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Conservation of Energy in Orbits

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data
    1.A rocket at the surface of the Earth is launched with a speed of 2500m/s. What is the orbit that the rocket will achieve?

    2.A meteorite travelling at 3500m/s approaches Earth from outer space. With what speed will the meteorite hit the surface of Earth?

    3.Also, in general does an object on the surface of Earth always have gravitational energy? If the problem doesn't state it's initial velocity, do I assume it has no kinetic energy initially on the surface of Earth?
    2. Relevant equations
    Ek = 1/2mv^2
    Eg = -Gmm/r
    Et = -Gmm/2r
    3. The attempt at a solution
    My main question is the third question. Do objects in orbit always have Eg? Ek? Because if I can understand that, then I'll know if my answers are right or not.

    1. 1/2mv^2 - Gmm/r = -Gmm/(r+r1)
    I believe the rocket will lose all it's kinetic energy at a point (slow down) and transferring all the energy to Eg. Solving for r1, I get about 3.36e5m

    2. 1/2mv^2 - Gmm/r = 1/2mv^2 - Gmm/r1
    Well, it is at a distance of r away from the planet before it travels towards Earth so it must have Eg right? Then when it returns to Earth's surface it will have an Eg of Gmm/r1, r1 is the earth's radius to center? But if this is right how do I know what is the initial distance?
     
  2. jcsd
  3. Feb 23, 2013 #2

    BruceW

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    For the first question, the rocket cannot be in orbit with zero kinetic energy. It is not obvious what they mean by "what orbit will it achieve" because the rocket might have one of many possible elliptical orbits. So I am guessing they want you to calculate the circular orbit which the rocket would achieve.

    For the second question, it says outer space. So this means the rocket would be very far from the earth. Since they don't give the initial distance, what do you think that you are meant to assume for the initial potential energy?
     
  4. Feb 23, 2013 #3
    I believe they are asking me how far (max altitude) it will go away from earth's center for the first question, which in that case I assume it will fall back down.

    And thanks. I understand what they mean now for the second question.
     
  5. Feb 23, 2013 #4

    BruceW

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    hmm. for the first question, I think it is unlikely that they want to know the max altitude, because they have used the word 'orbit'... Maybe they meant for you to classify the motion, i.e. is it elliptic, parabolic or hyperbolic (which one of the three 'types'). Or I still think that they most likely meant what is the circular orbit which it can take.

    Edit: uh, now I realise the classification of elliptic, parabolic or hyperbolic is essentially the same as the question of max altitude, since they don't tell us the angular momentum of the rocket, we could only assume it to be zero.
     
  6. Feb 23, 2013 #5
    If we hypothetically assume it was asking for max altitude, would Ek final be zero then like in my solution?
     
  7. Feb 23, 2013 #6
    The term orbit usually implies there is some motion around a gravitating body. Setting the final velocity to zero means the motion is a straight line from the center of the body up. Which one could hardly call an orbit.

    I think this question really asks whether a projectile launched with the given speed can really be in orbit about the Earth; mind the radius of the latter.
     
  8. Feb 23, 2013 #7
    So how exactly would you go towards solving it?
     
  9. Feb 23, 2013 #8
    The minimal orbit around the Earth would be circular with the radius of the Earth. What's the speed? How does it compare with the speed given?
     
  10. Feb 23, 2013 #9
    Vorb = 7907m/s, which is more than, Vgiven = 2500m/s. So it does not have enough velocity to orbit the Earth, is this right?

    But what if the question asked for maximum altitude?
     
  11. Feb 24, 2013 #10
    You have already found the max altitude. Since it is not clear what the problem is about, you could offer both results.
     
  12. Feb 24, 2013 #11

    BruceW

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    I get a different answer for the min velocity required to be in a circular orbit around the earth. (much less than your answer). Also, I think the question wants you to find the circular orbit which has the same energy as the rocket initially had. So this takes two steps: 1) calculate initial total energy. (and this can be used as the equation for final energy). 2) Think about what kind of relationship there must be between r and v when the rocket is in a circular orbit. From this, you have got 2 equations, and 2 unknowns (v and r in the circular orbit), so you can solve for the radius of the circular orbit.

    The answer of max altitude might be what they wanted. But I think it is not very likely, by the way the question is worded. To be honest, they should have been more specific in the question. But this will often happen, when the question is slightly vague, so you have to make the best guess at what they actually want you to solve.

    Edit: one last thing: in reality, there is no way for the rocket to 'go into' a circular orbit, given that it came out of earth's atmosphere with no fuel left. But still, I think that question 1) wants you to find the circular orbit which the rocket would go into, if there was some way to make this happen, while keeping the total energy of the rocket constant.
     
    Last edited: Feb 24, 2013
  13. Feb 24, 2013 #12
    Did you use Vorb = sqrt(GM/r) for the velocity?



    Are you referring to this system of equations?
    Et = -Gmm/r
    v^2/r = 4pi^2r/T^2

    Isn't period an unknown too?
     
    Last edited: Feb 24, 2013
  14. Feb 24, 2013 #13
    This is known as the "first cosmic velocity", and is widely known to about 8 km/s close to the Earth's surface. What value do you get, and how?
     
  15. Feb 24, 2013 #14

    BruceW

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    ah, now I calculate it again I do get that answer. Sorry about that. I must have forgotten to carry the something... So, the rocket can't achieve a circular orbit around the Earth. uhh, now I am starting to think there must have been a typo in the question. I guess the best thing you can do is to put the answer you have worked out for the max altitude. It seems like the best answer right now..
     
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