Basic Orbits and Mission Analysis

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EXTROVERT Space Propulsion 03 1 Basic Orbits and Mission Analysis

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Basic Orbits and Mission Analysis. Newton's Law of Gravitation To find the velocity increment required for various missions, we must calculate trajectories and orbits. This is done using Newton's Law of Gravitation: . - PowerPoint PPT Presentation

Transcript of Basic Orbits and Mission Analysis

Page 1: Basic Orbits and Mission Analysis

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Basic Orbits and Mission Analysis

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= 2-

rGMmFr

Newton's Law of GravitationTo find the velocity increment required for various missions, we must calculate trajectories and orbits. This is done using Newton's Law of Gravitation:

Here the lhs is the "radial force" of attraction due to gravitation, between two bodies; the big one of mass M, and the little one of mass m. The universal gravitational constant G is 6.670 E-11 Nm2/kg2.

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Keplerian Orbits

Johannes Kepler (1609: first law, 1619: third law) Kepler’s First Law: The orbit of each planet is an ellipse, with the sun at one focus. Second Law: The line joining the planet to the sun sweeps out equal areas in equal times.

Third Law: The square of the period of a planet is proportional to the cube of its mean distance from the sun

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EXTROVERT Space Propulsion 03Kepler's Laws applied to satellite of mass m, orbiting a much

bigger object of mass M. i.e, m << M):

1. The satellite travels in an elliptical path around its center of attraction, which is located at one of focii of the ellipse. The orbit must lie in a plane containing the center of attraction.

phy-061062.blogspot.com/

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Implication: The radius vector from the center of attraction sweeps equal areas of the orbit per unit time. As the satellite moves away, its speed decreases. As it nears the center of attraction, its speed increases.

Second Law: The line joining the planet to the sun sweeps out equal areas in equal times.

Found on the web at a secondary site. Primary developer of this application is unknown

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Third Law: The square of the period of a planet is proportional to the cube of its mean distance from the sun

Implication: The ratio of the squares of the orbital periods of any two satellites about the same body equals the ratio of the cubes of the semi-major axes of the respective orbits.

ττ

=2 3

1 12 3

2 2

aa

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EXTROVERT Space Propulsion 03Satellite Equations of Motion

Newton’s Second Law rmFr&&r

=⇒

 Law of Gravitation  F=magnitude of force due to gravity (N)

G=Universal constant of gravity =  M=mass of Earth (kg)M =satellite massR =distance from center of Earth to the satellite (km) = GM for Earth = 398,600.5  Equation for satellite acceleration (from the above two Newton laws)   (2-body equation of motion)

22 rm

rGMmF μ

−≡−

=

2171067.6KgNkmx −

μ 23 sec−kμ

03

rr&& =+ rr

r μ

The solution to this is the polar equation of a conic section.

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Source: http://www.bartleby.com/61/imagepages/A4conics.html

Courtesy Houghton-Miflin Co.

ε

Conic Energy Semi-major axis a

e

Circle <0 = radius 0

Ellipse <0 >0 0<e<1

Parabola 0 1

Hyperbola >0 <0 >1

 

Total specific mechanical Energy of an orbit: (mechanical energy per unit mass)

Why Conic Sections are called conic sections:

arV

22

2 μμε −=−=

Conic Section Orbits / Trajectories

where a is the semi major axis and V is magnitude of velocity.

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EXTROVERT Space Propulsion 03A Few Commonly-Used Results on Orbits

Circular orbit around Earth:rr

Rr

V Ec

3481.631905366.7 ≈⎟⎟⎠⎞

⎜⎜⎝⎛

≈= μ r in km is radius of orbit

rrR

rV Eescape

8611.89217988.112 =≈= μ V in km/s; r in kmEscape Velocity from Earth:

Speed at any point in an elliptical orbit: ⎛ ⎞= −⎜ ⎟⎝ ⎠2

0 02 1v g rr a

( )( )−

= =+

20 0 min

11apogeee

v g r ve

( )( )+

= =−

20 0 max

11perigeee

v g r ve

Time for one orbit is:πτ =

3

0 0

2 ar g

Specific angular momentum: is constant for the 2-body problem, per Kepler (no plane change)

Vxrhrrr

=

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EXTROVERT Space Propulsion 03Orbit Maneuvers

currentneed VVVrrr

−=Δwhere V is the velocity at that point in the corresponding orbit.

The assumption made usually in simplifying orbit maneuvers is that we are dealing with a 2-body problem. For transfers between earth orbit and say, Mars orbit, we assume initially an Earth-vehicle problem. When the vehicle leaves the vicinity of Earth, we assume a Sun-vehicle problem, and finally a Mars-vehicle problem

Hohmann transfer orbit:  Transfer orbit’s ellipse is tangent to initial and final orbits, at the transfer orbit’s pengee and apogee respectively Most efficient transfer between two circular orbits. Two-burn transfer between circular orbits at altitudes :

BA rr &

rA

rB

ΔVB

ΔVA

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EXTROVERT Space Propulsion 03Hohmann Transfer

BATotal VVV Δ+Δ≡Δ⎥⎥⎥

⎢⎢⎢

⎟⎟⎟

⎜⎜⎜

⎛⎟⎟⎠⎞

⎜⎜⎝⎛

−⎟⎟⎠⎞

⎜⎜⎝⎛

−+⎟⎟⎠⎞

⎜⎜⎝⎛

−⎟⎟⎠⎞

⎜⎜⎝⎛

−=21

21

21

21

112112

BtxBAtxA rarrarμ

Example:

skmVskmVskmV

kmrkmr

Total

BA

B

A

/95.3/49.1;/46.2

421606567

=Δ→=Δ=Δ⇒

==

ΔV for a two-burn transfer between circular orbits at altitudes . BA rr &

For Earth, with delta-v in km/s, μ =631.3481

( )+=

2A B

txr r

a

When the radii of the two orbits are close, the delta-v’s are approximately equal.

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EXTROVERT Space Propulsion 03One-Tangent Burn

This is a quicker orbit transfer than the Hohman transfer ellipse, but is more costly in delta-v.

ΔVA

ΔVB

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EXTROVERT Space Propulsion 03Orbit Plane Changes

Simple plane change through angle q at an orbital velocity of Vi.

q⎛ ⎞Δ = ⎜ ⎟⎝ ⎠2

2iv v Sin

Plane changes are often combined with orbit-change burns. Usual practice is to do a little of the plane change at the initial burn (at the lower orbit where velocity is higher) and most of the plane change at the final burn at the higher altitude where the velocity is lower. Plane change combined with Hohman transfer.

( )( )qΔ = + −2 2 2i f i fv v v v v Cos

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EXTROVERT Space Propulsion 03Launch from Earth

Note that the objective is to achieve the required tangential velocity for orbit – the need to gain height above Earth is mainly dictated by the need to get out of the atmosphere and minimize drag losses.  As the vehicle trajectory begins to slant over, gravity pulls the trajectory down. At low speeds, this requires a substantial pitch angle between the instantaneous velocity vector and the thrust direction to counter gravity.

0≠α

0≈α

Gravity Turn

1.    Begin ascent with short vertical rise, during which it starts usually and completes a roll from launch-pad azimuth to desired flight azimuth.

2.    Pitch-over (pitch because angular change of axis occurs faster than change of velocity vector from vertical)

3.    Adjust altitude to keep in trajectory through aerodynamic load region.

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EXTROVERT Space Propulsion 03Gravity turn

If the angle of the trajectory with respect to the local horizontal is gamma (g) then, this angle changes as

Local horizontal

V(t)V(t+dt)

g

-dg

gldt

glCos g dt

−( )F D dtm

rtermshigherordeVdtgd e

+=−

gg cos(rad.) or

cosegddt V

gg=− (rad/s)

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Controlled Angle of Attack – Linear Tangent Steering

Angle of attack or the tangent of the thrust angle with respect to the local horizontal, are controlled as functions of time to minimize losses.

Angle of attack can be increased as dynamic pressure decreases beyond the “max q” point, subject to limitations on vehicle bending moments. Tangent of the thrust angle usually starts at a high value and is reduced according to a linear or other specified schedule, i.e., the slope B is usually negative. .

q = +Tan A Bt

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Launch Trajectories

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Acceleration During Launch

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EXTROVERT Space Propulsion 03Interplanetary Trajectories

Earth-Mars Hohmann scheme

1. Escape velocity from Earth – to an orbit around the Sun, with zero velocity with respect to Earth. 2. Impulsive burn to get into transfer orbit. 3. Impulsive burn at Mars orbit to match velocity with Mars.

Δ = +Δ +ΔHoffman esc T Marsu u u u

( )ε ε

Δ= +

2

2T

Hoffman escu

( )Δ=

2

2Tu

Change in orbital energy per unit mass (w.r.t. Earth)

ΔUT

ΔUMars

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EXTROVERT Space Propulsion 03Data on Solar System Destinations

Body Relative Mass

Rel. Mean Dia.

Orbit radius, Millions of km

Eccentricity Period, Earth yrs

Sun 332,488. 109.15

Mercury 0.0543 0.38 57.9 0.2056 0.241

Venus 0.8136 0.967 108.1 0.0068 0.615

Earth 1.0000 1.000 149.5 0.0167 1.000

Mars 0.1069 0.523 227.8 0.0934 1.881

Jupiter 318.35 10.97 777.8 0.0484 11.862

Saturn 95.3 9.03 1426.1 0.0557 29.458

Uranus 14.58 3.72 2869.1 0.0472 84.015

Neptune 17.26 3.38 4495.6 0.0086 164.788

Plato <0.1? 0.45 5898.9 0.2485 247.697

Moon 0.0123 0.273 0.3844 0.0549 27.322 days

Mass of Earth = 5.975 E24 kg. Mean Diameter = 12742. 46km

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Earth-Mars Oberth Scheme

Escape from Earth while carrying enough propellant to produce the transfer velocity is wasteful in work against gravity. Can be avoided by achieving the right velocity magnitude and direction in the first maneuver.

The launch velocity is larger at Earth, directly placing the vehicle in the Hohman transfer ellipse. Thus

ε = −2

e2 ROberth e

Oberthu GM

Δ− =

22 2

2 2 2Oberth esc Tu u u

= +Δ 22escOberth Tu u u

< + Δesc Tu uΔ = +ΔOberth Oberth Marsu u u

Complications: 1) Atmospheric drag 2) Orbit desired @ Mars

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Faster Trajectories for Human-Carrying MissionsFour impulsive burns: 1. Escape from Earth towards Mars with an excess over the

Hohman transfer velocity. 2. Increment to get captured into Mars orbit: Most trajectories put

the vehicle in Mars orbit, moving slower around the Sun than Mars – need a burn to get captured.

3. Opposite of the above to escape Mars into a transfer trajectory towards Earth

4. Burn to decelerate to a safe Earth re-entry velocity.

Months Total delta-v, fps

5 120,000

8 55,000

12 25,000

15 19,000

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Values for Sample Missions: Garrison & StuhlerΔV m/s

Earth- LEO (270km above Earth’s surface) 7600LEO – GEO (42,227 km radius) 4200LEO – Earth escape 3200LEO – Lunar Orbit (7 days) 3900LEO – Mars orbit (0.7 yrs) 5700LEO – Mars orbit (40 days) 85000LEO – Neptune Orbit (50 yr) 13,400LEO – Neptune Orbit (29.9 yr) 8,700LEO – Solar escape 70,000LEO – 1000 AU (50yr)LEO – Alpha Centauri (50yr)

142,00030 Million

1 Astronautical Unit = mean distance from Earth to Sun = 149,558,000km Impulsive burns assumed in all the above.

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Spiral Transfer: Low, Continuous Thrust

Delta v required is greater than that for Hohman transfer, but a small engine can be used. Also, possibility of using solar energy as the heat source. Thrust applied along the circumferential directionis then given by

q q⎛ ⎞⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

2

2 2d dr dT m rdt dtdt

orq⎛ ⎞= ⎜ ⎟⎝ ⎠

2Tr d drm dt dt

rr r νrτ τ

⎛ ⎞+ =⎜ ⎟⎜ ⎟⎝ ⎠

1/ 223

2d dd d

with initial conditions at qτ=1d

drτ

=0dd

r =1τ =0

where r =0

rr τ = 3

0

GM tr

ν = =2

0Tr TGMm mg

Solution shows that the delta v required with a very small thrust-to-EarthWeight ratio is about 2.4 times the impulsive-burn delta v. As thrust-to-weight approaches 1, this comes down close to 1.

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Spiral Transfer: Earth-Moon

“SMART-1 to continue alternating between long and short thrust arcs until 15 Nov lunar capture: orbital period 4 days, 19 hrs; apogee altitude has increased from 35,880 km to 200,000 km; lunar commissioning, science operations begin in mid-Jan 2005” http://sci.esa.int/

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EXTROVERT Space Propulsion 03Gravity Assist Maneuvers

http://science.howstuffworks.com/framed.htm?parent=question102.htm&url=http://www.jpl.nasa.gov/basics/bsf4-1.html

Vehicle approaches another planet with velocity greater than escape velocity for that planet. Planet’s gravity adds momentum to the vehicle in desired direction.

With respect to the planet, vehicle gains speed on approach and decelerates as it moves away. But when the planet’s velocity is considered, there is net gain in angular momentum with respect to the Sun.

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EXTROVERT Space Propulsion 03Orbital Elements

a: semi-major axis.

e: eccentricity .

i: inclination of the angular momentum vector from the Z-axis. (Z is usually Earth’s axis of rotation.

W: right ascension of the ascending node: Angle from the vernal equinox to the “ascending node.” “node” : one of two points where the satellite orbit intersects the X-Y (equatorial) plane. Ascending node is the point where the satellite passes through the equatorial plane moving from south to north. Right ascension is measured in degrees of right-handed rotation about the pole (Z axis).

w: argument of perigee. The angle from the ascending node to the eccentricity vector, measured in degrees in the direction of the satellite’s position. The eccentricity vector points from the center of the Earth to the perigee with magnitude equal to the eccentricity of the orbit, in degrees.

ν : “true anomaly”. Angle from the eccentricity vector to satellite’s position vector, measured in the direction of the satellite motion. Can also use the time since perigee passage, in degrees.

( )21e v r r v vrμ

μ⎛ ⎞⎧ ⎫⎛ ⎞= − − •⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠⎩ ⎭⎝ ⎠

r r r r r