5.5.  Satellites in Circular Orbits

Today there are many satellites in orbit about the earth. The ones in circular orbits are examples of uniform circular motion. Like a model airplane on a guideline, each satellite is kept on its circular path by a centripetal force. The gravitational pull of the earth provides the centripetal force and acts like an invisible guideline for the satellite.

There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius. To see how this fundamental characteristic arises, consider the gravitational force acting on the satellite of mass m in Figure 5.12. Since the gravitational force is the only force acting on the satellite in the radial direction, it alone provides the centripetal force. Therefore, using Newton’s law of gravitation (Equation 4.3), we have

where G is the universal gravitational constant, M_E is the mass of the earth, and r is the distance from the center of the earth to the satellite. Solving for the speed v of the satellite gives

(5.5)

If the satellite is to remain in an orbit of radius r, the speed must have precisely this value. Note that the radius r of the orbit is in the denominator in Equation 5.5. This means that the closer the satellite is to the earth, the smaller is the value for r and the greater the orbital speed must be.

Figure 5.12  For a satellite in circular orbit around the earth, the gravitational force provides the centripetal force.

The mass m of the satellite does not appear in Equation 5.5, having been eliminated algebraically. Consequently, for a given orbit, a satellite with a large mass has exactly the same orbital speed as a satellite with a small mass. However, more effort is certainly required to lift the larger-mass satellite into orbit. The orbital speed of one famous artificial satellite is determined in the following example.

Determine the speed of the Hubble Space Telescope (see Figure 5.13) orbiting at a height of 598 km above the earth’s surface. Figure 5.13  The Hubble Space Telescope orbits the earth, after being released from the Space Shuttle Discovery. (Courtesy NASA) Reasoning  Before Equation 5.5 can be applied, the orbital radius r must be determined relative to the center of the earth. Since the radius of the earth is approximately 6.38×106 m, and the height of the telescope above the earth’s surface is 0.598×106 m, the orbital radius is r=6.98×106 m. Problem solving insight The orbital radius r that appears in the relation is the distance from the satellite to the center of the earth (not to the surface of the earth). Solution The orbital speed is

Many applications of satellite technology affect our lives. One is a network of 24 satellites called the Global Positioning System (GPS), which can be used to determine the position of an object to within 15 m or less. Figure 5.14 illustrates how the system works. Each GPS satellite carries a highly accurate atomic clock, whose time is transmitted to the ground continually by means of radio waves. In the drawing, a car carries a computerized GPS receiver that can detect the waves and is synchronized to the satellite clock. The receiver can, therefore, determine the distance between the car and a satellite from a knowledge of the travel time of the waves and the speed at which they move. This speed, as we will see in Chapter 24, is the speed of light and is known with great precision. A measurement using a single satellite locates the car somewhere on a circle, as Figure 5.14a shows, while a measurement using a second satellite locates the car on another circle. The intersection of the circles reveals two possible positions for the car, as in Figure 5.14b. With the aid of a third satellite, a third circle can be established, which intersects the other two and identifies the car’s exact position, as in Figure 5.14c. The use of ground-based radio beacons to provide additional reference points leads to a system called Differential GPS, which can locate objects even more accurately than the satellite-based system alone. Navigational systems for automobiles and portable systems that tell hikers and people with visual impairments where they are located are two of the many uses for the GPS technique. GPS applications are so numerous that they have developed into a multibillion dollar industry.

Figure 5.14  The Navstar Global Positioning System (GPS) of satellites can be used with a GPS receiver to locate an object, such as a car, on the earth. (a) One satellite identifies the car as being somewhere on a circle. (b) A second places it on another circle, which identifies two possibilities for the exact spot. (c) A third provides the means for deciding where the car is.

Equation 5.5 applies to man-made earth satellites or to natural satellites like the moon. It also applies to circular orbits about any astronomical object, provided M_E is replaced by the mass of the object on which the orbit is centered. Example 10, for instance, shows how scientists have applied this equation to conclude that a supermassive black hole is probably located at the center of the galaxy known as M87. This galaxy is located at a distance of about 50 million light-years away from the earth. (One light-year is the distance that light travels in a year, or 9.5×10¹⁵ m.)

The Hubble telescope has detected the light being emitted from different regions of galaxy M87, which is shown in Figure 5.15. The black circle identifies the center of the galaxy. From the characteristics of this light, astronomers have determined an orbiting speed of 7.5×105 m/s for matter located at a distance of 5.7×1017 m from the center. Find the mass M of the object located at the galactic center. Figure 5.15  This image of the ionized gas (yellow) at the heart of galaxy M87 was obtained by the Hubble Space Telescope. The circle identifies the center of the galaxy, at which a black hole is thought to exist. (Courtesy NASA and Space Telescope Science Institute) Reasoning and Solution Replacing ME in Equation 5.5 with M gives , which can be solved to show that The ratio of this incredibly large mass to the mass of our sun is (4.8×1039 kg)/(2.0×1030 kg)=2.4×109. Thus, matter equivalent to 2.4 billion suns is located at the center of galaxy M87. Considering that the volume of space in which this matter is located contains relatively few visible stars, researchers have concluded that the data provide strong evidence for the existence of a supermassive black hole. The term “black hole” is used because the tremendous mass prevents even light from escaping. The light that forms the image in Figure 5.15 comes not from the black hole itself, but from matter that surrounds it.

The period T of a satellite is the time required for one orbital revolution. As in any uniform circular motion, the period is related to the speed of the motion by v=2pr/T. Substituting v from Equation 5.5 shows that

Solving this expression for the period T gives

(5.6)

Although derived for earth orbits, Equation 5.6 can also be used for calculating the periods of those planets in nearly circular orbits about the sun, if M_E is replaced by the mass M_S of the sun and r is interpreted as the distance between the center of the planet and the center of the sun. The fact that the period is proportional to the three-halves power of the orbital radius is known as Kepler’s third law, and it is one of the laws discovered by Johannes Kepler (1571–1630) during his studies of planetary motion. Kepler’s third law also holds for elliptical orbits, which will be discussed in Chapter 9.

An important application of Equation 5.6 occurs in the field of communications, where “synchronous satellites” are put into a circular orbit that is in the plane of the equator, as Figure 5.16 shows. The orbital period is chosen to be one day, which is also the time it takes for the earth to turn once about its axis. Therefore, these satellites move around their orbits in a way that is synchronized with the rotation of the earth. For earth-based observers, synchronous satellites have the useful characteristic of appearing in fixed positions in the sky and can serve as “stationary” relay stations for communication signals sent up from the earth’s surface. This is exactly what is done in the digital satellite systems that are a popular alternative to cable TV. As the blowup in Figure 5.16 indicates, a small “dish” antenna on your house picks up the digital TV signals relayed to earth by the satellite. After being decoded, these signals are delivered to your TV set. All synchronous satellites are in orbit at the same height above the earth’s surface, as Example 11 shows.

Figure 5.16  A synchronous satellite orbits the earth once per day on a circular path that lies in the plane of the equator. Digital satellite system television uses such satellites as relay stations for TV signals that are sent up from the earth’s surface and then rebroadcast down toward your own small dish antenna.

What is the height H above the earth’s surface at which all synchronous satellites (regardless of mass) must be placed in orbit? Reasoning  The period T of a synchronous satellite is one day, so we can use Equation 5.6 to find the distance r from the center of the earth. To find the height H of the satellite above the earth’s surface we will have to take into account the fact that the earth itself has a radius of 6.38×106 m. Solution A period of one day corresponds to T=8.64×104 s. In using this value it is convenient to rearrange the equation as follows: By squaring and then taking the cube root, we find that r=4.23×107 m. Since the radius of the earth is approximately 6.38×106 m, the height of the satellite above the earth’s surface is

Two satellites are placed in orbit, one about Mars and the other about Jupiter, such that the orbital speeds are the same. Mars has the smaller mass. Is the radius of the satellite in orbit about Mars less than, greater than, or equal to the radius of the satellite orbiting jupiter? Background: Understanding the concepts of centripetal force and the gravitational force is be key to answering this question. For similar questions (including calculational counterparts), consult Self-Assessment Test 5.1. The test is described at the end of Section 5.7.

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