Many satellites are moving in a circle in the earth's...

Many satellites are moving in a circle in the earth's equatorial plane. They are at such a height above the earth's surface that they always remain above the same point. (a) Find the altitude of these satellites above the earth's surface. (Such an orbit is said to be $geosynchronous.$) (b) Explain, with a sketch, why the radio signals from these satellites cannot directly reach receivers on earth that are north of 81.3$^\circ$ N latitude.

Key Concepts

Geosynchronous Orbit

A geosynchronous orbit is one in which a satellite's orbital period equals the Earth's rotational period. In a geostationary configuration, where the orbit lies in the Earth's equatorial plane, the satellite appears to remain fixed over the same point on the surface. This requires the satellite to be at a specific altitude where the balance between gravitational attraction and the necessary centripetal force for circular motion is achieved.

Orbital Mechanics and Altitude Calculation

The altitude of a satellite in a geosynchronous orbit is determined by setting the gravitational force equal to the centripetal force needed for circular motion, using Newton's law of universal gravitation and the requirements of orbital mechanics. This calculation involves the Earth's mass, the gravitational constant, and the period of the orbit (24 hours), resulting in a specific orbital radius from the center of the Earth, from which the altitude above the Earth's surface is found by subtracting the Earth's radius.

Line-of-Sight Communication

Radio signal propagation in satellite communications relies on line-of-sight transmission. Due to the curvature of the Earth, a signal from a satellite in the equatorial orbit cannot reach all points on the Earth's surface because the planet itself obstructs the path between the satellite and receivers located beyond the horizon. This fundamental limitation impacts the regions that can have direct communication with a satellite.

Earth's Curvature and Horizon Limitations

The curvature of the Earth imposes a geometrical limitation on the area coverage of satellites in geosynchronous orbit. Receivers at high latitudes, such as those north of a certain latitude, may fall outside the direct line-of-sight path of the satellite's signal because the Earth blocks the signal. This is a consequence of the spherical shape of the Earth, where the available 'horizon' is significantly limited for locations far from the equator.

Transcript

00:01 We're asked to find what height above the surface of the earth, so h here, a satellite needs to be so that it's always above the same point on the earth.

00:10 So that's basically asking us then for the satellite to be in geosynchronous orbits so that it and the earth have the same period of rotation.

00:18 So if that's true, then there'll always be this constant line between the earth and the satellite, so it won't be moving with respect to the earth.

00:25 So that tells us that the period for this motion needs to be 24 hours.

00:32 So just one day, so that's how long it takes earth to rotate once.

00:36 And so we want to convert that to seconds for use in our calculations.

00:40 So doing that, we should get 8 .64 times 10 to the fourth seconds.

00:50 So now we want to figure out what this h needs to be.

00:53 So to start for circular motion, circular orbit, we know the velocity of the orbit should always be the square root of g, m over r.

01:03 So m here would be the mass of the earth, and then r would be this total distance from the center of the earth out to over here.

01:10 So re plus h in this case.

01:15 And so then we also know that the period of motion would be given by 2 pi r.

01:21 So again, the same r of this total distance here.

01:24 So the circumference of the motion divided by, the velocity.

01:29 And so if we can figure out r from these equations, then we'd be able to find what this h is.

01:35 So to start, we know the value of t here.

01:39 We know t is 8 .64 times 10 to the fourth.

01:42 So let's substitute in this expression for v into our equation for t.

01:46 So we'd have then t equals 2 pi r and then divided by v, so we just flip this.

01:54 So we'd have then square root of r over g m e and then we can simplify that to 2 pi r to the three -halfs over square root of g m e so now we can just solve for r and then we should be able to calculate because i think we'll know all the other terms then so if we square each side and then take the cube root and move all these terms around, we should get then that r will be t squared gme over 4 pi squared, and then that whole thing is to the one -third.

02:39 So, t -squared is just this t that we found up here.

02:43 We look up mass of the earth, we know g, so we compute this as 4 .23 times 10 to the 7th meters...

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