orbital mechanics: 3. keplerian orbits

3 Keplerian Orbits

3.1 Newton's Universal Law of Gravitation

Newton's law of gravitation states that any two particles of mass m 1 and m2, distance r apart, are attracted toward each other with a force

Gmlm2 F -- r2 (3.1)

where G is a universal constant called the gravitational constant. The value of G is (6.6726 ± 0.0005) x 10 -11 m3/kg-sZ(lcr), which is ~ o w n to about 1 part in 13,000.1 The value of /z = GMe, on the other hand, where Me is the Earth mass, is known to a much higher accuracy, and is therefore used for astrodynamical calculation. Accurate measurement of the gravitational attraction between small masses is limited by the accuracy to which G is known.

There are several laboratory techniques available for the determination of G. One of the most accurate experiments to determine G is the "time of swing method" of Heyl. This approach is illustrated in Fig. 3.1, where two large masses M and two small masses m are placed on coaxially suspended torsional springs.

When two balances are aligned in parallel, the period of oscillation is less than when they are aligned at right angles. In the former position, the gravitational attraction adds to the torsional spring whereas, in the latter position, the attraction subtracts from the spring. The periods are on the order of 0.5 h and can be measured to 0.1 s of accuracy.

A constant torque is thus maintained on the free balance. The angular displace- ment, after many hours, determines G.

Other methods for improving the accuracy of G include a dynamically resonant torsional balance system, with a possible accuracy of 1 part in 106. Miniature orbital systems with equal or unequal masses have also been proposed that could be established in space to determine G to a high accuracy.

The gravitational force equation (3.1) can be expressed in terms of the gradient V of a scalar gravitational potential function V of a body. Thus, one can write

F = VV (3.2)

where V = tz/r for a spherical body. For a nonspherical body, V contains additional terms that can be used to determine F, as will be shown later.

3.2 General and Restricted Two-Body Problem

A more useful and general expression of the gravitational law is in its vector formulation. Thus, referring to Fig. 3.2, the masses m and M are moving in an inertial reference frame El , E2, E3. It is desired to determine the motion of m relative to the larger mass M.

21

22 ORBITAL MECHANICS

T o r s i o n a l Spring

M

Fig. 3.1 Torsional balance system.

The general two-body problem results i f a ~ 0. Then, for the mass M,

G M m r M ~ - - - -

F 3 (3.3)

and, for the mass m,

m/~

Subtracting Eq. (3.4) from (3.3) yields

G M m r

r 3 (3.4)

G ( M + m)r

r 3

G ( M + m)r Y'+ - 0

1,.3

(3.5)

In a restricted two-body problem, the principal mass M is assumed fixed in inertial space. This implies that M >> m, so that m does not affect the motion of M. Letting M be at the origin O (i.e., a = O), then the force on the mass m is

G M m r F - -

r 3

or, in accordance with Eq. (3.4) with p = r,

G M m r F -- - m~

r 3

(3.6)

(3.7)

K E P L E R I A N O R B I T S 23

a

Fig. 3.2 Two-body system.

Therefore,

G M r i; + -- 0 (3.8)

?.3

where i ~ = d2r/dt 2 = acceleration of mass m relative to the inertial frame. Equation (3.8) represents the motion of mass m in a gravitational field of mass M, assumed spherically symmetric and, therefore, concentrated at the origin of the reference system.

Equation (3.8) differs from Eq. (3.5) only in the gravitational constant term. The motion of the restricted two-body problem is therefore similar to that of the general two-body system and is affected by the magnitude of the gravitational term. The latter is a negligible effect when m << M, which is true for the satellites of the Earth and other planetary bodies.

3.3 Conservation of Mechanical Energy Consider Eq. (3.8) in the form

/ z r =o

where/z = GM, and m is assumed negligibly small compared to M. Scalar multiplication of Eq. (3.9) by/" results in

(3.9)

o r

~ r • ~ . ~ + - - - - 0

F 3 (3.10)

d-t -b ~-~ ~-~ =0 (3.11)


since d(r • r)/dt = 2r#, etc. This equation can be integrated to yield

(#)2 # - - 8

2 r

= specific mechanical energy (3.12)

Here (#)2/2 = v2/2 = specific kinetic energy and -tz/r is the specific potential energy of the satellite. The specific potential energy is also equal to the gravitational potential function per unit mass.

3.4 Conservation of Angular Momentum The specific angular momentum H of a satellite (angular momentum per unit

mass) can be obtained by vector-multiplying Eq. (3.9) by r. Then,

]zr r x i a + r x ~ = 0 (3.13)

which shows that

d r x ~ = ~ ( r x k )

d ~ - - H

dt

= 0 (3.14)

Consequently, H = const. This means that r and/" are always in the same plane. The actual solution for the satellite motion can be obtained by cross-multiplying Eq. (3.9) by H. Then,

or

i; x H = ~ ( H x r) (3.15)

d / Z (r20)r0 @ x H)

dt r j

= ~0~

= # d ( ? ) (3.16)

where the magnitude of the specific angular momentum H = r20, 0 is a unit vector normal to the unit vector ? along the r vector, and 0 is the angular rate of the r vector.

Integrating Eq. (3.16), one obtains

k x H = / z ? + B (3.17)

where B is a constant of integration.

KEPLERIAN ORBITS 25

Furthermore, since

therefore,

where

r • (k x H ) = r • (#~ + B )

= (r x k ) . H

= H . H

: n 2

H 2 = # r + r B cos 0

H2/l~ r ~

1 + (B/ lz ) cos 0

H2/ l z = p = semilatus rec tum

2 e H 2 ,] 1/2 B/Iz = e = 1 + - 7 - ] = eccentrici ty

0 = true anomaly

The general equat ion for the radius r in Eq. (3.9) is therefore of the form

(3.18)

(3.19)

a = semimajor axis = (ra + rp) /2 b = semirninor axis

e = eccentrici ty = (ra - r p ) / ( r a -t- rp )

0 = true anomaly

ra = apogee radius = a(1 + e)

rp = perigee radius = a(1 - e)

p = a(1 - e 2) = b2/a = rp(1 + e) = ra(1 - e) = semilatus rec tum

y -- flight-path angle

= zr/2 - / ~

The radial veloci ty componen t Vr can be found as follows: d r d r dO dr •

-- 0 vr = ~ - d 0 d t dO

dr H

dO r 2 (3.21)

3 .5 Orbital P a r a m e t e r s o f a S a t e l l i t e

The orbit el l ipse geometry is shown in Fig. 3.3. The fol lowing notat ion is used:

P r - (3.20) 1 + e c o s 0

This is an equat ion for a conic section, an example of which is the ellipse illustrated in Fig. 3.3.


APOGEE

but, since

VELOCITY

J

r SEMIMINOR AXIS

EMPTY FOCUS GEOMETR

~ ae =

~a=a(1

APOGEE

FLIGHT PATH ANGLE

F

GEOMETRIC CENTER F~]CUS

= a l l ÷ e )

APGGEE RADIUS

"1" ' SEMIMAJDR AXIS

SEMILATUS RECTUM

1 ~ERIGEE

;L, R%u,

Fig. 3.3 Ellipse geometry (from Ref. 2).

dr d dO = d-O [p(1 + ec°sO)-l]

H e sin 0 U r - - _ _

P

= ~ / ~ e sin 0 (3.22)

The normal component vn is found as

=r(H) ~pp vn = rO ~-~ = (1 + e c o s O ) (3.23)

The flight-path angle y is given by

V = cos-1 = tan-1 (3.24)

where

v = (d + @,/2

= (1 + e 2 + 2e cos O) (3.25)

KEPLERIAN ORBITS 27

The velocity at perigee Vp is

v e = g ~ ( l + e ) (3.26)

and, at apogee,

va = ~ / - ~ ( 1 - e ) (3.27)

These are the maximum and minimum values of v in orbit, respectively. The velocity v at any position in orbit is found from the energy or vis-viva

equation (3.12) as follows:

v 2 /z e - (3.28)

2 r

as e --+ 1, va --~ 0, ra --+ 2a; therefore, e = - t z / 2 a since the energy e remains constant. From Eq. (3.28),

(3.29)

For a circular orbit r = a and, therefore, the velocity Vc = x / -~ / r . For the escape

trajectory, a = e~ (parabola) and the velocity Vesc = ~/2vc. The classification of the different possible conic sections of orbits in terms of the eccentricity e is as follows:

0

<1

1 >1

Orbit

Circle (a = r)

Ellipse (a > 0)

Parabola [a ~ w (undefined)]

Hyperbola (a < 0)

The period of an elliptic orbit is P = 2rr /n , where n = v/-~-/a 3 = mean motion. If the posit ion of a satellite is desired at a specified time t, then it can be found

from

M = n ( t - r ) (3.30)

where r ----- t ime of perigee passage, and M = mean anomaly, and from Kepler 's equation,

M = E - e s inE (3.31)

which can be solved for E.


F A0×,L,A.Y C,.CLE

~ PERIGEE

Fig. 3.4 Definition of eccentric anomaly (from Ref. 2).

The definition of the eccentric anomaly E is shown in Fig. 3.4. The true anomaly 0 can be determined from

0 I I + e ] 1/2 E = tan - - (3.32)

tan ~ L 1 - e J 2

The trigonometric arguments 0 /2 and El2 are not always in the same quadrant. Conversely, if the time t of travel from one point on the ellipse to another point

is desired, then it can be found from Eq. (3.30), where M is given in Eq. (3.31).

3.6 Orbital Elements

The motion of a satellite around the Earth may be described mathematically by three scalar second-order differential equations. The integration of these equations of motion yields six constants of integration. It is these constants of integration that are known as the orbital elements.

The Keplerian orbital elements are often referred to as classical or conventional elements and are the simplest and easiest to use. This set of orbital elements can be divided into two groups: the dimensional elements and the orientation elements.

The dimensional elements specify the size and shape of the orbit and relate the position in the orbit to time (Fig. 3.3). They are as follows:

a = semimajor axis, which specifies the size of the orbit. e = eccentricity, which specifies the shape of the orbit. r = time or perigee passage, which relates position in orbit to time (r is often

replaced by M, the mean anomaly at some arbitrary time t; the mean anomaly is a uniformly varying angle)

The orientation elements specify the orientation of the orbit in space (Fig. 3.5). They are as follows:

i = inclination of the orbit plane with respect to the reference plane, which is taken to be the Earth's equator plane for satellite orbits. 0 deg < i < 180 deg).

KEPLERIAN ORBITS 29

EARTH'S o = LONGITUDE OF THE ASCENDING NODE NORTH

o = OLE . . . . SATE,.+E

f . " - ' - - PE RIG E E / r

[ .---'-~-L'_~-~--~ :7".4"~=~-/'=Z - -.~ECUPTJC / / ~ ~ ~ " / " ~ ~ PLANE

Fig. 3.5 Orientation of orbit in space (from Ref. 2).

For 0 deg < i < 90 deg, the motion is "posigrade" or "direct"; for 90 deg < i < 180 deg, the motion is termed "retrograde"

f2 = right ascension of the ascending node (often shortened to simply "node"); f2 is measured counterclockwise in the equator plane, from the direction of the vernal equinox to the point at which the satellite makes its south-to-north crossing of the equator (0 deg < f2 < 360 deg)

09 = argument of perigee; ~o is measured in the orbit plane in the direction of motion, from the ascending node to perigee (0 deg < w < 360 deg)

The angles i and f2 specify the orientation of the orbit plane in space. The angle w then specifies the orientation of the orbit in its plane. The argument of latitude u defines the position of the satellite relative to the node line.

Still another system of specifying the satellite state vector involves the scalar quantities of

v = velocity

r ---- radius f2 -- node

Y = flight-path angle ---- geocentric latitude

Az = azimuth of v from true north

Various sets of elements are used in orbit determination, the inertial rectan- gular (x, y, z, .~, ~, ~) and the spherical (or, ~, fi, Az, r, v), where ot is longitude

KEPLERIAN ORBITS 31

a) b) NP

EQUATOR

NP

Fig. 3.6 Earth orbits: a) retrograde orbit: 9 0 < i < 1 8 0 deg, 180<Az<360 deg; b) posigrade orbit: 0 < i < 90 deg, 0 < A z < 180 deg.

and/3 = rr/2 - y are examples of such elements. Note that both of these sets of elements give the satellite's position as a point in space at any specific time. Several orbital element systems are listed in Table 3.1. Posigrade and retrograde orbits are illustrated in Fig. 3.6.

References 1Luther, G. G., and Tower, W. R., "Redetermination of the Newtonian Gravitational

Constant G," Physical Review Letters, Vol. 48, No. 3, 18 Jan. 1982, pp. 121-123. 2Cantafio, L. (ed.), Space-Based Radar Handbook, Artech House, Norwood, MA, 1989.

Selected Bibliography Battin, R. H., An Introduction to the Mathematics and Methods of Astrodynamics, AIAA

Education Series, AIAA, Washington, DC, 1987. Bate, R. R., Mueller, D., and White, J. E., Fundamentals of Astrodynamics, Dover, New

York, 1971. Flury, W., Vorlesung Raumfahrtmechanik, Technische Hochschule Damstadt, 1994. Kaplan, M. H., Modern Spacecraft Dynamics and Control, Wiley, New York, 1976. Prussing, J. E., and Conway, B. A., Orbital Mechanics, Oxford, 1993. Roy, A. E., Orbital Motion, 3rd ed., Adam Hilger, Bristol and Philadelphia, 1988. Taft, L. G., Celestial Mechanics, Wiley, New York, 1985. Wertz, J. R., and Larson, W. J., eds., Space Mission Analysis and Design, Kluwer

Academic Publishers, 1991.

Problems 3.1, The period of revolution of a satellite is 106 min. Find the apogee altitude if the perigee altitude is 200 km.

(/z = 3.986 × 105 km3/s 2)

3.2. Find the period of revolution of a satellite if the perigee and apogee altitudes are 250 and 300 km, respectively.

3,3. Find the maximum and minimum velocity of the Earth if the eccentricity of the Earth's orbit about the sun is 1/60. What is the mean velocity if the mean distance to the sun is 149.6 x 10 6 kill? (/zs = 1.327 × 1011 km3/s 3)


3.4. Show that for a satellite moving on an elliptical orbit the velocity at the time of passage through the minor axis is equal in magnitude to the local circular velocity (vc). (Hint: Vc = ~/-~/r.)

3.5. A satellite in a circular orbit at an altitude hc above the Earth's surface is given a velocity v0 with a flight-path angle y. What is the magnitude of v0 if the perigee altitude of the resultant orbit is to be equal to hp(hp < he)?

3.6. A sounding rocket is launched from a planet. Find initial velocity to reach height H above the surface. Assume a spherical planet of radius R and gravity g at surface.

3.7. Find the escape velocity from the moon's surface. Assume that

1 gmoon = ~gEarth

rmoon = ~rEarth

3.8. Assuming that the period of Mars about the sun is 687 Earth days, find the mean distance of Mars to sun if the Earth distance to sun is 149.5 x 106 kin.

3.9. Find the mass of the sun using Kepler's third law.

3.10. For a Keplerian orbit with period P = 205 min, eccentricity e = 0.40, and true anomaly 0 = 60 deg, find the time t since passage of perigee.

3.11. Given the following orbit:

h p = 200 km

h a = 600 km

a) What is the time interval over which the satellite remains above an altitude of 400 km? Assume a spherical Earth with radius = 6378 km and/z = 3.986 x 105 knl3/s 2 .

b) What additional information is needed to solve part a if the Earth is assumed to be nonspherical?

3.12. A spaceship is in a 200-km circular orbit above a spherical Earth. At t = 0, it retrofires its engine, reducing its velocity by 600 m/s. How long (in minutes of time) does it take to impact the Earth?

3.13. An object was observed at a distance of (1.05) 2/3 Earth radii from the center of the Earth. Sixteen minutes later, the same object was observed at a position 60 deg (measured at the Earth center) from the original position. Show whether the object is in a circular orbit. Assume that/z = 0.00553 (ER3/min2).

KEPLERIAN ORBITS 33

3.1.

3.2.

3.3.

3.10.

1882 km

1.5 h

29.29, 30.28 km/s

14 min, 56 s

Selected Solutions

orbital mechanics: 3. keplerian orbits

Technology