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Find the minimum velocity

  1. Jan 29, 2005 #1
    Hi ppl, could someone just clarify this for me

    i need to find the minimum velocity that the satellite meteosat (time period is 23h56min and altitude is 35800km + radius of earth(6370km)) must be given to escape from the force of attraction of the Earth.

    I started by saying that the total energy needed to effect this is GMm/2(R) where M is mass of earth, m is mass of satellite and R is orbit. Then I equated this to mv^2/2

    My final answer was 3078ms^-1 Could someone verify my working or at least show me where I made an error if there is one?

    Thanks, Joe
     
  2. jcsd
  3. Jan 29, 2005 #2

    dextercioby

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    Yes,your answer is correct,as well as the method to get it...

    Daniel.
     
  4. Jan 29, 2005 #3

    Andrew Mason

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    Why 2(R) in the denominator? I think it should be:

    [tex]\frac{1}{2}v^2 = \frac{GM}{R}[/tex]

    [tex]v = \sqrt{2GM/R}[/tex]

    [tex]v = (2*6.67e(-11)*5.98e24/4.217e7)^.5 = 4349 m/sec.[/tex]

    AM
     
  5. Jan 30, 2005 #4
    I think so too.
     
  6. Jan 30, 2005 #5
    The reason I took 2r as the denominator is that GMm/2r is the total energy of the satellite at that position in orbit, whereas GMm/r is just the potential....

    Is this correct?
     
  7. Jan 30, 2005 #6

    dextercioby

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    Yes,my guess is that u were right in the first place,as the satellite certainly has KE,not only PE.If my memory doesn't cheat me,it is your formula...

    However,i may be wrong...

    We need some other opinion.

    Daniel.
     
  8. Jan 30, 2005 #7
    Can you explain why you think the satelite's energy is -GMm/2r ? Does this include it's kinetic energy?
     
  9. Jan 30, 2005 #8

    dextercioby

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    Yes.It's the only possible reason.

    Daniel.
     
  10. Jan 30, 2005 #9
    Assuming circular motin the satelites potential energy is:
    [tex]-G\frac{Mm}{r}[/tex]
    It's kinetic energy is:
    [tex]\frac{1}{2}m\frac{(2 \pi r)^2}{T^2}[/tex].

    Keplers law gives

    [tex]T^2=\frac{(2 \pi r)^2 r}{GM}[/tex]

    so the kinetic energy can be written
    [tex]G \frac{Mm}{2r}[/tex].
    The total energy is kinetic plus potential:

    [tex]-G\frac{Mm}{2r}[/tex], so you're correct.

    At infinity it's total energy wil be zero, so you'll have to add:
    [tex]\frac{1}{2}mv^2 = G\frac{Mm}{2r}[/tex].
    So the escape velocity is:

    [tex]v_{esc}=\sqrt{G\frac{M}{r}}=3075ms^{-1}[/tex]

    So you're totally right.

    EDIT: As Andrew Mason pointed out, you (and I) calculated the extra velocity required to escape, not the escape velocity...
     
    Last edited: Jan 30, 2005
  11. Jan 30, 2005 #10

    Andrew Mason

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    Not quite. It is -GMm/r. It is important to include the - sign in the potential.

    The question asks for the minimum velocity needed to escape so it doesn't matter what velocity it has while orbiting. But if you were asked what the additional speed is required, it would be the difference between the escape velocity and the orbital velocity. You calculated the orbital velocity, not the escape velocity.

    ie. The energy of the satellite in orbit is:

    (1)[tex]E_{orbit} = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{r}[/tex]

    Since:
    [tex]\frac{mv^2}{r} = F = \frac{GMm}{r^2}[/tex]

    (1) reduces to:

    [tex]E_{orbit} = \frac{1}{2}\frac{GMm}{r} - \frac{GMm}{r}[/tex]

    [tex]E_{orbit} = - \frac{1}{2}\frac{GMm}{r}[/tex]

    Since the condition for escape is that KE + PE = 0, the kinetic energy needed for escape is twice the orbital kinetic energy. This requires increasing the kinetic energy by factor of 2 or its speed by [itex]\sqrt{2}[/itex]

    AM
     
  12. Jan 30, 2005 #11

    Gokul43201

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    Since the question asks for the escape velocity rather than the root sqaure excess velocity, you must use R and not 2R. I concur with AM.

    [tex] KE(in) + PE(in) = E(fin) = 0 [/tex]

    [tex]\frac{1}{2}mv_{esc}^2 - \frac{GMm}{R} = 0 [/tex]

    That's all there is to it.
    If you add the existing KE to the PE to get _GMm/2R, you are then only finding the extra KE required. This is not what's asked.

    How did you go from additional KE to escape velocity ?
     
  13. Jan 30, 2005 #12

    dextercioby

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    Okay,Gokul,which part didn't u understand??This one:

    Please explain to me how would you keep a geostationary satellite in orbit 42000Km (from the Earth's center) MOTIONLESS... :uhh:

    Daniel.
     
  14. Jan 30, 2005 #13
    Read the EDIT one line below...
     
  15. Jan 30, 2005 #14

    Andrew Mason

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    Not exactly. The additional velocity is [itex](\sqrt{2}-1) \times 3075 = 1274 m/s.[/itex]

    AM
     
  16. Jan 30, 2005 #15

    Andrew Mason

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    No one said it is motionless. Its orbital speed is 3075 m/s. It needs to have a speed of 4349 m/s (or twice the kinetic energy than 3075 m/s) in order to escape.

    AM
     
  17. Jan 30, 2005 #16

    Gokul43201

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    dexter, if you see a mistake in my post, point it out. And take it easy with the large fonts !

    Your calculation does not even find the correct additional velocity required, as AM pointed out above. It only calculates what I refer to as the "root excess square velocity", which is quite a meaningless number for the given problem.
     
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