Trip to Mars How do we get there? OAPT May ’08 By John Berrigan

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 The Hohmann transfer orbit involves a low energy transfer. It only requires two boosts of energy or delta-v’s to change orbits. The red orbit is the smallest transfer orbit from the lower orbit to the higher orbit. This is called the Hohmann transfer orbit We want to go from the inner orbit to the outer orbit.

The Delta-v’s Δv 1 gets you into transfer orbit Δv 2 gets you into destination orbit Both Δv’s involve change in speed not direction since velocities are tangential to the orbit. Δv1Δv1 Δv2Δv2 Destination orbit Transfer orbit Δv needed Destination orbit Transfer orbit Δv needed

Larger Energies 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 2 nd Δv both changes the magnitude AND direction. It is a faster route but “more expensive” due to more fuel. Δv2 The two orbits may actually have the same speed at that point…but the Direction change is the main factor.

The Physics of it All!!

What do we need to know? Ellipse properties F net =F centripetal Energy conservation –Kinetic Energy (½mv 2 ) –Gravitational Potential Energy(-GMm/r) Orbital Velocity Equation Relative motion Kepler’s 3 rd Law

The Ellipse…. Review on ellipses Objects orbit in ellipses. Central body at one of the focus points V P = Velocity at periapsis V A = Velocity at apoapsis vpvp vAvA r p = Periapsis r A = Apoapsis Major axis = 2a = r p + r A a = semimajor axis = ½ (r p + r A ) v p > v A

The Ellipse continued… Equation x 2 /a 2 + y 2 /b 2 = 1 -a b a - b aeae 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 E T = E K + E P E T = E T ’ ½mv 2 -GMm/r = 0 Therefore, v 2 = 2GM/r For Earth v escape = ~11.1 km/s R’ = infinity V’ = 0 Therefore E T ’ = 0

Relative motion From previous slide we found the escape velocity. This means at infinity, the velocity is zero relative to the Earth. 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. Earth R = “infinity”

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. Probe Sun

So What do we need to do? As Buzz Light-year has famously said, we need to go… “To infinity and beyond!!!” When we get to “infinity”, we need to have a velocity in order to change orbits!! But how much faster?

Circular Orbits: Orbital Energy To solve for the trajectory we need to review orbital energy. FgFg E T = E K + E P = ½mv 2 +( -GMm/R) = ½m(GM/R) – GMm/R = ½(-GMm/R) = ½ E P F net = F g F c = F g mv 2 /R = GMm/R 2 v 2 = GM/R 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 E T = -½GMm/R. The radius is really the semi-major axis so E T = -½GMm/a Where “a” is the semi-major axis.

Elliptical Orbital Velocities We know energy is conserved so E T = E p = E A E T = E K + E P -½GMm/a = ½mv 2 – GMm/r Rearranging and solving for v we get v 2 = GM(2/r – 1/a) vpvp vAvA EPEP EAEA

Advanced solution If you introduce angular momentum, R x V, at periapsis and apoapsis, R and V are perpendicular. Therefore, r p v p = r A v A, we can then derive the equation. We know E A = E p. Therefore, ½mv A 2 – GMm/r A = ½mv p 2 – GMm/r p Substitute for v p and simplify. After a bunch of math we get V A 2 = GM(2/r A – 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 F net = F g. (We will assume they are circular orbits.) Determine r A, r p, “a” of the transfer orbit. (An extension, find eccentricity of the orbit.) Determine v A and v p. 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) r p = 1.50e11 m, r A = 2.25e11 m, a = 1.885e11 m v p = 32.6 km/s, v p = 21.6 km/s

The Delta v’s Therefore delta v’s needed are Δv 1 = |V p – V Earth | = 2.9 km/s Δv 2 = |V A – V mars | = 2.6 km/s Δv1Δv1 Δv2Δv2 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 Δv 1 to be 2.9 km/s. Therefore the probe needs to travel 2.9 km/s faster than the Earth is traveling. Δv1Δv1 Δv2Δv2 So, the probe, after launching from the Earth, must have a velocity of 2.9 km/s when it gets to “infinity”.

How fast must you launch from Earth’s surface to get into transfer orbit? V = ? R = r E T = E K + E P E T = E T ’ ½mv 2 -GMm/r = ½mv infinity 2 v launch = ~11.6 km/s R’ = infinity E T ’ = E K ’ V’ = 2.9 km/s Orbit Transfer 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?

How much should you slow down when you arrive at the Martian’s surface? E T ’ = E K ’ + E P ’ The calculation: E T ’ = E T ½mv 2 -GMm/r = ½mv infinity 2 v = ~5.7 km/s V’ = ? R’ = r R = infinity E T = E K V infinity = 2.6 km/s Landing on Mars 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 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, v 2 = GM/R, so orbital velocity is 3.5 km/s.

Now to find the velocity as probe approaches from infinity. If no Δv, probe does a “fly by”. E T ’ = E K ’ + E P ’ The calculation: E T ’ = E T ½mv’ 2 -GMm/r = ½mv infinity 2 v’ = ~5.6 km/s V’ = ? R’ = r Orbiting of Mars To orbit you need to a Δv of 5.6 km/s km/s Or 2.1 km/s R = infinity E T = E K V infinity = 2.6 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.

If Δv = 2.1 km/s orbit is a circle. Orbit Energy 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 T 2 = K R 3, 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, T 2 = K R 3 becomes T 2 = K a 3 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 m. To find K for the sun, use Earth data. Earth T earth = 1 year, a earth = 1.5 so K sun = Transfer a transfer = T 2 = K sun (1.885) 2 T = 1.41 years Mars a Mars = 2.27 T 2 = K sun (2.27) 2 T = 1.87 years

Almost done….. Transfer orbit takes 1.41 years to do a full orbit. So it takes years or 8.46 months for half that orbit. How far does Mars Travel during the transit time? Simple ratio Degrees = 360 o = __x__ Period X= 136 o. So Mars travels 136 o while probe heads towards Mars.

FINALLY!!! Since the probe arrives at Mars 180 o from where Earth was at launch. Mars must be 180 o – 134 o = 46 o in front of the Earth at launch. 46 o

When can we do it again? Angular Velocity of Earth = 360 o /1 year Angular Velocity of Mars = 360 o /1.86 year. Difference is 166 o per year or 360 o 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 V infinity = 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 23 rd

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 T 2 = K a 3, the only factor is the semi- major axis. How “oval” it is, is irrelevant.

Orbital Velocity Velocity is independent of eccentricity. Since v 2 = 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 E T = -GMm/a. V initial ΔVΔV V final

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.

Conclusion Robert A. Heinlein, "Once you make it to orbit, you're half-way to anywhere." E T = E p E T ’ = ½E p ’ E T ’’ = 0 ΔE = ~ ½E p

Space tidbits Spacex 2 nd 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 24 th..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