Trip to Mars

1. Trip to Mars How do we get there? OAPT May ’08 By John Berrigan Berriganj@hdsb.ca

2. The Theory How do we get from Earth to Mars?

3. 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)

4. The Solution We will change the multi-body problem down to a series of two body problems. • Earth/probe • Sun/probe • Mars/probe The result gives a pretty accurate representation of what needs to be done.

5. 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.

6. Hohmann Transfer Orbit We want to go from the inner orbit to the outer 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

7. Δv1 Δv2 The Delta-v’s 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. Destination orbit Transfer orbit Δv needed

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

9. The Physics of it All!!

10. 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

11. rp = Periapsis rA = Apoapsis vA vp Major axis = 2a = rp + rA a = semimajor axis = ½ (rp + rA) 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 vp > vA

12. Equation x2/a2 + y2/b2 = 1 ae The Ellipse continued… b -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

13. 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

14. V = ? R = r Energy Conservation How fast must you go to JUST escape the Earth? ET= ET’ ½mv2 -GMm/r = 0 Therefore, v2 = 2GM/r For Earth vescape = ~11.1 km/s R’ = infinity V’ = 0 Therefore ET’ = 0 ET = EK + EP

15. Earth 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. R = “infinity”

16. Probe 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. Earth Sun

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

18. Fg Circular Orbits: Orbital Energy To solve for the trajectory we need to review orbital energy. ET = EK + EP = ½mv2 +( -GMm/R) = ½m(GM/R) – GMm/R = ½(-GMm/R) = ½ EP Fnet = Fg Fc = Fg mv2/R = GMm/R2 v2 = GM/R This means that in a circular orbit the total Energy is equal to one half the potential Energy at that radius.

19. 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.

20. EP EA vA vp Elliptical Orbital Velocities 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)

21. 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!!)

22. 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.

23. 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

24. Δv1 Δv2 The Delta v’s 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.

25. 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

26. Δv1 Δv2 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 launching from the Earth, must have a velocity of 2.9 km/s when it gets to “infinity”.

27. R’ = infinity ET’ = EK’ 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= ET’ ½mv2 -GMm/r = ½mvinfinity2 vlaunch = ~11.6 km/s ET = EK + EP Note: there is small difference (400 m/s) in launch velocity for JUST escaping and having final velocity of 2.9 km/s.

28. 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?

29. Vinfinity = 2.6 km/s V’ = ? R’ = r R = infinity ET = EK Landing on Mars How much should you slow down when you arrive at the Martian’s surface? The calculation: ET’ = ET ½mv2 -GMm/r = ½mvinfinity2 v = ~5.7 km/s ET’ = EK’ + EP’ 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.

30. 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!!

31. 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.

32. Vinfinity = 2.6 km/s R = infinity ET = EK 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’ 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

33. 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.

34. 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??

35. 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)

36. Back to our problem….

37. 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?

38. 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.

39. 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 Tearth = 1 year, aearth = 1.5 so Ksun = 1.5-3 Transfer atransfer= 1.885 T2 = Ksun (1.885)2 T = 1.41 years Mars aMars= 2.27 T2 = Ksun (2.27)2 T = 1.87 years

40. 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.

41. 46o 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.

42. 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…

43. 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?

44. 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.)

45. 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.

46. 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.

47. Lagrange points Earth Sun Lagrange points or libation points

48. Lagrange points Gravitational “topographical” force map

49. Phoenix Landerfrom April 25th

50. May 23rd