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Universal gravitational word problem test tomorrow!

  1. Aug 6, 2014 #1
    1. What is the strength of the gravitational field (in N/kg) at an orbit of 1.0 x106 m above the planet Zoklopgniald with mass of 3.45 x1026 kg and mean radius of 7.80 x107m? ANS:3.69 N/kg What is the gravitational force exerted on a 2840 kg satellite orbiting Zoklopgniald at this altitude? ANS:10 500 N

    2. What is the force of gravity between two electrons (mass = 9.11 x 10-31 kg) that are 1.0 m apart? ANS: 5.54 x 10-71 N


    I HAVE NO IDEA HOW TO DO THIS! i tried using the two formulas for universal gravitation but my answers are soo off! and our teacher said the questions are kind of similar to this review package. can someone walk me through these! thank you :):confused:
     
  2. jcsd
  3. Aug 6, 2014 #2
    g=Gm/r^2
    used this formula
     
  4. Aug 6, 2014 #3
    g= Gm/r^2

    =(1.0 x 10^6)(3.45 x 10^26) / (7.80 x 10^7)^2
    this is what i did for question 1 of the 1st part
     
  5. Aug 6, 2014 #4

    Nathanael

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    For the first part of question 1, this is the correct formula.

    m=mass of the planet

    r=distance from the center of the planet


    Edit:
    G is a constant (I think it's called "the universal constant of gravity" or something)
    You might want to google the value of G for this problem.

    You used
    r = 7.80 x 10^7 = the radius of the planet

    Is the radius of the planet equal to the distance of the satellite to the center of the planet?
     
  6. Aug 6, 2014 #5
    are my substitutions correct because i caluculated by answer and it wasn't the same as 3.69 N/kg
     
  7. Aug 6, 2014 #6

    Nathanael

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    I edited my last post.
     
  8. Aug 6, 2014 #7
    so would the radius be 2840
     
  9. Aug 6, 2014 #8

    Nathanael

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    No how did you get that?


    The problem says that the distance from the center to the surface of the planet (the radius) is 78 million meters. It also says that the satellite is 1 million meters above the surface of the planet.

    So how far is the satellite from the center of the planet?

    That is what you use for "r"
     
  10. Aug 6, 2014 #9
    1 million metres above the surface
     
  11. Aug 6, 2014 #10

    Nathanael

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    That is the distance from the satellite to the surface of the planet.

    That is not the distance from the satellite to the center of the planet.

    It will help if you draw a picture.
     
  12. Aug 6, 2014 #11
    so is it like 78million-1million=77million
     
  13. Aug 6, 2014 #12

    Nathanael

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    How did you get that answer?

    Sorry if I sound rude but it seems like you are guessing. In physics, you should be able to explain each step you take.
    So what makes you think it is 77 million? Maybe that's right, but what makes you think it's right?


    I really encourage you to draw a picture of the situation.
     
  14. Aug 6, 2014 #13
    i don't really know how to draw the picture because i don't get the question properly
     
  15. Aug 6, 2014 #14

    Nathanael

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    This is what the problem says:

    You have a planet of a certain radius (it's approximately a sphere) with a satellite that is a certain height above the surface. What is the gravitational acceleration (a.k.a. "gravitational field strength") of the satellite?

    Can you draw a picture of that now? Is anything unclear? (It really helps if you can tell me what is unclear)


    So do you know what the variables "M" and "r" represent in the equation that you posted? ([itex]G\frac{M}{r^2}[/itex])
     
  16. Aug 6, 2014 #15
    m=mass which is 3.45 x 10^26 but i don't get how you find the radius because i thought it was 7.80 x 10^7? can u show how you find the radius
     
  17. Aug 6, 2014 #16

    Nathanael

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    That is the radius, of the planet.

    But in the equation you posted you do not use the radius of the planet for "r"

    "r" is the distance from the center of the planet to the satellite.
     
  18. Aug 6, 2014 #17
    i did g= Gm/r^2
     
  19. Aug 6, 2014 #18

    Nathanael

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    What is r?
     
  20. Aug 6, 2014 #19
    7.80 x 10^7
     
  21. Aug 6, 2014 #20

    Nathanael

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    That is the r of an object on the surface of the planet. That is not the r of an object 1million meters above the surface of the planet.


    I'm sorry but I just don't know how else to put it.
     
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