The Two Body Problem and Keplerian Motion - … › ~ooabdelk › Space Mechanics … · Web...

1

2 The Two Body Problem and Keplerian Motion

I can calculate the motion of heavenly bodies, but not the madness of people!

Sir Isaac Newton (1642-1727)

2.1 IntroductionBefore we introduce the two-body problem, we provide the following short review of necessary concepts and definitions from kinematics and dynamics of particles. We strongly recommend the reader to read these sections first before tackling the two-body problem.

2.1.1 Particle Kinematics

The motion of any particle P can be tracked in a Euclidian space with the help of a Cartesian coordinate system and a clock! In the frame of

reference XYZ, we can define the particle position r ( t )as

r ( t )=x( t ) i+ y ( t ) j+ z( t )k (2-1)

where i , j , and k are unit vectors in the X, Y, and Z directions.

r=‖r‖=(r⋅r )1/2=√x2+ y2+z2 (2-2)

Then, the particle velocity is

v (t )=drdt

=r= x i+ y j+ z k

=v xi+v y j+vz k(2-3)

22

XY

Z r

V

o

P

O

a

rC

s

Fig. 2-1. Particle kinematics.

2 C H A P T E R 2 T H E T W O B O D Y P R O B L E M

and its acceleration will be

a ( t )=dvdt

=v= r

= v x i+ v y j+ vz k¿ax i+a y j+az k

(2-4)

Particle Trajectory

The trajectory or path of a particle is the locus of points the particle occupies as it moves through space. Since a velocity vector describes the direction of motion (or the future position of the particle), it is always tangent to the trajectory.

The velocity vector of a particle is always tangent to its trajectory.

Let us define the unit vectors ut and un which are tangent and normal to the particle trajectory at its local position respectively. Since the velocity is always tangent to the trajectory, then we can write it as

v=vut

v=‖v‖=√v⋅v(2-5)

The distance traveled by the particle along its trajectory, s is related to the particle speed (magnitude of velocity) through

ds=v .dtv=s

(2-6)

Note that

s=v≠ r , or

‖r‖≠ ddt (‖r‖)

(2-7)

You can check this by taking, for example, r=3 t i+2t j+5 t k and comparing the two sides of the above inequality. The acceleration of the particle can be expressed in the osculating plane (the plane of motion) in

terms of the unit vectors ut andun as follows

XY

Z r

V

o

P

O

a

rC

s

un

ut

Fig. 2-2. Particle trajectory and osculating plane.

C H A P T E R 2 T H E T W O B O D Y P R O B L E M 3

a=a tut+anun

a t=v=s , an=v2

ρ

(2-8)

where ρ is the radius of curvature of the trajectory at the particle position, which is the distance from the particle position to the center of curvature C as illustrated in Error: Reference source not found.

2.1.2 Particle Dynamics

Angular Momentum

The angular momentum of a particle about a point is the moment of momentum (or more specifically, linear momentum) of the particle about that point. For the particle P shown in Error: Reference source not found, which has mass (m), the angular momentum H about O is given by

(H )O=r×(mv ) (2-9)

Then, for constant m, we can find the rate of change of angular momentum

( H )O=ddt [r×(mv )]

¿ r×(mv )+r×(ma)(2-10)

The first term on the right hand side will cancel by vector identity (Note

thatr×v=v×v=0 ). If the net force acting on the particle is Fnet , then from Newton’s second law (conservation of linear momentum), we can write, for constant m,

Fnet=ma (2-11)

Therefore, equation Error: Reference source not found can be written as

( H )O=r×Fnet (2-12)

XY

Z r

V

m

P

O

Fnet

Fig. 2-3. Particle kinetics.


Now, r×Fnet is exactly the moment of the net force Fnet about O, or

(M net )O .

Then, we can write

( H )O=( M net)O (2-13)

The above equation is analogues to Newton’s second law for linear motion, and is called Newton’s second law for angular motion or the conservation of angular momentum.

2.2 The Two-Body Problem2.2.1 Problem Description

The two-body problem is the dynamic problem to find the trajectory of motion of a system composed of two body masses M and m , for instance, in the absence of any effect other than the mutual gravitational force – given some initial condition on the positions and velocities of these body masses. From this description, we notice that an actual two-body system does not exist in reality, but as we will see later, the trajectory of motion of many body pairs in space can be approximated, to a sufficiently high-degree of accuracy, by a two-body motion.

2.2.2 Problem Formulation

In order to mathematically formulate the problem, let us consider the sys-tem of two body masses M and m (as shown in Error: Reference sourcenot found). Assume X’Y’Z’ is an inertial frame of reference (frame of reference which is neither accelerating nor rotating as illustrated in Error:Reference source not found). Let XYZ be a non-rotating frame of reference parallel to X’Y’Z’ with its origin O coincident with the center of mass M.

Fg=−G Mmr2

rr (2-14)

2.2.3 Equation of Motion

The position vectors of M and m with respect to X’Y’Z’ are rM and rm

respectively. Then, the position of m relative to M will be

Fg m

Mr

Fg

Fig. 2-4. Earth and a rotating satellite is a good approximation of two-body system.

Mm

XY

Z

O

X’Y’

Z’

Fg

Fig. 2-5. Formulation of the two-body problem.

r

XY

Z

O

Inertial frame

Moving frame

Fig. 2-6. Inertial frame and moving frame.


r=rm−rM (2-15)

Apply Newton’s second law of motion to m and M to get

m rm=−G Mmr2

rr

M rM=G Mmr2

rr

(2-16)

Or

rm=−G Mr2

rr (2-17)

rM=G mr2

rr (2-18)

If we subtract (2-17) from (2-18), we get

rm−rM=r=−G (M+m)r3 r (2-19)

Equation (2-18) is the vector differential equation of the relative motion.

Now, if we assume m << M, then G(M+m)≈GM and equation (2-19) will become

r=−G Mr3

r (2-20)

If we compare equation (2-18) and equation (2-20), we notice that r and r will measure the same magnitude and direction whether in XYZ or X’Y’Z’ (Recall that the frame XYZ is non-rotating and parallel to the inertial frame X’Y’Z’.) also, when m << M (e.g. a satellite with respect to Earth), we can write

G(M +m)≈GM=μ (2-21)


is called the gravitational parameter which can be found from the astronomical data of most known celestial objects. Hence, the equation of motion will become

r+ μr3

r=0 (2-22)

We notice that the equation of motion (2-22) is a 2nd order differential equation of position in time which needs to be integrated twice to obtain the trajectory of a spacecraft for example. Then, two constant vectors are needed for the solution of the equation of motion, which may be taken as the initial position and velocity vectors.

2.2.4 Constants of Motion

If we take the dot product of the equation of motion (2-22) with r we get

r⋅r+ μr 3

r⋅r=0 (2-23)

The first term can be mathematically manipulated as

r⋅r=12

ddt

( r⋅r )=12

ddt

(v⋅v )= ddt ( v

2

2 ) (2-24)

We know thatddt

(r⋅r )=2 r⋅r (2-25)

butddt

(r⋅r )= ddt

(r2)=2 r r (2-26)

Then, we get the important relation

r⋅r=r r (2-27)

Now, if we take the time derivative of(− μ

r ), we get

ddt (− μ

r )= μr 2 r=

μr3 (r r )= μ

r3 r⋅r (2-28)

Now, if we substitute from (2-24) and (2-28) into (2-23) to get


ddt ( v

2

2−μ

r )=0 (2-29)

By integration, the quantity in brackets will be a constant. This quantity is also the total mechanical energy per unit mass of spacecraft, . The total mechanical energy of spacecraft is the summation of the kinetic energy,

12 mv2

and potential energy,−m μ

r . We know from dynamics that the potential energy of an object depends upon the selection of the datum at which the potential energy vanishes. Here, we opt to have the datum for gravitational potential at infinity. For other selections of datum, at the

center of M, for example, a constant should be added to−m μ

r .

Therefore, for the two-body motion, we can write

ε= v2

2− μ

r=const (2-30)

Equation (2-30) is known as the energy equation which is also referred to as the vis viva equation (vis viva, in Latin, means the living-force). This is the first constant of motion which complies with the conservation of energy principle. According to which, a body moving in a conservative field, such as the gravitational field, has constant mechanical energy.

Now, let us take the cross product of the equation of motion (2-22) with rfrom left to get

r×r+ μr3

r×r=0 (2-31)

The second term on left hand side vanishes by vector identity, sincer×r=0 . So, we have

r×r=0 (2-32)

Now, if we take the time derivative ofr×r , we get

ddt

(r×r )=r×r+r×r=r× r=0 (2-33)


Note that r×r=0 . By integration, the quantity in brackets will be a constant. This quantity is nothing but the angular momentum of

spacecraft per unit mass,h . So, we can write

h=r× r=r×v=const (2-34)

This is the second constant of motion which complies with the conservation of angular momentum principle. According to equation (2-13), since the gravitational force acting on spacecraft always passes through the center of the central body (O), it will have zero moment about O. Hence, the angular momentum of the particle about point O will be

invariant. We also notice that since h=r×v=const , the plane of motion which contains r and v (the osculating plane) will be fixed in space.

For convenience, we will usually refer to angular momentum per unit mass as angular momentum and mechanical energy per unit mass as mechanical energy.

2.2.5 Trajectory Equation

To find the trajectory, we need to integrate the equation of motion. The nonlinear equation of motion cannot be integrated directly. Instead, we will use the following procedure to integrate the equation of motion indirectly. First, let us take the cross product of equation of motion (2-22)

with h from right, we get

r×h+ μr3

r×h=0 (2-35)

We can show that

ddt

( r×h )=r×h+r×h=r×h (2-36)

The term r×hwill vanish since h is constant. Also since

r×h=r×( r×v )=r (r⋅v )−v (r⋅r )=r r r−r2 v

(2-37)

If we substitute into equation , we get

g f

M

m

XY

Z

O

Vh

Fig. 2-7. Angular momentum.


ddt

( r×h )+μ [ vr − rr2

r ]=0 (2-38)

ddt (μ r

r )=μ ( rr− rr 2 r )=μ( vr −

rr 2 r ) (2-39)

Now, if we substitute into equation, we get

ddt [ r×h−μ r

r ]=0 (2-40)

By integration we get

r×h−μ rr=B (2-41)

where B is a constant vector. Take the dot product of equation with r , to get

r⋅( r×h)= μrr⋅r+r⋅B (2-42)

From vector identity, the left hand side can written as

r⋅( r×h)=h⋅(r×r )=h⋅h=h2 (2-43)

Then,

h2=μr+rB cos ν (2-44)

is the angle between B and r which is called true anomaly. The true anomaly of spacecraft may be considered as an angular coordinate of the spacecraft position measured from a fixed direction in the plane of motion. Now, we can solve equation r , to get

r= h2/ μ1+(B /μ )cosν

(2-45)

Defining

e=Bμ

(2-46)

Then, we can write the trajectory equation as

m

MO

V

B

Fig. 2-8. True anomaly.

1 0 C H A P T E R 2 T H E T W O B O D Y P R O B L E M

r= h2/ μ1+e cosν

(2-47)

2.2.6 Orbital Elements

The position of a spacecraft can be specified using six parameters (recall r, r at t=0). Another set of six parameters are commonly used in specifying spacecraft positions in space, the orbital elements. In order to specify a spacecraft position in space completely, we need to specify:

The Orbit Plane

In general a plane can be defined in space using 2 parameters. These 2 parameters could be two components of a unit vector normal to the plane, or two angles measured from a reference frame. In the standard orbital elements, the orbit plane is defined using two angles:

i) The inclination, i, of the orbit plane to the equatorial plane

ii) The right ascension of ascending node, Ω.

The Shape of Orbit

For an ellipse,

a : semi-major axis

b : semi-minor axis

e : eccentricity

a2=b2+(ae )2 (2-48)

The shape of orbit is specified by any 2 of the above 3 parameters. The equation above can be used to compute the 3rd parameter. In standard orbital elements, a and e are used to describe the shape of orbit.

1. The orientation of orbit in plane

C H A P T E R 2 T H E T W O B O D Y P R O B L E M 1 1

Once we have specified the ellipse shape and the orbit plane, then we need to describe how to place the ellipse in the plane. This is done by specifying the angleω, argument of perigee, between the line of nodes and the perigee point vector.

The Position of Spacecraft on the Orbit

Finally, the spacecraft position in orbit is specified by the true anomaly angle, ν, measured from perigee direction

Note that h , μ and e are all constants.

e : eccentricity

θ : true anomaly

The above function is solution for r as a function of true anomaly, θ.

2.2.7 Orbit Equations

r= h2/ μ1+e cosν

(2-49)

This is called the orbit equation. It defines the path of a spacecraft with respect to the central body. The velocity that is perpendicular to r is:

V ¿=r ν (2-50)

and

V⊥=hr= h

( h2

μ ) 11+ecos ν

( 2-99 )

∴V⊥=μh(1+e cosν ) (2-100)

V r=r= ddt [ h

2

μ1

1+ecos ν ]= h2

μe sin ν

(1+e ν )2hr2 (2-101)


∴V r=μhe sin ν (2-102)

From the orbital equation, r is only minimum when ν=0. This is called a periapsis.

∴r p=h2

μ1

1+e(2-103)

At periapsis,

V r=0 ;V⊥= μh

(1+e)

Flight Path Angle

tan γ=V r

V⊥= e sin ν

1+ecos ν(2-104)

Since cosine is an even function, therefore the trajectory described by the orbit equation is symmetric about the apse line.

p≡ semi latus rectum ≡ orbit parameter

p=h2

μ(2-105)

v2

2−μ

r=ε ≡constant (2-112)

Energy at Periapsis,

ε p=v p

2

2− μ

r p(2-113)

But at perigee, V r=0. Therefore V p=V ⊥= hr p

∴ ε=12h2

r p2 − μ

r p(2-114)

and


r p=h2

μ1

1+e(2-115)

ε=−12

μ2

h2 (1−e2 ) (2-116)

2.2.8 Summary of the Two-Body Problem

1. The only possible path for an orbiting satellite in a two-body system is a conic section (circle, ellipse, parabola, or hyperbola).

2. The focus of the conic orbit is located at the center of the central body.

3. The mechanical energy () of a satellite (sum of kinetic and potential energies) does not change as the satellite moves along its conic orbit. However, the kinetic and potential energies may exchange.

4. The orbital motion takes place in a plane which is fixed in inertial space.

5. The specific angular momentum (h) of a satellite about the central body remains constant. Hence, as r and v changes, the flight path angle (g) must change as well to keep h = r ´ v constant.

3 Keplerian OrbitsWe have in the previous section that the solution of the two-body system results in a planar trajectory which has the shape of a conic section. Since such orbits agrees with the Kepler’s laws of planetary motion, the two-body orbits are usually referred to as Keplerian orbits. A Keplerian orbit can be circular, elliptic, parabolic, or hyperbolic.

5.1.1 Circular Orbits

If we let e=0 in the orbit equation,

∴r=h2

μ≡constant

V

Fig. 2-9. Circular orbit.


This proves that the orbit is circular. Since r=0, therefore V r=0 and V=V ⊥.

h=r V⊥=rV

∴r=h2

μ=

(rv )2

μ(2-117)

∴V circular=√ μr

(2-118)

The time, T required for one orbit is known as the period. Since speed is constant,

T= circumferencespeed

=2πr

√ μr

(2-119)

∴T circular=2π√μ

r32=2π

n;n=√ μ

r3 (2-120)

Specific energy,

ε=−12

μ2

h2 (2-121)

Since r=h2

μ,

ε circular=−μ2 r (2-122)

5.1.2 Elliptical Orbits

If 0<e<1, then the orbit equation is bounded. The initial orbit equation describes an ellipse if it is bounded from 0<e<1.

The minimum value of r is

r p=h2

μ1

1+e(2-123)

and its maximum value is

Vg

periapsisapoapsis

Fig. 2-10. Elliptic orbit.

F’ F

ae

ra rp

C

rB

a

b

a

Fig. 2-11. Geometry of an ellipse.


ra=h2

μ1

1−e(2-124)

2a=ra+r p (2-125)

∴2a=h2

μ ( 11−e

+ 11+e ) (2-126)

∴a=h2

μ1

1−e2 →h2

μ=a(1−e2) (2-127)

∴r= a(1−e2)1+ecos ν

(2-128)

r B=a(1−e2)1+ecos β

(2-129)

Projection of r B on apse line is ae.

∴ae=rB cos (180−β )=−r B cos β

¿− a(1−e2)1+ecos β

cos β (2-130)

Solving for e,

∴ e=−cos β

∴r B=a

∴b2=r B2 −(ae )2=a2(1−e2)

∴b=a √1−e2 (2-131)

ε=−12

μ2

h2 (1−e2 )

But for an elliptic orbit, h2=μa (1−e2)


∴ ε=−μ2a (2-132)

∴ v2

2− μ

r=−μ

2a(2-133)

Kepler’s law,

dAdt

=h2

Missing figure

∴∆ A=h2∆ t

Area of ellipse ¿ πab

∴πab=h2T→T=2 π ab

h (2-134)

∴T= 2πh

a2 √1−e2=2 πh ( h2

μ1

1−e2 )√1−e2 (2-135)

∴T= 2πμ2 ( h

√1−e2 )3

(2-136)

Since h2=μa (1−e2),

∴T=2πn

;n=√ μa3 (2-137)

r p

ra=1−e

1+e

Solving for e

∴ e=r a−r p

r a+r p(2-138)

Since ra+r p=2a, therefore a is the average of ra and r p. The average value for r is:


r ν=1

2 π∫02π

r (ν )d ν= 12π

a(1−e2)∫0

2π d ν1+e cosν

(2-139)

∴r θ=a√1−e2=b=√r p ra (2-140)

5.1.3 Parabolic Orbits

If e=1,

∴r=h2

μ1

1+cosν(2-141)

∴ ε=−12

μ2

h2 (1−e2 )=0 (2-142)

As ν→180° , r→∞.

∴ v2

2− μ

r=0

∴ v=√ 2μr

(2-143)

If a spacecraft is on a parabolic orbit, it will reach to ∞ with zero velocity. Parabolic orbits are called escape trajectories. At a distance r from Earth,

the escape velocity is: vesc=√ 2μr

If spacecraft is in a circular orbit,

∴ vesc=√2vcircular (2-144)

This indicates a required velocity boost of 41.4%

Missing figure

tan γ= sin ν1+cos ν (2-145)

From trigonometric identities,


sin ν=2sin ν2

cos ν2

cos ν=cos2 ν2−sin2 ν

2=2 cos2 ν

2−1

∴ tan γ=tan ν2

∴ γ= ν2 (2-146)

5.1.4 Hyperbolic Orbits

If e>1, then the orbit equation

r=h2

μ1

1+e cosν(2-147)

describes a hyperbola. Two symmetric curves, one is occupied by the spacecraft and the other one is empty.

Missing figure

The true anomaly of asymptotes,

ν∞=cos−1(−1e );90 °<ν∞<180 °

Where ν∞ corresponds to r→∞.

From trigonometry,

sin ν∞=√e2−1

e

For −ν∞<ν<ν∞, spacecraft is in hyperbola I while for ν∞<ν<(360°−ν∞), vacant orbit in hyperbola II is traced.

For ν=0,

r p=h2

μ1

1+e(2-148)

For θ=π ,


ra=h2

μ1

1−e;ra<0 (2-149)

2a=|ra|−r p=−ra−r p

∴a=h2

μ1

e2−1(2-150)

∴r= a (e2−1 )1+ecos ν

(2-151)

r p=a (e−1 ) ;ra=−a(e+1)

ε=−12

μ2

h2 (1−e2 )

∴ ε= μ2a (2-152)

Hyperbolic excess speed, v∞ occurs when spacecraft is at ∞, given that:

v2

2−μ

r= μ

2a→v2=2 μ

r+ μa

∴ v∞=√ μa

(2-153)

Recall that vesc=√ 2μr

,

∴ v2=v2esc+v2

∞ (2-154)

Characteristic energy ≡C3=v2∞

v∞ represents excess kinetic energy over that which is required to simply escape from the center of attraction.

Example 2.2


At a given point of a spacecraft’s geocentric trajectory, r=14600 km, v=8.6km /s and γ=50°, show that the path is hyperbolic.

vesc=√ 2μr

=7.389 km /s

Since v>vesc, therefore it is hyperbolic.

a) Compute C3

C3=v2∞=v2−v2

esc=19.36 k m2/s2

b) Compute h

h=r v⊥

v⊥=v cos γ=8.6 cos50 °=5.528 km /s

∴h=80710 k m2/s

c) Compute ν

r=h2

μ1

1+e cosν

∴ ecos ν=0.1193

vr=μhe sin ν→e sin ν=6.588 km / s

∴ ecos νesin γ

=tan ν=11.18

∴ ν=84.89 °

d) Compute e

e=1.339

e) Compute r p


r p=h2

μ1

1+e=6986 km

f) Compute a

a=h2

μ1

e2−1=20590km

5.2 ReferencesValado (2007). Fundamentals of Astrodynamics with Applications.

Curtis (2005). Orbital Mechanics for Engineers.

Ulrich Walter (2008). Astronautics.

Unsöld, A., Baschek, B., & Brewer, W. (2001). The New Cosmos: An Introduction to Astronomy and Astrophysics. Berlin: Springer.

Satellite Tool Kit Tutorial. STK version 8.0

Date post:	05-Jul-2020
Category:	Documents
Upload:	others
View:	1 times
Download:	0 times

The Two Body Problem and Keplerian Motion - … › ~ooabdelk › Space Mechanics … · Web...

Documents