Review. A global positioning system (GPS) satellite moves in...

Review. A global positioning system (GPS) satellite moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The nonmilitary GPS signal is broadcast at a frequency of 1 575.42 MHz in the reference frame of the satellite. When it is received on the Earth’s surface by a GPS receiver (Fig. P39.55), 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. It is called a blueshift to indicate a change to a higher frequency.The magnitude of that fractional change is given by $$\frac{\Delta f}{f}=\frac{\Delta U_{g}}{m c^{2}}$$ where $U_{g}$ is the change in gravitational potential energy of an object-Earth system when the object of mass $m$ is moved between the two points where the signal is observed. Calculate this fractional change in frequency due to the change in position of the satellite from the Earth's surface to its orbital position. (e) What is the overall fractional change in frequency due to both time dilation and gravitational blueshift?

Review. A global positioning system (GPS) satellite moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The nonmilitary GPS signal is broadcast at a frequency of 1 575.42 MHz in the reference frame of the satellite. When
it is received on the Earth’s surface by a GPS receiver (Fig. P39.55), 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. It is called a blueshift to indicate a change to a higher frequency.The magnitude of that fractional change is given by
$$\frac{\Delta f}{f}=\frac{\Delta U_{g}}{m c^{2}}$$
where $U_{g}$ is the change in gravitational potential energy of an object-Earth system when the object of mass $m$ is moved between the two points where the signal is observed. Calculate this fractional change in frequency due to the change in position of the satellite from the Earth's surface to its orbital position. (e) What is the overall fractional change in frequency due to both time dilation and gravitational blueshift?

Key Concepts

Special Relativity Time Dilation

Time dilation, a result of special relativity, indicates that time elapses at different rates for observers in relative motion. For satellites moving at high speeds relative to observers on Earth, the effect causes a measurable fractional change in the frequency of timing signals, as the passage of time onboard the satellite differs from that on the surface.

Gravitational Blueshift

Gravitational blueshift is a prediction of general relativity whereby signals emitted from a region of lower gravitational potential (such as near a massive body like Earth) are shifted to higher frequencies when observed from a region of higher gravitational potential (such as a satellite in orbit). This shift is quantitatively related to the change in gravitational potential energy relative to the rest mass energy of the object or signal-carrying particles.

Orbital Period and Radius Relationship

The orbital period of a satellite is intrinsically linked to the radius of its circular orbit. By setting the gravitational force equal to the centripetal force, one can derive a formula that connects the orbital period to the orbit’s radius, allowing calculation of one when the other is known.

Circular Orbital Motion

Circular orbital motion describes an object moving along a circular path under the influence of a centripetal force that continuously changes the object’s direction. In the context of satellites, this concept is used to relate the necessary centripetal acceleration to the gravitational pull exerted by a planet, ensuring that the satellite remains in a stable orbit.

Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation provides the framework for understanding the gravitational force between two masses. For orbiting bodies, this law is used to equate the gravitational force acting on the satellite with the required centripetal force, thereby linking the satellite’s mass, the mass of the planet, and the distance between their centers.

Transcript

00:01 Satellite is 11 hours 50 minutes which is equal to 43 ,080 seconds and gravitational constant here is g is equal to 6 .67 into 10 to 10 to 10 to 1 .11 newton meter square per kilogram square.

00:36 Mass of the earth is m .e.

00:40 Is equal to 5 .972 into 2 .972 into 10 to power 24 kilograms.

01:09 Now the time period t is equal to, we know that circumference of the orbit in our case which is 2 pi r upon orbital velocity which is v0.

01:30 We know that the orbital velocity v0 is equal to square root of gm e and e upon r so let's put this v0 in above equation so we will have again t is equal to 2 pi r square root of r upon gm e now to find the square on both sides so we can get the final equation of t or r so let's take the square both the side on the above equation so we will get t squared is equal to 4 pi square r cube upon gm e now solve for the r therefore the radius we can write r is equal to g which is our 6 .67 into 10 to power minus 11 into m e which is 5.

03:13 972 into 10 to power 24 into t square which is 43 ,080 square upon 4 pi square and the whole bracket will we have the power 1 1 .3 .3 .0.

03:52 And the solution of this r will give the answer 2 .66 into 20 .6 .6.

04:01 10 to power 7 meter.

04:05 So therefore, we can say that the radius of the satellite orbit is this one.

04:15 Now moving further to the second part, speed of the satellite v0 can be calculated as circumference of the orbit, which is 2 pi r upon time period which is t.

04:34 Now let's put the values here.

04:36 So it's a 2 pi radius that already we have calculated 2 .66 into 10 to power 7 upon time period is 43 ,080 seconds.

04:56 So the answer of this speed will be 3 .879 km per second, which is our speed of the satellite now time dilation can be calculated as delta t is equal to gamma delta tp let us find the value for gamma so gamma is equal to 1 upon square root of 1 minus v square upon c square so let's put the values.

05:53 So 1 upon square root of 1 minus b is 3 ,879 square that already we have calculated just above.

06:05 And the c is 3 into 10 to power 8 square.

06:12 So the answer of the gamma will be 1 .9 times 0 seconds .8...

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