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Mission to Mars

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A manned expedition is planned to be sent to Mars in 2023, one way. If the colony module was 100.0 tonnes in mass. Determine the work that must be done by the rockets to carry this colony from Earth to the surface of Mars.

    2. Relevant equations
    ET = PE +KE

    PE = -GmM/r

    KE = 1/2mv^2


    3. The attempt at a solution
    Escape speed on earth is 11 200 m/s
    Escape speed of mars is 5000 m/s

    Do i just do energy required to break free of earth's gravity and add in energy gained from Mar's gravity, since it'll pull you in once you're close to it?

    1/2 * 1000 * (11200)^2 - 1/2 * 1000 * (5000)^2 = 5.02 * 10 ^10 J

    However, my friend said that the sun plays a role in this question as well and i don't exactly see how it does, if the sun is involved, could someone please explain why and how to account for it please?
     
  2. jcsd
  3. Dec 18, 2014 #2
    Got a better understanding now, apparently there is a point at which Fg earth + Fg sun = Fg mars, the sun is needed since its a massive body and its gravitatio0nal feild will have a effect on you. Anyways, we need to find the energy needed to get to that point so that Mar's gravity will be strong enough to pull you in, and also some energy to slow down 100 tonnes of mass, so you don't crash and burn on Mars.
     
  4. Dec 18, 2014 #3

    haruspex

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    You can't really make use of that.
    Yes, but also take into account that Earth and Mars aren't sitting still.
    More precisely, how will the effect of Earth's gravity diminish along the way? How does that compare with the way that the sun's pull will diminish?
    That's an interesting one. In principle, it doesn't require energy to slow down, but it does require momentum.
    Are you supposed to take into account the weight of the fuel?
     
  5. Dec 18, 2014 #4
    No we're not suppose to take into account fuel. My other friend said something like this: Total energy on mars - Total energy on earth = energy required
    And Total energy on earth consists of: PE - earth on rocket, PE of sun on rocket, KE of earth's rotation, KE of rocket traveling around the sun on earth. And the Total energy on mars would just be the same as earth except replace everything from earth to mars. Do you think is will work? I could see how it would work, but im afraid there might be some flaws as well.
     
  6. Dec 18, 2014 #5

    ehild

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    I think it will work.
     
  7. Dec 18, 2014 #6
    Alright cool, but unfortunately i got a negative answer when i did Mars subtract Earth.

    For Earth i got: - G*(Mass of rocket)*(Mass earth)/ radius of earth - G*(Mass of rocket)*(Mass of Sun)/ distance between sun and earth + 1/2*(mass of earth)*(2pi*distance between earth and sun)^2 + 1/2*(mass of rocket)*(2pi*distance between earth and sun)^2
    Which comes out to be: -6.25E12 - 8.87E13 +3.55E38 +5.45E18 ... 1
    can you see anything wrong with this?
    With Mars its the same except with mass of mars and distance between mars and sun instead
    Comes out to be: -1.38E24 - -5.82E13 +8.3E37 +1.3E19 ... 2

    Then 2 -1 i got - 2.72E38 J
     
  8. Dec 19, 2014 #7

    ehild

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    How did you calculated the kinetic energy?
     
  9. Dec 19, 2014 #8
    1/2mv^2, v = 2pi*r/T, opps i forgot to write in T before (sry), but i did include T in my calculations
     
  10. Dec 19, 2014 #9

    ehild

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    Something is wrong with your calculation. The total energy when orbiting along a circular orbit is half the potential energy, and negative with respect to the energy at infinity. The kinetic energy can not be magnitudes higher than the potential energy.
     
  11. Dec 19, 2014 #10

    Bandersnatch

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    1.Why did you add the kinetic energy of Earth in it's orbit? (same with Mars) What bearing does it have on the energy the rocket needs?
    2.You don't need to climb the Sun's gravitational potential well to infinity.
     
  12. Dec 19, 2014 #11

    haruspex

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    I hate to disagree with ehild, but I don't believe that is right. In order to get to Mars you may have to negotiate a point where the total energy is higher than at either origin or destination. The energy you need is the energy to get to that point. You can't borrow energy to get there and pay it back on the downhill run to Mars' surface.
    I say "may have" because at that critical point you do not need any KE. There is no requirement to be in orbit. But on reaching Mars you will need to be matched to Mars' orbital speed. I think it very likely that the energy needed at the surface of Mars will nonetheless be lower than at the peak. So, as you mentioned, there's still the problem of landing gently. I need to give that more thought; perhaps you are supposed to ignore it.
    In principle, the KE (of the vessel in Earth's orbit) is available to assist in the journey. If a satellite in orbit fires a thruster to reach a higher orbit, its linear speed at the higher orbit is less, so its KE is less, but the lost KE doesn't just disappear - it helps in raising the orbit. Of course, if it fires the thruster the wrong way ....
    Europa did not do that. The part from Mars to infinity cancels out in the PE subtraction.
     
  13. Dec 19, 2014 #12

    Bandersnatch

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    I'm not sure if we're talking about the same thing. KE of Earth is relevant?

    Agreed on the second point.
     
  14. Dec 19, 2014 #13

    haruspex

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    Hmm... I see what you mean:
     
  15. Dec 19, 2014 #14

    ehild

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    You are right, I answered too hastily.
    the lowest energy needed is via a Hohmann orbit http://en.wikipedia.org/wiki/Hohmann_transfer_orbit
    That means to launch the spacecraft with velocity to enter at elliptic orbit around the Sun with closest point at the Earth and farthest point near the Mars. Detailed explanation for orbit from Earth to Mars is here http://www.phy6.org/stargaze/Smars1.htm
    Reaching the Mars on that orbit, it is also needed to match the orbital velocity to that of the Mars, along its nearly circular orbit.
     
    Last edited: Dec 19, 2014
  16. Dec 20, 2014 #15

    ehild

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    You have to subtract the energy at Earth from that at Mars instead of the opposite way.
    Let be m the mass of the spaceship, Ve the escape velocity from Earth, Vm the same for Mars, Re the radius of Earth, Rm radius of Mars, re radius of the orbit of Earth, rm radius of orbit of the Mars.
    First you have to escape from Earth gravity, that needs 1/2 m Ve2 energy.
    You have to switch over to the Hohmann orbit, by changing the speed of the vessel, so it becomes an ellipse that reaches the orbit of Mars. It can be calculated, that the spaceship reaches the orbit with less speed than that of Mars, so it needs to increase the speed again. The work done by the rockets must be at least the energy difference between the orbits around Mars and Earth.
    The orbits of both Earth and Mars are nearly circular. On circular orbits, the kinetic energy is half of the potential energy, so the total energy is -0.5 GmM(Sun)/r.
    So the energy needed to transfer from the Earth orbit to the Mars orbit is 0.5 G M(sun) m (1/re -1/rm).
    When the spaceship travels on the same orbit as the Mars, close enough to it, the gravity of Mars pulls it in, and the rockets have to do work against the gravitational pull so as the spaceship do not crash. Without the rockets, that speed would become Vm near the Mars surface, so 1/2 m Vm2work has to be done to slow down the spaceship.

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