Transcript Trip to Mars

Trip to Mars
How do we get there?
OAPT May ’08
By John Berrigan
[email protected]
The Theory
How do we get from Earth to Mars?
The Problem
The trip to Mars is a complicated “multibody”
problem.
The main players are:
–
–
–
–
–
Probe
Earth
Mars
Sun
Jupiter (not a major player but for long trips can move
you off course)
The Solution
We will change the multi-body problem down
to a series of two body problems.
1) Earth/probe
2) Sun/probe
3) Mars/probe
The result gives a pretty accurate
representation of what needs to be done.
How should we get there?
Traveling in space is expensive. At present,
depending upon the source, it costs around
$10,000 / kilogram to put into low Earth orbit, so
fuel savings is important.
Less fuel needed for trip = less cost.
As well, launching payloads to orbit can mean
large launch increases. If the payload reaches a
certain mass, a more expensive launcher is
needed.
The Hohmann transfer orbit is one way to
minimize the costs.
Hohmann Transfer Orbit
We want to go from the inner orbit to the
outer orbit.
The red orbit is the
smallest transfer orbit
from the lower orbit to
the higher orbit. This is
called the Hohmann
transfer orbit
The Hohmann transfer orbit involves a low energy
transfer. It only requires two boosts of energy or
delta-v’s to change orbits.
Δv1
The Delta-v’s
Δv2
Destination orbit
Transfer orbit
Δv needed
Destination orbit
Transfer orbit
Δv needed
Δv1 gets you into transfer orbit
Δv2 gets you into destination orbit
Both Δv’s involve change in speed not direction
since velocities are tangential to the orbit.
Larger Energies
Δv2
The two orbits may actually
have the same speed at
that point…but the
Direction change is the
main factor.
You can do the transfer using a larger Δv on the first burn.
This means a larger Δv is needed at the other side.
The 2nd Δv both changes the magnitude AND direction.
It is a faster route but “more expensive” due to more fuel.
The Physics of it All!!
What do we need to know?
• Ellipse properties
• Fnet=Fcentripetal
• Energy conservation
– Kinetic Energy (½mv2)
– Gravitational Potential Energy(-GMm/r)
• Orbital Velocity Equation
• Relative motion
• Kepler’s 3rd Law
The Ellipse….
Review on ellipses
• Objects orbit in ellipses.
• Central body at one of the focus points
VP = Velocity at periapsis
VA = Velocity at apoapsis
rp = Periapsis
rA = Apoapsis
vp > vA
vA
vp
Major axis = 2a = rp + rA
a = semimajor axis
= ½ (rp + rA)
The Ellipse continued…
Equation
x2/a2 + y2/b2 = 1
b
ae
-a
a
-b
e is the eccentricity. Simply how oval it is.
Changes position of focus relative to the “x-int”
e = 0, circle
e < 1 ellipse
e = 1 parabola
e > 1 hyperbola
Additional Jargon
• Periapsis is the closest point from a focus.
• Apoapsis is the farthest point from a focus.
These names can be modified to the body
being orbited:
Sun (helion) = Perihelion and aphelion
Earth (gee) = Perigee and apogee
Moon (lune) = Perilune and apolune
Mars (areion) = Periareion and Apoareion
Energy Conservation
How fast must you go to JUST escape the
Earth?
V=?
R=r
ET = EK + EP
ET= ET’
½mv2 -GMm/r = 0
Therefore,
v2 = 2GM/r
For Earth
vescape = ~11.1 km/s
R’ = infinity
V’ = 0
Therefore ET’ = 0
Relative motion
From previous slide we found the escape velocity.
This means at infinity, the velocity is zero relative
to the Earth.
Earth
R = “infinity”
If we change the frame of reference to the Sun, the
Earth has a velocity. That means when the probe
“gets to infinity”, the probe has the same speed as
the Earth.
Probe
Sun
Earth
Even though the probe never gets an infinite
distance away, we can argue that the probe is in
the same orbit as the Earth (since it has the same
speed) but it is outside of the Earth’s Gravitational
influence.
So we obviously can’t get to Mars with just the
escape speed.
So What do we need to do?
When we get to “infinity”, we need to have a
velocity in order to change orbits!!
But how much faster?
As Buzz Light-year has famously said, we need to
go…
“To infinity and beyond!!!”
Circular Orbits: Orbital Energy
To solve for the trajectory we need to review
orbital energy.
Fnet = Fg
F c = Fg
mv2/R = GMm/R2
v2 = GM/R
Fg
ET = EK + EP
= ½mv2 +( -GMm/R)
= ½m(GM/R) – GMm/R
= ½(-GMm/R)
= ½ EP
This means that in a circular orbit the total Energy is equal
to one half the potential Energy at that radius.
Elliptical Orbits
Since a Circle is a type of ellipse we can
modify the Total Energy equation
ET = -½GMm/R.
The radius is really the semi-major axis so
ET = -½GMm/a
Where “a” is the semi-major axis.
Elliptical Orbital Velocities
vA
EP
vp
We know energy is conserved so
ET = Ep = EA
ET = EK + EP
-½GMm/a = ½mv2 – GMm/r
Rearranging and solving for v we get
v2 = GM(2/r – 1/a)
EA
Advanced solution
If you introduce angular momentum, R x V,
at periapsis and apoapsis, R and V are
perpendicular. Therefore,
rpvp = rAvA , we can then derive the equation.
We know EA = Ep. Therefore,
½mvA2 – GMm/rA = ½mvp2 – GMm/rp
Substitute for vp and simplify. After a bunch
of math we get
VA2 = GM(2/rA – 1/a)
(this is a GREAT exercise for the stronger math students in the class!!)
What can we do now?
We now can solve a good chunk of the problem!
• Find the velocity of the Earth and Mars by
using Fnet = Fg. (We will assume they are
circular orbits.)
• Determine rA , rp , “a” of the transfer orbit. (An
extension, find eccentricity of the orbit.)
• Determine vA and vp.
This data can now be used to determine the
Δv’s needed for the transfer orbits.
The Orbit data and our results
Earth (Circular orbit)
r = 1.50e11 m, v = 29.7 km/s
Mars (Circular orbit)
r = 2.27e11 m, v = 24.2 km/s
Transfer orbit, (Elliptical orbit)
rp = 1.50e11 m, rA = 2.25e11 m, a = 1.885e11 m
vp = 32.6 km/s, vp = 21.6 km/s
The Delta v’s
Δv1
Δv2
Therefore delta v’s needed are
Δv1 = |Vp – VEarth| = 2.9 km/s
Δv2 = |VA – Vmars| = 2.6 km/s
These delta v’s are the values for the two
body problem of the probe and the Sun.
Now to leave and arrive!!
Now that we have figured out the transfer
orbit, we now need to worry about how Mars
and Earth affect the values.
Using relative motion, we will now address
the two body problem of the probe and
Earth, as well as, the probe and Mars
Earth launch speed
We found that Δv1 to be 2.9 km/s. Therefore
the probe needs to travel 2.9 km/s faster
than the Earth is traveling.
So, the probe, after
Δv
launching from the Earth,
must have a velocity of 2.9
km/s when it gets to
“infinity”.
1
Δv2
Orbit Transfer
How fast must you launch from Earth’s
surface to get into transfer orbit?
V’ = 2.9 km/s
V=?
R=r
ET = EK + EP
ET= ET’
½mv2 -GMm/r = ½mvinfinity2
vlaunch = ~11.6 km/s
R’ = infinity
ET’ = EK’
Note: there is small difference (400 m/s) in launch velocity for JUST
escaping and having final velocity of 2.9 km/s.
Arriving at Mars
Arriving at Mars is a little different.
We found that Mars is traveling 2.6 km/s faster
than the probe at the transfer point. (So Mars is
actually catching the probe.)
This means relative to Mars, at “inifinity” the probe
is approaching Mars at a speed of 2.6 km/s.
What Δv is needed to arrive at the planet?
Depends!!!!!
Do you want to land or orbit?
Landing on Mars
How much should you slow down when you
arrive at the Martian’s surface?
Vinfinity = 2.6 km/s
V’ = ?
R’ = r
ET’ = EK’ + EP’
The calculation:
ET’ = ET
½mv2 -GMm/r = ½mvinfinity2
v = ~5.7 km/s
R = infinity
ET = EK
So to land you need a Δv of 5.7 km/s**
**You are going to get this delta V regardless.... Trick is doing it safely.
Just ask the Mars Polar Lander of 1999.. cross fingers for tomorrows
landing of the Lander's Sister, Phoenix.
What is the real answer?
A quote from the FAQ from the Phoenix Lander site.
“Entry, Descent and Landing
The intense period from three hours before the spacecraft
enters Mars’ atmosphere until it reaches the ground safely
is the mission phase called entry, descent and landing. The
craft will hit the top of the atmosphere at a speed of 5.7
kilometers per second (12,750 miles per hour). Within the
next six and a half minutes, it will use heat-generating
atmospheric friction, then a parachute, then firings of
descent thrusters, to bring that velocity down to about 2.4
meters per second (5.4 miles per hour) just before
touchdown.”
Not too bad for some approximations!!
Orbiting Mars
To find the delta V, we first need to find the
orbital velocity in the final orbit. Lets
assume at an altitude of 200 km.
From earlier, v2 = GM/R,
so orbital velocity is 3.5 km/s.
Orbiting of Mars
Now to find the velocity as probe approaches
from infinity. If no Δv, probe does a “fly by”.
V’ = ?
R’ = r
ET’ = EK’ + EP’
Vinfinity = 2.6 km/s
R = infinity
ET = EK
The calculation:
ET’ = ET
½mv’2 -GMm/r = ½mvinfinity2
v’ = ~5.6 km/s
To orbit you need to a Δv of 5.6 km/s - 3.5 km/s
Or 2.1 km/s
Quick quiz
Lets see who is awake..
Q: What happens if you want to go into a
200 km circular orbit and the Δv is smaller
or larger than the 2.1 km/s needed?
A: Since you are really taking energy away
from the orbit when using the Δv, you are
changing the type of conic section the final
orbit will be in.
Orbit Energy
If Δv = 2.1 km/s orbit is a circle.
If Δv > 2.1 km/s, final orbit
energy is less.
If Δv is a little < 2.1 km/s
Quiz #2
What Δv is too small or too large??
Energy
If Δv is larger than 2.1 km/s and the
Periareion takes us into the atmosphere.
If Δv is smaller than 2.1 km/s and the total
Energy relative to Mars is:
•
Negative: ellipse (the larger the
negative, the smaller the semi major axis,
smaller the orbital period)
•
Zero: parabolic orbit (escapes)
•
Positive: Hyperbolic orbit (escapes)
Back to our problem….
Mom, we there yet?
Not quite…
So far we know:
• Earth Δv = 11.6 km/s
• Mars Δv is
• 5.6 km/s to land
• 2.1 km/s to circular orbit
Now we have to make sure Mars is there
when we get there!!
Where should Mars be when we launch?
When do we Launch?
Now for Kepler’s law!!
Remember T2 = K R3,
we can use this to find how long it takes to
get to Mars and how long Mars travels in
that time.
Once again, we can modify Kepler’s law to
any ellipse.
So, T2 = K R3 becomes T2 = K a3 where “a”
is the semi major axis.
Working with Kepler’s law
The K value can be of any units. For ease
of use, T is in years and “a” is in 1011 m.
To find K for the sun, use Earth data.
Earth
Transfer
Mars
Tearth = 1 year, atransfer = 1.885
aMars = 2.27
T2 = Ksun (1.885)2 T2 = Ksun (2.27)2
aearth = 1.5 so
T = 1.41 years T = 1.87 years
-3
Ksun = 1.5
Almost done…..
Transfer orbit takes 1.41 years to do a full orbit. So
it takes 0.705 years or 8.46 months for half that
orbit.
How far does Mars Travel during the transit time?
Simple ratio
Degrees = 360o = __x__
Period
1.86 0.705
X= 136o.
So Mars travels 136o while probe heads towards
Mars.
FINALLY!!!
Since the probe arrives at Mars 180o from
where Earth was at launch. Mars must be
180o – 134o = 46o in front of the Earth at
launch.
46o
When can we do it again?
Angular Velocity of Earth = 360o/1 year
Angular Velocity of Mars = 360o/1.86 year.
Difference is 166o per year
or
360o change in 2.16 years or 26 months.
Which is why we try go to Mars Every 26
months…
Ok, what now??
With the basics covered you can have lots of
extensions.
In real launches, most times the rocket puts the
probe into a circular orbit around the Earth first,
does a self check to see if all is well and then a
delta v takes it to the transfer orbit.
What Δv is needed
to get a Vinfinity =
2.9 km/s?
Design a mission
•
•
•
•
•
•
•
To have the arrival orbit as an ellipse.
To land on an asteroid.
To Orbit an asteroid.
To the moon.
To change orbit altitude around Earth.
To dock to Space Station once in orbit.
Calculate Delta V to land the shuttle
Note: Keep the objects orbits circular for ease of
calculation. Ellipses make it harder to figure out where
the planet is at a given time. (That can be another
presentation.)
How can you mark it???
• Answers can be “easily” created in excel.
• Give each group data for a “planet”.
• Minimizes copying. But encourages
discussion among groups
• You just check if they are right or not.
• I have a program that I get the kids to plug
numbers into to check if they are right.
Multibody problem method
•
•
•
•
•
Can Involve “Weak Stability Boundary”
No empirical solution
Can involve chaotic effects
Uses MUCH less fuel
Langrange points can be used
Golf putting analogy
Two body problem ignores little dips and valleys on
the green. Power the putt over the breaks.
Multibody problem can take the dips into account,
putt more slowly, ball JUST drops into the cup.
Lagrange points
Earth Sun Lagrange points or libation points
Lagrange points
Gravitational “topographical” force map
Phoenix Lander
from April 25th
May 23rd
Phoenix's Trajectory
Phoenix's Landing
What time will Phoenix land on Mars? What
time will the first signal be received from
Phoenix?
Phoenix will land at approximately 4:36pm Pacific
Daylight Time (7:36pm Eastern Daylight Time). We
hope to receive the first signal from the lander
approximately 17 minutes later at 4:53pm PDT
(7:53pm EDT).
Discovery channel has live feed at 7:00 pm on
Sunday.
Live NASA coverage starts at 4:45..go to their web
site
How can we get to the moon?
Resources
– Fundamentals of Astrodynamics by Bate,
Mueller and White
– Fly Me to the Moon: An Insider's Guide to the
New Science of Space Travel by Edward
Belbruno
– “Orbiter: Spaceflight simulator” by Martin
Schweiger. A FREE program. A STEEP
learning curve but fun. NOT a game!!
– SpaceX.com some cool goings on….
Some Orbit misconceptions:
Orbital Period
Period is independent of eccentricity.
Since T2 = K a3, the only factor is the semimajor axis. How “oval” it is, is irrelevant.
Orbital Velocity
Velocity is independent of eccentricity.
Since v2 = GM(2/r – 1/a), this shows that the
velocity of the object is only a function
radius if the semi major axis is the same.
Orbit Change
Δv towards the “ground” does not lower the
satellite.
It would put it in a higher orbit since the final
velocity would be higher then the start so the
overall energy is higher (less negative) which
means larger semimajor axis since ET= -GMm/a.
Vinitial
ΔV
Vfinal
Docking
If you are behind an object, you “slow down”
to dock with it.
Slightly Counter intuitive.
But, if you speed up significantly to try
docking you would actually drift away.
Faster speed. Larger semi-major axis.
Higher you go, slower your speed, object
gets farther in front.
ET = Ep
Conclusion
ET’ = ½Ep’
ET’’ = 0
ΔE = ~ ½Ep
Robert A. Heinlein, "Once you make it to
orbit, you're half-way to anywhere."
Space tidbits
• Spacex 2nd launch shut down
– Pressure was too low on first attempt so scrubbed..warmed up
fuel..launched 1 hour later
– Slight Bias..Lets hope SpaceX is successful… Next launch June
24th..hopefully
•
Off topic
– Teslamotors
• Bigelow.. Two orbiting “stations”
• Virigin Galactic..
– First two space craft are:
– VSS Enterprise
– VSS Voyager
• Google lunar X-prize
– 14 teams now..$30 million dollar prize
• Mars Science Laboratory (MSL)
– Launch ~Sept 2009, may be last Mars probe for a while…
• Lunar reconnaissance Orbiter
– November launch
MSL. It is BIG