ORBITAL MECHANICS TWO-BODY MOTIONDr. Ghulam Jaffer

Orbital PlaneIn the orbital plane, the position vector R and the velocity vector V specify the motion of the satellite,

Two-body Motion: Circular MotionAll of the celestial bodies from a fleck of dust to a supernova are attracted to each other in accordance with Newton's law of universal gravitation Fg = GMm/R2 The gravitational force on the Earth orbiting satellite where

(3-1)

Fg = universal gravitational force between bodies M, m = mass of the two bodies G = universal gravitational constant (6.67*10-11 Nm2/kg2) r = distance between the center of masses of the two bodies Fc = mV2/R (3-2)

The centrifugal force on the spacecraft is V = velocity of the spacecraft m = mass of the spacecraft

r = radius from the spacecraft center of mass to the central body center of mass For steady motion both forces must be balanced, therefore For circular orbits, Centrifugal force is balanced by Gravitational force, therefore V = (MG/R); MG = =398,600 km3/s2 Period (3-3), So V = (/R) P = 2 (R3/) (3-4) (3-5)

General Solution (1)Kepler's laws of planetary motion are confirmed and generalized to allow orbits

of any conic section, not just elliptical orbits. (Twobody motion is often called Keplerian motion) The sum of the potential energy and kinetic energy of the orbiting body per unit mass is a constant at all points in the orbit and is V2/2 - /R

General Solution (2)

General Solution (3): Specific Energyis the total mechanical energy per unit mass or specific energy of an object in any orbit about a central body. The kinetic energy term is V2/2 and the potential energy term is / r. Potential energy is considered to be zero at infinity () and negative at radii less than . The total energy of any orbit depends on the semimajor axis (a) of the orbit only For a circular orbit a = r and specific energy is ve, also for ellipse. Hence for all closed orbits (not trajectories), the energy is ve.

General Solution (4): Relative Energy of Orbit TypesFor parabolic orbits a = , and specific energy is zero; as we will see a parabolic orbit is a boundary condition between hyperbolas and ellipses For hyperbolic orbits. a is negative and specific energy is positive At a given radius velocity and specific energy increase in the following order: circular, elliptical. parabolic. hyperbolic; total spacecraft energy increases in the same order Additional energy must be added to a spacecraft to change an orbit from circular to elliptical. Energy must be removed to change from an elliptical to a circular orbit Both adding and removing energy requires a force on the vehicle and in general that means consumption of propellant

General Solution (5): Angular MomentumTotal angular momentum of the orbiting body is a constant equal to the cross product of the radius and the velocity vectors: H=rxV Where H is the angular momentum per unit mass (or specific momentum) and is a vector quantity. From vector mechanics. the magnitude of H can be determined by H = rVcos y (3-13) where H = magnitude of the specific momentum r = magnitude of the radius vector (the distance from the spacecraft to the center of mass of the central body) V = magnitude of the velocity vector y = flight path angle (the angle between the local horizontal and the velocity vector

General Solutions (6)

General Solutions (7)Specific energy and eccentricity are related The energy integral (also called the visviva integral) yields the general relation for the velocity at any point on any conic orbit) of an orbiting body -------------For a circle, a = r and -----------------------For an ellipse, a > 0 and --------------For a parabola, a = and -----------------For a hyperbola, a < 0 and------------(3-14)

(3-15) (3-16) (3-17) (3-18) (3-19)

General Solutions (8)

General Solutions (9): StepsDefine an orbit and discover its type given only r. V. and y at a point. The steps required are as follows:1.

Given r and V. the specific energy can be calculated from Eq. (3-7). With specific energy. the semimajor axis (a) can be obtained from Eq. (3-8). Given r, V and y, the magnitude of specific momentum (h) can be obtained from Eq. (313). With h and a, e can be obtained from Eq. (3-14). From the characteristics of the e, the orbit type can be determined (the Table). e and a define a conic orbit. Knowing these two elements and the orbit type, any other element can be obtained using relevant relations.

2.

3.

4.

5.

Exercises (1)An Earth satellite is in an orbit with a perigee altitude of 400 km and an eccentricity of 0.6. Find the following: (a) the perigee velocity (b) the apogee radius (c) the apogee velocity (d) the orbit period (e) the satellite velocity when its altitude is 3622 km (f) the true anomaly at altitude 3622 km (g) the path angle at altitude 3622 km

Exercises (2)The LANDSAT C Earth resources satellite is in a near-polar, near-circular orbit with a perigee altitude of 917 km, an eccentricity of 0.00 132 and an inclination of 89.1 deg. What are the apogee altitude, the orbit period and the perigee velocity? Solution: 436.3 km, 1.726 h, 7.397 km/s. ============================================== Two radar fixes on an unidentified Earth orbiter yield the following positions: Altitude = 1545 km at a true anomaly of 126 deg Altitude = 852 km at a true anomaly of 58 deg What are the eccentricity, altitude of perigee and semi major axis of the spacecraft orbit?

Exercises (3)The Magellan spacecraft was placed in an elliptical orbit around Venus with a periapsis altitude of 250 km and a period of 3.1 h. What is the apoapsis altitude? Solution: h = 7810 km. ================================== Consider an elliptical Earth orbit with a semi major axis of 12.500 km and an eccentricity of 0.472. What is the time from periapsis passage to a position with a true anomaly of 198 deg?

ORBITAL MECHANICS CALCULATIONS

Elliptical Orbit Parameter Relationships