### Orbit Meccanics:

- 1) Conic Sections
- 2) Orbital Elements
- 3) Types of Orbits
- 4) Newton’s Laws of Motion and Universal Gravitation
- 5) Uniform Circular Motion
- 6) Motions of Planets and Satellites
- 7) Launch of a Space Vehicle
- 8) Position in an Elliptical Orbit
- 9) Orbit Perturbations
- 10) Orbit Maneuvers
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In celestial mechanics where we are dealing with planetary or stellar sized bodies, it is often the case that the mass of the secondary body is significant in relation to the mass of the primary, as with the Moon and Earth. In this case the size of the secondary cannot be ignored. The distance *R*is no longer negligible compared to*r*and, therefore, must be carried through the derivation. Equation (4.9) becomesMore commonly the equation is written in the equivalent form where*a*is the semi-major axis. The semi-major axis used in astronomy is always the primary-to-secondary distance, or the*geocentric*semi-major axis. For example, the Moon's mean geocentric distance from Earth (*a*) is 384,403 kilometers. On the other hand, the Moon's distance from the*barycenter*(*r*) is 379,732 km, with Earth's counter-orbit (*R*) taking up the difference of 4,671 km.*r*are constant so that equal areas are swept out in equal times by the line joining a planet and the sun. For elliptical orbits, however, both and*r*will vary with time. Let's now consider this case. Figure 4.5 shows a particle revolving around*C*along some arbitrary path. The area swept out by the radius vector in a short time interval*t*is shown shaded. This area, neglecting the small triangular region at the end, is one-half the base times the height or approximately*r(rt)/2*. This expression becomes more exact as*t*approaches zero, i.e. the small triangle goes to zero more rapidly than the large one. The rate at which area is being swept out instantaneously is therefore For any given body moving under the influence of a central force, the value*r*is constant. Let's now consider two points^{2}*P*and_{1}*P*in an orbit with radii_{2}*r*and_{1}*r*, and velocities_{2}*v*and_{1}*v*. Since the velocity is always tangent to the path, it can be seen that if is the angle between_{2}*r*and*v*, then where*vsin*is the transverse component of*v*. Multiplying through by*r*, we have or, for two points*P*and_{1}*P*on the orbital path Note that at periapsis and apoapsis, = 90 degrees. Thus, letting_{2}*P*and_{1}*P*be these two points we get Let's now look at the energy of the above particle at points_{2}*P*and_{1}*P*._{2}*Conservation of energy*states that the sum of the kinetic energy and the potential energy of a particle remains constant. The kinetic energy*T*of a particle is given by*mv*while the potential energy of gravity^{2}/2*V*is calculated by the equation*-GMm/r*. Applying conservation of energy we have From equations (4.14) and (4.15) we obtain Rearranging terms we get ----------------------------------------------------------------------------------------------------------------------------------------------An artificial Earth satellite is in an elliptical orbit which brings it to an altitude of 250 km at perigee and out to an altitude of 500 km at apogee. Calculate the velocity of the satellite at both perigee and apogee. SOLUTION, Given: Rp = (6,378.14 + 250) × 1,000 = 6,628,140 m Ra = (6,378.14 + 500) × 1,000 = 6,878,140 m Equations (4.16) and (4.17), Vp = SQRT[ 2 × GM × Ra / (Rp × (Ra + Rp)) ] Vp = SQRT[ 2 × 3.986005×10**PROBLEM 3**^{14}× 6,878,140 / (6,628,140 × (6,878,140 + 6,628,140)) ] Vp = 7,826 m/s Va = SQRT[ 2 × GM × Rp / (Ra × (Ra + Rp)) ] Va = SQRT[ 2 × 3.986005×10^{14}× 6,628,140 / (6,878,140 × (6,878,140 + 6,628,140)) ] Va = 7,542 m/s

---------------------------------------------------------------------------------------------------------------------------------------------- The eccentricityA satellite in Earth orbit passes through its perigee point at an altitude of 200 km above the Earth's surface and at a velocity of 7,850 m/s. Calculate the apogee altitude of the satellite. SOLUTION, Given: Rp = (6,378.14 + 200) × 1,000 = 6,578,140 m Vp = 7,850 m/s Equation (4.18), Ra = Rp / [2 × GM / (Rp × Vp**PROBLEM 4**^{2}) - 1] Ra = 6,578,140/[2×3.986005×10^{14}/(6,578,140×7,850^{2})-1] Ra = 6,805,140 m Altitude @ apogee = 6,805,140 / 1,000 - 6,378.14 = 427.0 km*e*of an orbit is given by ----------------------------------------------------------------------------------------------------------------------------------------------

---------------------------------------------------------------------------------------------------------------------------------------------- If the semi-major axisCalculate the eccentricity of the orbit for the satellite in problem 4. SOLUTION, Given: Rp = 6,578,140 m Vp = 7,850 m/s Equation (4.20), e = Rp × Vp**PROBLEM 5**^{2}/ GM - 1 e = 6,578,140 × 7,850^{2}/ 3.986005×10^{14}- 1 e = 0.01696*a*and the eccentricity*e*of an orbit are known, then the periapsis and apoapsis distances can be calculated by ----------------------------------------------------------------------------------------------------------------------------------------------A satellite in Earth orbit has a semi-major axis of 6,700 km and an eccentricity of 0.01. Calculate the satellite's altitude at both perigee and apogee. SOLUTION, Given: a = 6,700 km e = 0.01 Equation (4.21) and (4.22), Rp = a × (1 - e) Rp = 6,700 × (1 - .01) Rp = 6,633 km Altitude @ perigee = 6,633 - 6,378.14 = 254.9 km Ra = a × (1 + e) Ra = 6,700 × (1 + .01) Ra = 6,767 km Altitude @ apogee = 6,767 - 6,378.14 = 388.9 km**PROBLEM 6**