, a satellite in a circular equatorialgeosynchronous
orbit appears to hover motionless over a single location on theequator. The
total delay in hearing your correspondent's reply, due to thefinite speed of
light, is due to two round trips. If you and your correspondentwere located
nearby, and on the equator, the only distance involved is thedistance from
you to the satellite, the total distance corresponding to the delayis 4 x 35,785
km. Calculate this delay, assuming the radio signal travels at thespeed of
light in a vacuum. Report your answer in units of seconds, anddiscuss in one
sentence whether this delay is long enough for you to notice.Identify at least
one factor that could cause the actual delay to be longer than whatyou
calculated.

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