Space Travel Overview (From 2004)
Space Travel Overview: You Can Get There from Here! John F Santarius
Lecture 5 Resources from Space NEEP 533/ Geology 533 / Astronomy 533 / EMA 601 University of Wisconsin
January 30, 2004 Outline
• Leaving Earth ¾ Laws of gravitation and motion ¾ Rocket physics • Near-Earth space ¾ Celestial coordinates ¾ Orbital dynamics ¾ Lagrange points • Traveling to the planets and asteroids ¾ Minimum-energy (Hohmann) transfers ¾ Gravity assists ¾ Low-thrust trajectories
JFS 2004 Fusion Technology Institute, University of Wisconsin 2 Newton’s Law of Gravitation
• Every particle of matter attracts every other particle of matter with a force directly proportional to the product of the masses and inversely proportional to the square of the distance between them. −GMm Fr= ˆ r 2
G = 6.67×10-11 m3 s-2 kg-1 is the gravitational constant.
JFS 2004 Fusion Technology Institute, University of Wisconsin 3 Newton’s Laws of Motion
• The fundamental laws of mechanical motion were first formulated by Sir Isaac Newton (1643-1727), and were published in his Philosophia Naturalis Principia Mathematica. • Calculus, invented independently by Newton and Gottfried Leibniz (1646-1716), plus Newton's laws of gravitation and motion are the tools needed to understand rocket motion.
JFS 2004 Fusion Technology Institute, University of Wisconsin 4 Newton’s Laws of Motion
• Every body continues in its state of rest or of uniform motion in a straight line except insofar as it is compelled to change that state by an external impressed force.
• The rate of change of momentum of the body is proportional to the dp impressed force and takes place F = in the direction in which the force acts. dt
• To every action there is an equal and opposite reaction.
JFS 2004 Fusion Technology Institute, University of Wisconsin 5 Tsiolkowsky’s Rocket Equation
• First derived by Konstantin Tsiolkowsky Konstantin in 1895 for straight-line rocket motion Tsiolkowsky with constant exhaust velocity, it also applies to elliptical trajectories with only initial and final impulses. • Conservation of momentum leads to the so-called rocket equation, which shows the dependence of payload fraction on exhaust velocity. • It assumes short impulses with coast phases between them, such as used for chemical and nuclear-thermal rockets.
JFS 2004 Fusion Technology Institute, University of Wisconsin 6 Tsiolkowsky’s Rocket Equation
• Assuming constant exhaust velocity, conservation of momentum for a rocket and its exhaust leads to dp d()mv dm dv ==vm+=0 dt dt dt dt dm −dv ⇒= mvex
Rocket equation mf ≡ final mass
mi ≡ initial mass m f −∆v ⇒=exp∆v ≡ velocity increment mv iexvex ≡ exhaust velocity
JFS 2004 Fusion Technology Institute, University of Wisconsin 7 High Exhaust Velocity Gives Large Payloads
• This plot of the rocket equation shows why high exhaust velocity historically drives rocket design: payload fractions depend strongly upon the ratio of the exhaust velocity to the ∆v. 1 • Vchem≤5 km/s
0.8 • Earth-escape ∆v~11.2 km/s o i t
c 0.6 a • Earth-orbit to r f
d Mars-orbit a o l 0.4
y ∆v~5.6 km/s a Pn Earth-orbit to 0.2 • Jupiter-orbit ∆v~14.4 km/s 0 0 1 2 3 4 5 Exhaust velocity Dv
JFS 2004 Fusion Technology Institute, University of Wisconsin 8 ê Many Factors Affect a Rocket’s Launch
• Number of stages
¾ Staging gains efficiency by eliminating mass of lower stages when accelerating a payload.
¾ Analyze by applying the rocket equation in series to the stages. • Gravitational loss
¾ Due to finite launch time; (~g ∆t), typically~25%. • Atmospheric drag
¾ Typically ~5%. • Divergence of exhaust from nozzle
JFS 2004 Fusion Technology Institute, University of Wisconsin 9 Near-Earth Space
• Distances are in Earth radii: ~6400 km
6 GEO 60 4 Moon 2 40 LEO
-6 -4 -2 2 4 6
20 -2 GEO Earth -4 -60 -40 -20 20 40 60 -6 -20 Earth -40 LEO ≡ Low Earth Orbit -60 GEO ≡ Geosynchronous Earth Orbit
JFS 2004 Fusion Technology Institute, University of Wisconsin 10 Escape from Earth
• Earth’s mass = 6 x 1024 kg • Earth’s average radius = 6380 km = 6.38 x 106 m • G=6.67×10-11 m3 s-2 kg-1
1/2 1/2 GM 26××.6710−11 ×6×1024 vesc ==2 6 r 6.38×10 =11.2 km/s
• The actual velocity increment required to escape from Earth is somewhat higher, because of the finite time of flight and atmospheric friction.
JFS 2004 Fusion Technology Institute, University of Wisconsin 11 Some Useful Orbital Dynamics Formulas
Circular velocity Escape velocity 1/2 1/2 GM GM vcir = vesc = 2 r r Velocity of a body on Energy (potential plus an elliptical orbit kinetic) of a body on 1/2 any conic-section orbit 21 (E<0 ⇒ trapped orbit) vGellipse =−M ra GM E = conic 2a G = 6.67×10-11 m3 s-2 kg-1 a ≡ semi-major axis
JFS 2004 Fusion Technology Institute, University of Wisconsin 12 Defining a Celestial Body’s Position Requires Six Parameters N • Two parameters orbital are not shown: plane
a, semimajor axis satellite motion e, eccentricity equatorial f true plane • Four of the six anomaly parameters appear apogee ii vernal inclination in purple at right. line of perigee inclination equinox nodes ω γ argument α right of perigee ascension ascending • These are equatorial node coordinates--appropriate for an Earth satellite. S JFS 2004 Fusion Technology Institute, University of Wisconsin 13 Lagrange Points are Equilibrium Positions in a Three-Body System
• The points L1, L2, and L3 lie on L4 a straight line through the two main bodies and are points of unstable equilibrium. That is, a small perturbation will cause the third body to drift away. L3 L1 L2 • The L4 and L5 points are at the third vertex of an equilateral triangle formed with the other Earth two bodies; they are points of Sun stable equilibrium. • In general, the three bodies can L5 have arbitrary masses.
JFS 2004 Fusion Technology Institute, University of Wisconsin 14 L1, L2, and L3 Lagrange Points for the Sun-Earth and Earth-Moon Systems
Sun-Earth
L3 L1 L2 X XX
-1.5 -1 -0.5 0 0.5 1 1.5 Distance AU
Earth-Moon H L
L3 L1 L2 X XX
-1.5 -1 -0.5 0 0.5 1 1.5 Distance Ear th -Moon units
JFS 2004 Fusion Technology Institute, University of Wisconsin 15 H L The Solar System
3 30 2 Asteroids
20 Earth 1 Earth Mercury Neptune 10 -3 -2 -1 1 2 3 Jupiter -1 Venus Mars -40 -20 20 -2 Saturn -10 -3 Pluto Uranus -20
-30
• Distances in astronomical units: 1 AU=1.5x108 km JFS 2004 Fusion Technology Institute, University of Wisconsin 16 Kepler’s Laws of Planetary Motion
Conic sections 3 Hyperbola
• The planets move in ellipses with the Parabola 2 sun at one focus. (escape) • Areas swept out by the radius vector Ellipse 1 Circle from the sun to a planet in equal times are equal. -2-1.5-1-0.5 0.5 1 • The square of the period of revolution is proportional to the cube -1 of the semimajor axis. T2 ∝ a3 -2 -3
JFS 2004 Fusion Technology Institute, University of Wisconsin 17 Hohmann’s Minimum-Energy Interplanetary Transfer
Earth-Mars • The minimum-energy, two-impulse transfer between circular orbits is an Hohmann transfer elliptical trajectory called the Hohmann Mars trajectory, shown at right for the Earth- Mars case. • A Hohmann transfer consists of three phases: Sun 1. Large impulse to leave circular orbit 2. Long coast phase on an elliptical orbit 3. Large impulse to match target planet’s Earth velocity
JFS 2004 Fusion Technology Institute, University of Wisconsin 18 Hohmann’s Minimum-Energy Interplanetary Transfer Can Be Calculated Easily
• The Hohmann trajectory appears at ∆v=2.63 km/s right for the Earth-Mars case, where Mars v =24.1 km/s the minimum total delta-v expended is M 5.6 km/s. • Shown are the velocities on Earth and Mars orbits plus the ∆v values required 9-month Sun travel time to put the rocket onto the elliptical
Hohmann trajectory at perihelion ∆v=2.92 km/s (closest to Sun) and aphelion (farthest Earth from Sun). vE=29.7 km/s ¾ Assumes circular, co-planar orbits.
• The sum of the velocity increments at Earth and Mars is the total delta-v value needed in the rocket equation.
JFS 2004 Fusion Technology Institute, University of Wisconsin 19 Hohmann Transfer Time
• Kepler's third law, T2 ∝ a3, can be used to calculate the time required to traverse a Hohmann trajectory by raising to the 3/2 power the ratio of the semimajor axis of the elliptical Hohmann orbit to the circular radius of the Earth's orbit and dividing by two (for one-way travel): 3/2 Ta • For example, call the travel time for an 11= Earth-Mars trip T and the semimajor 1 Ta22 axis of the Hohmann ellipse a1: Use Earth as reference: a1 = (1 AU + 1.5 AU)/2 = 1.25 AU a = 1 AU 3/2 2 T1 = 0.5 (a / 1 AU) years T = 1 year = 0.7 years = 8.4 months 2
JFS 2004 Fusion Technology Institute, University of Wisconsin 20 Many Spacecraft Trajectories Can Be Approximated by Hohmann Orbits Galileo Spacecraft’s Trajectory a = (1 AU + 5.2 AU)/2 = 3.1 AU T = 0.5 (a / 1 AU)3/2 years = ~2.7 years
The final phase (nearly elliptical) of Galileo’s trip to Jupiter took ~3 years.
JFS 2004 Fusion Technology Institute, University of Wisconsin 21 Gravity Assists Enable or Facilitate Many Missions
• In the planet’s frame of reference, a spacecraft enters and leaves the sphere of influence with a so-called hyperbolic excess velocity, vh, equal to the vector sum of its incoming velocity and the planet's velocity. 10 Motion in Jupiter’s Vh
i frame of reference d
a 5 r r e t i p u
Ji 0 e c n a t
s -5 i
D, Vh
-10 -20 -15 -10 -5 0 5 Distance, Jupiter radii
JFS 2004 Fusion Technology Institute, University of Wisconsin 22 Gravity Assists Enable or Facilitate Many Missions
• In the planet's frame of reference, the direction of the spacecraft's velocity changes, but not its magnitude. In the spacecraft's frame of reference, the net result of this trade-off of momentum is a small change in the planet's velocity and a very large delta-v for the spacecraft.
Vi
VJupiter Vi • Starting from an Earth-Jupiter Hohmann trajectory, a flyby of Jupiter at one Jovian radius has these parameters:
¾Hyperbolic excess (beyond escape) velocity vh~5.6 km/s ¾Flyby gives an angular change in direction of ~160o if the rocket skims Jupiter’s atmosphere. ¾Flyby gives a ∆v of ~13 km/s
¾Final velocity = 13+5.6 km/s=18.6 km/s (greater than Solar-System vesc)
JFS 2004 Fusion Technology Institute, University of Wisconsin 23 Missions to Mars and Beyond Require Larger Exhaust Velocities than Chemical Rockets Can Produce
Earth-Mars one-way trip: ∆v~5.6 km/s (Hohmann) 1 0.8 o i t c
a 0.6 r f d
a 0.4
o Chemical l
y rocket a
Pn 0.2
0 0 10 20 30 40 50 Exhaust velocity, km s
JFS 2004 Fusion Technology Institute, University of Wisconsin 24 Separately Powered Systems Differ Significantly from Chemical Rockets
• Propellant not the power source; power system adds mass. • High exhaust velocity (≥105 m/s). Low thrust (≤10-2 m/s} ≡10-3 Earth • Mars gravity) in most cases. Sun • Thrusters operate for a large fraction
of the mission duration. Earth • High-exhaust-velocity trajectories differ fundamentally from chemical- rocket trajectories (figure at right). Note: Trajectory is schematic, not calculated.
Separately powered rocket equation Ml=payload mass M =power and propulsion system MM+ −∆v w lw= exp mass MMlw++Mp vex Mp=propellant mass
JFS 2004 Fusion Technology Institute, University of Wisconsin 25 Figures of Merit for Separately Powered Systems are Exhaust Velocity (m/s) and Specific Power (kW/kg)
107 Fusion )
s 6 / 10 10 kW/kg m (
y 1 kW/kg t i c o l 5 0.1 kW/kg e 10 v t Nuclear
us (fission) Ga s-core fission
ha electric x 4 E 10 Nuclear thermal
Chemical 103 10-5 10-4 10-3 10-2 10-1 1 10 Thrust-to-weight ratio
JFS 2004 Fusion Technology Institute, University of Wisconsin 26 Where Do We Go from Here?
• Chemical propulsion (Mike Griffin, lecture (26) • Plasma propulsion (lecture 27) • Fusion propulsion (lecture 28) • Getting to the asteroids—and back (lecture 29)
Artwork by Pat Rawlings
JFS 2004 Fusion Technology Institute, University of Wisconsin 27