Satellites

U NMANNED F LYING O BJECTS

7.2 Satellites

In a baseball game, a pitcher throws the ball to a catcher. Depending on the pitcher’s level of competency and the batter’s skill, the batter may hit or miss the ball. The ultimate goal for the batter is to hit the ball and run the four bases counterclockwise to score runs. Assume the batter makes good contact with the ball, and the ball takes off and lands on the ground due to gravitational acceleration (g  9.8 m/s²), finding a gap in the outfield. The batter may choose to stay on first base or go to second or third base, or even to home plate. But no matter how hard the ball was hit, even in record-breaking cases, the ball eventually came back to the Earth, as shocking as it seems, fulfilling the prophecy of what goes up must come down according to Newton’s third law and the band Blood, Sweat, and Tears [145,146,147,148,149].

The previous example shows what happens to objects launched with insufficient velocity to escape the Earth’s pull. With increasing launch velocity of an object, the orbit, also known as the trajectory, approaches an ellipse based on Kepler’s first law of planetary motion, with the Earth being located at one of the foci. Kepler’s second law of planetary motion states that the area of the orbit’s segment for the object orbiting the heavenly body is the same for equal time intervals, as in Figure 58. Each segment of the elliptical orbit (a slice of the elliptical pie) (Figure 58) corresponds to 1/16 of the orbit’s period (let us say around an imaginary planet called DRLSM [150]). Assuming that the period of this heavenly body is 301 days, each slice of the ellipse represents 18.8 Earth days (451.5 hrs, 27,090 min, or 1,625,400 s). If the angles shown in Figure 58 are incremented in small

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intervals such that the increased area at each interval is equal, and this angle is plotted versus the cumulative area, the curve shown in Figure 59 is obtained. Taking a derivative of this angle with respect to the area, one gets the curve in Figure 59. Interestingly, a Gaussian curve with a confidence

FI GURE 58 Kepler’s second law of planetary motion, for an imaginary planet DRLSM–dimensions in AU (drawings created using Solid Edge CAD tool).

FIG URE 59 Angle versus the area for an ellipse trajectory presented to orbit DRLSM.

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level above 99 percent may be fitted to this curve, with the fit presented in Figure 60.

Equ ation (111) and equation (112), where uA4255.772 /1268.806, are the integral of this Gaussian curve, which should yield a function similar to the one presented in the previous figure.

2) 0

0.008 0.04664301

A

e^u dA

 

 



  ^×  ⁽¹¹¹⁾

0.008 52.44 A erf A 

    (112)

For the imaginary planet (DRLSM) shown in Figure 58, the relationship between the area covered (Area) at 1/16 of the orbit time and the associated angle () is presented by equation (113). Regression analysis has been employed to model this behavior, and the model is presented with a confidence level above 93 percent.

31.9 0.028758 Area

   (113)

Kepler’s third law suggests that the ratio of the square of the heavenly body (planet) period (T ²) to the cube of the distance of the orbiting object from the center of the heavenly body—(R  d)³—is constant and equal for all heavenly bodies: T ²/(R  d)³  cte. Using this formula, you are able to find either the orbit distance from the surface of the planet (d), the radius of the heavenly body (R), or the period of the heavenly body (T) based on the known data for the Earth. Inserting the radius of the Earth in the previous

FIGURE 60 Deriva tive of ellipse angle with respect to the area versus the area.

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relation and assuming that the radius of the orbit may be approximated as the radius of the Earth, the constant universal value of 2.87723 × 10^11 is obtained—assuming 86,400 s as the period of the Earth’s rotation about its axis (also known as a civil day) and the Earth diameter of of 12,756 km.

Generally speaking, if you throw a ball, you can expect it to return to the Earth. The question is whether there is a way for the ball to spin around the Earth instead of getting back to it. To overcome the attraction of the Earth—

to be able to escape from it—the ball needs to reach the escape velocity. In this situation the conservation of energy, which states that the total energy consisting of the kinetic and potential energies remains constant, may be useful. The total initial energy consists of the kinetic (0.5m_ov²) energy and potential energy (m_o g_r r), where (m_o) is the mass of the object, r is the distance between the centers of the heavenly body and the object, v is the launch velocity of the object, and g_r is the gravity of the planet at distance r. When the object is launched, it is to overcome the resistance, and it will eventually get to zero velocity as the minimal velocity required to escape gravity. Additionally, the gravity reduces to near zero for the extremely large distance from the Earth, since it is inversely related to the second power of the distance based on Newton’s universal law of gravity. Since the eventual kinetic energy is to approach zero, the initial kinetic energy must equal the difference in potential energy between the launch point and the final location. Note that the potential energy is calculated relative to the center of the Earth, since it is associated with gravity. In other words, the kinetic and potential energies are to be the same—the launch velocity that is predicted based on this logic (the escape velocity) is v_e 2g r_r . Based on this formula, the approximate escape velocity of 11.17 km/s is found, which would apply to an object at the surface of the Earth at the Equator (ignoring the Earth’s angular speed). If the object were to be launched from the North Pole, the escape velocity would increase by 0.34 percent, since the gravity at the North Pole is 1.35 percent less than that at the Equator. At a distance of 100 km from the surface of the Earth, the escape velocity is 11.09 km/s. At the uppermost part of atmosphere, the base of the exosphere (400 km AGL), the escape velocity is about 10.85 km/s. The last two velocities are obtained by assuming that the launch is performed from the Equator.

Increasing the launch velocity of the flying object (or a satellite) makes it enter into the orbit—also known as the trajectory—which approaches an ellipse, based on Kepler’s first law of planetary motion, with the Earth being located at one of the foci. For a satellite to follow an orbit around any heavenly body, gravitational force provides the centripetal force required

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to make the object follow the curved trajectory. Newton’s law of universal gravitation states that any two bodies are attracted to each other with a force which is proportional to their masses and inversely related to the square of their distance. Based on this law, any two objects attract each other by a gravitational force existing between them. For that force to be strong enough, either the objects are to be massive, for example like the Earth; the force of gravitation is to be very high; or they are to be located in close proximity to each other. The analogy is completed by varying its parameters. If either the distance (r) between the objects is increased or the masses of the objects (m_o) or heavenly body (m_h) are decreased, the force (F) will decrease. The distance must be the only parameter with a significant effect on the force of attraction. You decrease the attraction by increasing the distance between the objects based on the universal law of gravitation. With that analogy in mind, equation (114) is presented herein, where G is the gravitational constant (6.674 × 10⁻¹¹ Nkg^–2m²).

2 o h

F Gm m

 r (114)

It is said that Newton obtained the gravitational constant by assuming that the density of the Earth is five to six times greater than that of water—

this number is very close to the value presented previously. Using the relation Gg/ 8



 _w r_Earth



,assuming that the density of the Earth is six times that of water (solid state—w (915 kg/m³) and the Earth’s mass may be obtained by multiplying its density by its volume (assuming a sphere of radius r_Earth), the gravitational constant (G) within 0.42 percent of the currently known accurate value may be obtained.

Since the mass of a planet (like the Earth) can be assumed to be constant, it is convenient to calculate the object’s weight by multiplying its mass by the gravitational acceleration (F  m_og). Substituting this gravitational force in equation (114), you can express the gravitational acceleration (g) in terms of the rest of the parameters in equation (114)—mass of the heavenly body (m_h), gravitational constant (G), and the distance between the centers of the two objects (r)—equation (115). This equation is easier to apply to the common task of calculating the gravitational force on the object of a known mass.

2

mh

g G

 r (115)

Equation (115) states that the closer you are to the COG, the higher the gravitational force is. As you get farther, like when you are flying at high altitudes, the gravity decreases. Assuming that the mass and radius of any heavenly body are known, you can calculate the gravitational acceleration on

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its surface. Table 25 presents heavenly body data along with the calculated values for gravitational acceleration, the drop in an object’s elevation after one second of freefall on the planet’s surface, and the planet’s orbital velocity and period. Note the duration of the local day (one rotation about its axis) associated with each heavenly body and expressed in Earth days.

You would have to complete your daily activities in 0.24 fraction of the time available to you if you were to live on Mercury, while you would have about 248 times more time available to complete the same tasks if you lived on Pluto, followed by Neptune (165), Uranus (84), Saturn (30), and Jupiter (12)—you may decide to either get more done or rest more (see Table 25).

Venus (0.62) and Mars (1.88) seem to be the least disrupting scenarios.

Note that the solar day is not to be mistaken with the sidereal day, which is the time it takes for the Earth to complete one rotation relative to the background stars (23 hrs 56 min 4 s). For the Earth orbiting around the Sun as well, an additional (3’56”) is required to complete a solar day (civil day—from the sunrise to the next rise). Equation (116) presents the relationship between the orbital period (t), distance from the heavenly body (r), and mass of the parent (heavenly body or planet) (m_h) whose object (particle) is orbiting around it. Equation (117) presents the relation between the orbital velocity (v), mass of the parent (heavenly body) whose object is orbiting around (m_h), and R, the radius of the heavenly body. The period of the heavenly body is obtained from Kepler’s third law, T²/(R  d)³  cte, where (r  R  d) is the distance from the center of the orbiting object to the center of its parent. The circular orbital velocity is obtained from

 

oc hp/

v  Gm Rd , where m_hp is the mass of the heavenly parent (e.g., the Earth is the parent of the Moon). The elliptical orbital velocity is similar to that of the circular orbital velocity that has been corrected to include the semi-major axis of the heavenly body, v_oe Gm_hp(2/r1/ ),a where a is the semi-major axis.

Table 25 shows the relationship between the escape velocity (v_e) and the distance from the center of the heavenly body (r) presented by equation (118), where () is the density of the heavenly body. The alternative expression for the escape velocity presented by equation (118) is for the uniformly distributed matter (i.e., homogeneous) scenarios.

3 3

2

hp

t r

Gm G

   

 (116)

4 3

Gmhp G

v r

r

    (117)

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2 8 3

hp e

Gm G

v r

r

    (118)

Note that equations (117) and (118) are obtained based on the assumption that the atmosphere does not introduce resistance (i.e., drag) to the object’s motion. It also assumes that the orbit is circular. If the orbit is elliptical, the vis viva (meaning living force in Latin) relation—also known as orbital-energy-invariance law—is employed to describe the motion of the orbiting body. To obtain this relation, energy conservation and conservation of angular momentum laws are applied, similar to the previous scenario using apogee and perigee—the farthest and nearest points from the heavenly body or planet reached by an object or particle orbiting it. The total energy consists of kinetic (0.5m v_{o e}²) and potential (m_og_r r) energies, where m_o is the object mass, r is the distance between the centers of the planet and the object, v_e is the escape velocity of the particle, and g_r is the gravity of the planet at distance (r). The total energy is to remain the same for all locations, including those two (0.5m v_{o e}²m g r_{o r} cte), where cte represents a constant value. The conservation of angular momentum results in rv_e  cte. Therefore, by equating the two relations for the two locations (apogee and perigee), a more general form of the relationship presented by equation (119) is derived, which introduces a correction factor based on the semi-major axis into equation (118)—note that in 2a  r_a  r_p and b²

 r_ar_p, subscripts “a” and “p” are associated with the apogee and perigee, respectively.

2 1

e h

v Gm

r a

 

    (119)

Based on equation (119), the specific orbital energy (i.e., the total energy of the particle per unit mass) is presented by (Gm_h /2a). Note the negative sign for the gravitational force applied to the particle is toward the planet.

The escape velocities at the apogee (v_ea) and perigee (v_ep) are obtained using equation (120)—where r_a and r_p are the distances from the center of the particle to the center of the planet at the apogee and perigee, respectively.

Note that variable r_{(a, p)} is presented as an alternative to indicate the same relationship is valid if either subscript “a” or “p” is required and is not to be mistaken with the plane coordinates. The specific angular momentum—the cross product of a particle position and momentum vectors per unit mass of the particle—r_{a p,  }v_{e a p, }b Gm a_h/ —is another important variable that along with the linear momentum and energy of the particle are defined in quantum mechanics—recall the translational, rotational, and vibrational

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modes of energy transfer in molecules. The orbital period remains the same as that of the circular orbit, as in equation (116).

 ^,  , h e a p

a p

b Gm

v r a (120)

Equation (120) is applied to a ballistic object (e.g., a cannonball) not aided by a propulsion system (e.g., a rocket). The equation gives the speed required by the object to be launched from the Earth or an altitude in order to overcome gravity. This parameter in addition to the chosen trajectory, the angle at which the object is erected with respect to the horizon of the launching location, are the factors determining if the object orbits the planet (e.g., satellite launched from the Earth), runs out of energy and hits a surface (e.g., missiles), or continues its journey forever (e.g., meteorites). It is to be noted that atmospheric conditions and the rotation of the Earth are not included in these calculations; therefore, as before, you are to keep in mind the direction of the rotation of the Earth (counterclockwise if you are looking down on the North Pole) and deduct or add the velocity to obtain relative velocities based on easterly or westerly launches when performing launch calculations. Because of the jet stream—winds aloft traveling to the east with the flow fluctuating to the north and south—it usually takes longer to fly west in the Northern Hemisphere—considering the Earth’s counterclockwise rotation—than traveling to the east. For instance, flying west from London to Toronto, against the jet stream, causes the plane’s ground speed to decrease and as a result the flight time to increase compared to flying this route to the east. Note that the Earth’s rotation has a partial role in the generation of the jet stream.

Nevertheless, the orbiting objects equipped with propulsion systems are supported by the additional force available to them to reach the escape velocity required. There are circumstances in which the particles (orbiting objects) are launched first to a lower orbit as an intermediate step. Within this orbit, they may travel as many complete turns as planned before reaching the escape velocity required to move to the upper orbit belonging to another planet. An example is a Mars satellite that is launched from the Earth that first follows the Earth orbit, and then is transferred to the Mars orbit. The transfer path, which is a straight line, is also known as the Hohmann transfer orbit.

The physics and mathematics of an orbiting satellite when entering a higher orbit are complex. Nevertheless, a brief example is discussed here for enthusiasts. Let us assume that the Mars explorer satellite mission is an

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orbiting DAQ system that collects information related to the Sun’s Coronal Mass Ejection (CME) activities and relays this information back to the Earth. The explorer is to be placed in the Mars orbit, with Mars being located at one of its foci based on Kepler’s first law of planetary motion. As the satellite travels the path, there are times when it is directly exposed to the Sun’s radiation. In radiation heat transfer this is known as a shape factor of one—meaning that it is capable of receiving the Sun’s uninterrupted radiation. Because of this direct radiation exposure, the satellite’s structure needs to be resistant to high temperatures, and so measures need to be taken to minimize radiation impact. Thus, the aluminum alloy exterior surfaces of satellites are coated with special white paints (e.g., AZ-93) to reflect about 85 percent of the Sun’s radiation as a means of thermal management [151]. Use of metamaterial Optical Solar Reflectors (OSR) as a surface coating on the exterior surfaces of the spacecraft in order to emit solar heat is suggested to thermally control heat exposure. These surfaces are made of quartz tiles, which are relatively heavy, costly to manufacture and launch, difficult to assemble, and cannot be easily implemented on surfaces with complex geometries [152]. Nevertheless, they are resistant to thermal radiation effects. The metal oxide embedded in these coatings results in a metamaterial that has very high emissivity and small absorptivity to the solar energy.

There are instances, however, when Mars comes in between the satellite and the Sun—for example, at the farthest distance from the Sun, where an eclipse is formed. In this scenario, the radiation information is recorded before entering the eclipse and right after leaving the eclipse with a time interval that is smaller than that of the regular DAQ for further processing.

The Mars explorer not only records the radiation information and surface activities of the Sun but also absorbs the Sun’s energy through its solar panels that are automatically adjusted for the highest energy absorption to generate the power required to operate its mechanical and electrical systems. These are examples of the engineering problems that need to be addressed before launching a satellite. There are dedicated mathematical programs which predict temperature variations as a function of the orbital locations (e.g., ESATAN-TMS—Thermal Modelling Suit), where the system exposed to the thermal radiation is modeled either using a lumped capacitance technique, a Finite Element Method (FEM), or a combination of both.

Clearly, it is desired to make the launch under the most suitable conditions possible. That includes minimizing the drag due to adverse atmospheric conditions, minimizing gravity, maximizing the initial launch

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