## Satellite Orbits

In our text, theninthedition of Paul Hewitt's, this topic ofConceptual PhysicsSatellite Orbitsappears in thenextchapter,. I think it fits better here. You can read about it in chapter 10 of Hewitt's text, however.Projectile Motion and Satellite OrbitsThe

onlyforce on a satellite in orbit around our Earth is theforce of gravity. Since this is theonlyforce, it is certainly thenet force. Thenet forceon any object moving in a circle is also called thecentripetal force, F. This centripetal force is_{c}directed toward the centerand is equal toF_{c}= m v^{2}/ rThe

onlyforce on a satellite is theforce of gravity.A satellite remains in orbit because this force of gravity provides thecentripetal forcenecessary for it to move in a circle. Of course satellite orbits may be ellipses, but we will restrict our attention to satellites incircularorbits. This keeps the radius and the speed constant.F We will look at two particular orbits. The Space Shuttle uses a_{g}= F_{c}F

_{g}= m v^{2}/ rin which it travels just above Earth's atmosphere. As you watch the Olympic Games, the signal is sent from China to a communications satellite and then re-transmitted to receivers in the United States. Such communications satellites appear to be stationary because they are in orbits with periods of 24 hours; these are calledlow-Earth orbitorgeosynchronousorbits. These geostationary orbits are also directly over the equator.geostationary

Low-Earth Orbit

The

Space Shuttleis an excellent example of a satellite in alow-Earth orbit. The Space Shuttle orbits about 300 km to 500 km above Earth's surface. Earth's radius is about 6 000 km so this is an increase of only about 5% or 8%. That means the force of gravity is only about 10% to 15% less thanat Earth's surface.Therefore, we can use the approximation that the force of gravity is the

sameon the orbiting satellite as it is at Earth's surface. That is,F _{g}= m g

F

_{g}= m v^{2}/ rm g = m v

^{2}/ rg = v

^{2}/ rv

^{2}= g rWe know the radius; to be in a low-Earth orbit, the radius of the orbit must be nearly equal to Earth's radius of about 6 000 km or 6 x 10

^{6}m. g is the acceleration of gravityat the surface of the Earthso we know g is about 10 m/s^{2}. Of course, we could use 9.8 m/s^{2}but our entire calculation is a reasonable approximation so we will use 10 m/s^{2}.We could simply calculate the orbital velocity and we will do that. But it is also interesting -- and fairly easy -- to extend this and calculate the orbital

period-- the time necessary for one orbit. But first, the orbital velocity.v^{2}= g rv

^{2}= ( 10 m/s^{2}) ( 6 x 10^{6}m ) = 6 x 10^{7}(m/s)^{2}v = 7.7 x 10

^{3}m / sThat's a fine answer but most of us have little or no "intuitive feel" for how fast that really is. How fast is a thousand meters per second? How fast is ten thousand meters per second? Let's convert that to km/h -- or even to mph.

v = 7.7 x 10^{3}m / s [^{km}/_{1000 m}] [^{3600 s}/_{h}] = 28 000 km / hv = 28 000 km / h [

^{mi}/_{1.61 km}] = 17 000^{mi}/_{h}What about the

period? How long does it take to make one orbit?v = D / TT = D / v

T = 2 r / v

T = 2 ( 3.14 ) ( 6 x 10

^{6}m ) / ( 7.7 x 10^{3}m / s )T = 4 900 s [

^{1 min}/_{60 s}] = 81 min(Remember, we've made some approximations; the

periodof a low-Earth satellite is about an hour and a half).

From a NASA web site:How fast does a Space Shuttle travel? What is its altitude? How much fuel does it use?

Like any other object in low Earth orbit, a Shuttle must reach speeds of about 17,500 mph (28,000 kilometers per hour) to remain in orbit. The exact speed depends on the Shuttle's orbital

altitude, which normally ranges from190 milesto330 miles(304 kilometers to 528 kilometers) above sea level, depending on its mission.Many

communications satellitesuse a booster rocket to go to an orbit 22,300 miles [35,680 kilometers] high. Known asgeosynchronousorbit, this height is special because a satellite orbiting here circles the Earth at the same rate as a point on the equator, allowing it to hover over a ground station located at that point.Each of the two Solid Rocket Boosters on the Shuttle carries more than one million pounds of solid propellant. The Shuttle's large External Tank is loaded with more than 500,000 gallons of supercold liquid oxygen and liquid hydrogen, which are mixed and burned together to form the fuel for the Shuttle's three main rocket engines.

Geo-synchronous Orbit"Tracking" a communications satellite docked in an equatorial orbit with a

period of 24 his far, far easier than tracking a satellite in a low-Earth orbit. "Tracking" such a communications satellite simply means pointing an antenna at the satellite and leaving it alone. A satellite in such an orbit appears to bestationary. Such an orbit is calledgeostationaryorgeosynchronous.What

orbital radiusis required for such a geostationary satellite?F_{g}= v^{2}/ rHowever, we can no longer use an approximation like F

_{g}= mg. Now we must use the "real" gravitational force,F_{g}= G M m / r^{2}F

_{g}= m v^{2}/ rG M m / r

^{2}= m v^{2}/ rG M / r

^{2}= v^{2}/ rv = 2 r / T

G M / r

^{2}= ( 2 r / T )^{2}/ rG M / r

^{2}= ( 4^{2}r^{2}/ T^{2 }) / rG M / r

^{2}= 4^{2}r^{2}/ T^{2}rG M / r

^{2}= 4^{2}r / T^{2}G M T

^{2}/ 4^{2 }= r^{3}r

^{3}= G M T^{2}/ 4^{2}r

^{3}= [ G M / 4^{2}] T^{2}First, notice that this

isKepler's Third Law, r^{3}/ T^{2}= constant.Now we can plug in numbers and evaluate r

T = 24 h [^{3600 s}/_{h}] = 86 400 s = 8.64 x 10^{4}sM = M

_{Earth}= 6 x 10^{24}kgG = 6.67 x 10

^{-11}N kg^{2}/ m^{2}r

^{3}= [ ( 6.67 x 10^{-11}N kg^{2}/ m^{2}) ( 6 x 10^{24}kg ) / ( 4 ) ( 3.14 )^{2}] [ 8.64 x 10^{4}s ]^{2}r

^{3}= 7.57 x 10^{22}m^{3}r = 4.23 x 10

^{7}m

r = 42 300 kmr = 42 300 km [

^{0.61 mi}/_{1 km}] = 25 800 mi[ You might ask why we did all this? You might ask why we went to this much trouble?

You have all the tools -- all the knowledge and skill -- needed to understand these two satellite orbits in detail. The next time you watch a television broadcast about the Space Shuttle or a broadcast from the other side of the Earth, smile a little. You understand the details. You don't really need to be a rocket scientist to understand this aspect of Rocket Science! ]

(c) Doug Davis, 2003; all rights reserved

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