﻿ Motions of Planets and Satellites | Aerospace Engineering

## Motions of Planets and Satellites

### 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 ----------------------------------------------------------------------------------------------------------------------------------------------  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) becomes More 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.
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(r  t)/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 r2 is constant. Let's now consider two points P1 and P2 in an orbit with radii r1 and r2, and velocities v1 and v2. Since the velocity is always tangent to the path, it can be seen that if is the angle between r and v, then  where vsin is the transverse component of v. Multiplying through by r, we have or, for two points P1 and P2 on the orbital path Note that at periapsis and apoapsis, = 90 degrees. Thus, letting P1 and P2 be these two points we get Let's now look at the energy of the above particle at points P1 and P2Conservation 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 mv2/2 while the potential energy of gravity 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 ----------------------------------------------------------------------------------------------------------------------------------------------
```PROBLEM 3

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 × 3.986005×1014 × 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 × 3.986005×1014 × 6,628,140 / (6,878,140 ×
(6,878,140 + 6,628,140)) ]
Va = 7,542 m/s```
```PROBLEM 4

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 × Vp2) - 1]
Ra = 6,578,140/[2×3.986005×1014/(6,578,140×7,8502)-1]
Ra = 6,805,140 m

Altitude @ apogee = 6,805,140 / 1,000 - 6,378.14 = 427.0 km```
---------------------------------------------------------------------------------------------------------------------------------------------- The eccentricity e of an orbit is given by ----------------------------------------------------------------------------------------------------------------------------------------------
```PROBLEM 5

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 × Vp2 / GM - 1
e = 6,578,140 × 7,8502 / 3.986005×1014 - 1
e = 0.01696```
---------------------------------------------------------------------------------------------------------------------------------------------- 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 ----------------------------------------------------------------------------------------------------------------------------------------------
```PROBLEM 6

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```