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Final velocity of satellite that falls from orbit

  1. Dec 15, 2015 #1
    Hey, how's it going! Just a little confused about how to do the problem below. Any help would be appreciated!
    1. The problem statement, all variables and given/known data

    A satellite of mass m=100,000 kg is moving around the earth in a circular orbit at an altitude of 100,000 feet above the surface.
    If the satellite were stopped and released from rest, what would its final velocity be before it hit the earth?
    Assume no dampening force.

    2. Relevant equations
    mv^2/r=G*M1*M1/r^2
    mv^2/2=G*M1*M2/r
    PE=mgh
    KE=.5*m1*v^2

    3. The attempt at a solution
    I used equation 2 on the list, plugged all the variables in, and got 11,094 m/s. I have no idea if it is correct however!

    Regards,

    JT

     
  2. jcsd
  3. Dec 15, 2015 #2

    Orodruin

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    Try not to just insert things into formulas. This is not what physics is about. Physics is about using reasoning and logic to arrive at a result.

    What does the second formula tell you? Why do you think it is applicable to this situation? What physical principles are you using?
     
  4. Dec 15, 2015 #3
    Equation 2 relates the speed and the radius for a satellite in a circular orbit, but we're told the satellite is released from rest.
    What happens when the satellite falls, is that all its potential energy is converted to kinetic energy.
     
  5. Dec 15, 2015 #4

    Orodruin

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    No it doesn't. It is off by a factor of two. Equation 1 does so (by equating the centripetal force required with the gravitational force).
     
  6. Dec 15, 2015 #5
    So...I will try to think about this in the most logical way, I guess. So, earth and the satellite are part of the system. I am not sure if we can treat gravity's acceleration as equaling 9.81 m/s/s here...can we just use 9.8 as an approximation? Anyway, the satellite has PE at 100,000 meters proportional to the distance between the satellite and the center of mass of the earth. PE=G*M1*M2/r as stated in my physics textbook...this has to all be translated into kinetic energy at the earth's core, (r=Radius of earth+altitlude of satellite) the formula being equal to .5m*v^2/2. however, using this formula this way assume that there is still PE when the satellite arrives at the earth. So, PE=KEf+mgh (mgh being at the earths surface; h=radius of the earth).

    Sooooo...after plugging and chugging, I get 11,005.86 m/s as an approximation. That is really frigging fast, holy S***. Anyway, Halp? Orodruin, what do you think? Does this make sense?
     
  7. Dec 15, 2015 #6

    SteamKing

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    Is the satellite orbiting at 100,000 feet (from the problem statement) or 100,000 meters? One altitude is in space, the other is within the upper atmosphere.
     
  8. Dec 15, 2015 #7
    100,000 meters NOT feet! Sorry, I made a transcription error! Yup, this baby is in space, Streamking!
     
  9. Dec 15, 2015 #8
    *Steamking...xD
     
  10. Dec 15, 2015 #9

    Orodruin

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    This is only an approximation which is valid if the gravitational field is homogeneous and of the same strength as at the Earth's surface. You cannot apply it generally.

    But the satellite is not reaching the core, it crashes into the surface first! Also, the potential with this definition is relative to infinity, not with respect to the center of the Earth! As a result of this, you are getting a number which is simply very close to the escape velocity, i.e., the velocity required to escape the Earth's gravitational field.
     
  11. Dec 15, 2015 #10
    Exactly! Did you read all of what I said? I said that it didn't.
    Anyway, if you think that I am wrong, please give me your opinion. I have done all that I can do for now.
     
  12. Dec 15, 2015 #11

    haruspex

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    Read the last part of Orodruin's post. Your equation 2 is for an object falling from infinity to radius r. There's no point memorising equations if you don't remember the conditions under which they apply and what the variables represent.
    Think about the change in PE as it falls 100km.

    Of course, 100km is quite small compared with the radius of the Earth, so gravitational acceleration won't change much over that drop. Using mgh would at least give you a fairly tight upper bound on the answer, useful as a check.
     
  13. Dec 15, 2015 #12
    Yo, I think I got this figured out. I would use energy conservation.
    Ei=Efinal.
    The satellite only has potential energy when it is stopped given by-(G*m1*m2)/(r1). r1 equals radius of earth + altitude.
    If we ignore drag, part of this PE is converted into KE, .5*m*v^2, and the other goes into "potential energy" or more precisely, the work required to move a stationary object an infinite distance away from earth at the earth's radius r, so PEfinal=-(G*m1*m2)/r2. r2=radius of earth.
    Plugging in my numbers and solving for V:

    We geeeeet: 1390.1 m/s

    Yata!!! There we go! :) I think this sounds more reasonable...right?
     
  14. Dec 15, 2015 #13
    Sorry, my final equation is: -(GM1)/r1=.5*v^2-G(M1)/r2
     
  15. Dec 15, 2015 #14

    haruspex

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    Check it yourself using the tip I gave you in post #11.
     
  16. Dec 15, 2015 #15
    Using mgh=.5*m*v^2
    vf=1400 m/s.

    Pretty close to 1390.1 :)
     
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