Communications satellites are used to bounce radio waves from the earth\'s surfa

Question

Communications satellites are used to bounce radio waves from the earth's surface to send messages around the curvature of the earth. In order to be available all the time, they must remain above the same point on the earth's surface and must move in a circle above the equator

How long must it take for a communications satellite to make one complete orbit around the earth? (Such an orbit is said to be geosynchronous.)
Apply Newton's second law to the satellite and find its altitude above the earth's surface

Explanation / Answer

In any circular orbit, the centripetal force required to maintain the orbit (Fc) is provided by the gravitational force on the satellite (Fg). To calculate the geostationary orbit altitude, one begins with this equivalence: By Newton's second law of motion, we can replace the forces F with the mass m of the object multiplied by the acceleration felt by the object due to that force: We note that the mass of the satellite m appears on both sides — geostationary orbit is independent of the mass of the satellite.[13] So calculating the altitude simplifies into calculating the point where the magnitudes of the centripetal acceleration required for orbital motion and the gravitational acceleration provided by Earth's gravity are equal. The centripetal acceleration's magnitude is: where ? is the angular speed, and r is the orbital radius as measured from the Earth's center of mass. The magnitude of the gravitational acceleration is: where M is the mass of Earth, 5.9736 × 1024 kg, and G is the gravitational constant, 6.67428 ± 0.00067 × 10-11 m3 kg-1 s-2. Equating the two accelerations gives: The product GM is known with much greater precision than either factor alone; it is known as the geocentric gravitational constant µ = 398,600.4418 ± 0.0008 km3 s-2: The angular speed ? is found by dividing the angle travelled in one revolution (360° = 2p rad) by the orbital period (the time it takes to make one full revolution). In the case of a geostationary orbit, the orbital period is one sidereal day, or 86,164.09054 seconds).[14] This gives: The resulting orbital radius is 42,164 kilometres (26,199 mi). Subtracting the Earth's equatorial radius, 6,378 kilometres (3,963 mi), gives the altitude of 35,786 kilometres (22,236 mi). Orbital speed (how fast the satellite is moving through space) is calculated by multiplying the angular speed by the orbital radius: By the same formula we can find the geostationary-type orbit of an object in relation to Mars (this type of orbit above is referred to as an areostationary orbit if it is above Mars). The geocentric gravitational constant GM (which is µ) for Mars has the value of 42,828 km3s-2, and the known rotational period (T) of Mars is 88,642.66 seconds. Since ? = 2p/T, using the formula above, the value of ? is found to be approx 7.088218×10-5 s-1. Thus, r3 = 8.5243×1012 km3, whose cube root is 20,427 km; subtracting the equatorial radius of Mars (3396.2 km) we have 17,031 km.

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