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Orbital speed

  1. Oct 20, 2013 #1

    462chevelle

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    1. The problem statement, all variables and given/known data

    what is the speed of MRO in its orbit around mars.
    M mars=6.4X10^23
    m MRO=907 Kg
    d between the two=200 km
    R of mars=3400 km
    G=6.7X10^-11

    2. Relevant equations
    D=(mMG)/r^2=(mV^2)/R=(MG/R)^1/2



    3. The attempt at a solution
    I attached a pic of some of my work. if I need to add more detail or its hard to read let me know. im just curious where the mistake is.
    thanks
    Lonnie
     

    Attached Files:

  2. jcsd
  3. Oct 20, 2013 #2

    462chevelle

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    to clarify since I noticed its hard to interpret. the answer I got was appx 27 km/s and the correct answer is 3.5km/s
    thanks
    Lonnie
     
  4. Oct 20, 2013 #3

    Dick

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    You are certainly right that I can't interpret that. Hence I have no idea how you got 27km/s instead of the correct answer. Can you explain clearly how you got 27km/s?
     
  5. Oct 20, 2013 #4

    462chevelle

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    if I use D=mV^2/R
    I get 200km = (907(V^2))/3400
    multiply 3400*200 then divide by 907
    to get 749=V^2
    then (749)^1/2=appx. 27
    that's the only equation from lecture that has velocity in it is the reason im using it. I would think I would need to use an equation with G or M in it. but I don't have one written down here
     
  6. Oct 20, 2013 #5

    Dick

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    You don't understand what your equations mean. mV^2/R is the centripetal force exerted on the satellite by its rotation around the planet. It's not equal the distance from the planet. R is the distance to the center of the planet from the satellite, I hope you know what the rest mean. Now GMm/R^2 is the gravitational force. If the satellite is in a stable circular orbit then the two forces are equal. Set them equal and solve for V.
     
    Last edited: Oct 20, 2013
  7. Oct 21, 2013 #6

    462chevelle

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    youre correct. I wasn't understanding what I was doing. I guess I just started plugging stuff out of frustration.
    I started over and set them equal to each other like so
    mV^2/R=GMm/R^2
    factored out useless variables and rewrote as
    V=(GM/r)^1/2
    solved It again and got crap numbers again. I think I need to get some sleep and try it tomorrow with a clear mind. I know im doing something wrong that is very simple. I have to be. because in order to get 3.5 as velocity I will have to get somewhere around 12 as my answer inside of the parenthesis and that seems impossible with my lack of sleep or focus right now.
     
  8. Oct 21, 2013 #7

    Dick

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    Yes, try it again. Now you've got the right formula, but keep units in mind. G is probably given in units of meters, put everything in the same units and expect to get ~3500 m/s as your answer.
     
  9. Oct 21, 2013 #8

    462chevelle

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    ahh crap.. i think you're right. i always screw up the units. after class ill have another stab at it and let you know how it goes.
    thanks
     
  10. Oct 21, 2013 #9

    462chevelle

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    the only way i can get it to work out right is if i convert
    6.7X10^-11 to 6.7X10^-20
    and i figured that out by leaving G as a variable and plugging 3.5 into v and solving for G
    i wouldnt think that i would have to move my units as far as 9 places. but as many values are in the constant g its hard for me to interpret how to convert it
    my lesson says G=6.7X10^-11m^3kg^-1s^-2
    i guess i should understand what all of those units are interpreted as.
     
  11. Oct 21, 2013 #10

    Dick

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    The units on G and M are fine! kg,m,s. The only ones that aren't fine are 200km and 3400km. Just change those to meters.
     
  12. Oct 22, 2013 #11

    462chevelle

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    awesome. that worked out perfect. thanks man. i would have to say units is something that messes me up a lot.
     
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