### 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

Through a lifelong study of the motions of bodies in the solar system, Johannes Kepler (1571-1630) was able to derive three basic laws known as *Kepler’s laws of planetary motion*. Using the data compiled by his mentor Tycho Brahe (1546-1601), Kepler found the following regularities after years of laborious calculations:

1. All planets move in elliptical orbits with the sun at one focus.

2. A line joining any planet to the sun sweeps out equal areas in equal times.

3. The square of the period of any planet about the sun is proportional to the cube of the planet’s mean distance from the sun.

These laws can be deduced from Newton’s laws of motion and law of universal gravitation. Indeed, Newton used Kepler’s work as basic information in the formulation of his gravitational theory.

As Kepler pointed out, all planets move in elliptical orbits, however, we can learn much about planetary motion by considering the special case of circular orbits. We shall neglect the forces between planets, considering only a planet’s interaction with the sun. These considerations apply equally well to the motion of a satellite about a planet.

Let’s examine the case of two bodies of masses *M* and *m* moving in circular orbits under the influence of each other’s gravitational attraction. The center of mass of this system of two bodies lies along the line joining them at a point *C* such that *mr = MR*. The large body of mass *M* moves in an orbit of constant radius *R* and the small body of mass *m* in an orbit of constant radius *r*, both having the same angular velocity . For this to happen, the gravitational force acting on each body must provide the necessary centripetal acceleration. Since these gravitational forces are a simple action-reaction pair, the centripetal forces must be equal but opposite in direction. That is, *m ^{2}r* must equal

*M*. The specific requirement, then, is that the gravitational force acting on either body must equal the centripetal force needed to keep it moving in its circular orbit, that is

^{2}RIf one body has a much greater mass than the other, as is the case of the sun and a planet or the Earth and a satellite, its distance from the center of mass is much smaller than that of the other body. If we assume that *m* is negligible compared to *M*, then *R* is negligible compared to *r*. Thus, equation (4.7) then becomes

If we express the angular velocity in terms of the period of revolution, * = 2/P*, we obtain

where *P* is the period of revolution. This is a basic equation of planetary and satellite motion. It also holds for elliptical orbits if we define *r* to be the semi-major axis (*a*) of the orbit.

A significant consequence of this equation is that it predicts Kepler’s third law of planetary motion, that is *P ^{2}~r^{3}*.

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Calculate the period of revolution of a the satellite SOLUTION: Given: r = 6,578,140 m Equation (4.9), PPROBLEM 1^{2}= 4 ×^{2}× r^{3}/ GM P = SQRT[ 4 ×^{2}× r^{3}/ GM ] P = SQRT[ 4 ×^{2}× 6,578,140^{3}/ 3.986005×10^{14}] P = 5,310 s

Calculate the radius of orbit for a Earth satellite in a geosynchronous orbit, where the Earth's rotational period is 86,164.1 seconds. SOLUTION: Given: P = 86,164.1 s Equation (4.9), PPROBLEM 2^{2}= 4 ×^{2}× r^{3}/ GM r = [ P^{2}× GM / (4 ×^{2}) ]^{1/3}r = [ 86,164.1^{2}× 3.986005×10^{14}/ (4 ×^{2}) ]^{1/3}r = 42,164,170 m

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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 |

Kepler’s second law of planetary motion must, of course, hold true for circular orbits. In such orbits both and *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 ^{2}* is constant.

Let’s now consider two points *P _{1}* and

*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 _{1}* and

*P*on the orbital path

_{2}Note that at periapsis and apoapsis, = 90 degrees. Thus, letting *P _{1}* and

*P*be these two points we get

_{2}Let’s now look at the energy of the above particle at points *P _{1}* and

*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

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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 × 3.986005×10PROBLEM 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

A 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 × VpPROBLEM 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

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The eccentricity *e* of an orbit is given by

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Calculate 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 × VpPROBLEM 5^{2}/ GM - 1 e = 6,578,140 × 7,850^{2}/ 3.986005×10^{14}- 1 e = 0.01696

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If the semi-major axis *a* and the eccentricity *e* of an orbit are known, then the periapsis and apoapsis distances can be calculated by

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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 kmPROBLEM 6