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Force of Earth's gravity on a spacecraft

  1. Oct 18, 2009 #1
    Calculate the force (in Newtons) of Earth's gravity on a spacecraft 12900 km ABOVE the Earth's surface if its mass is 1410 kg.Give 3 sig figs.Be careful with your units!

    Again, here I know force is mass times acceleration, so would it just be 1410 times 9.8? The thing that worries me is this seems way too simple for the type of problems we've been doing, haha, I feel like I'm missing something?
     
  2. jcsd
  3. Oct 18, 2009 #2
    And where does the fact that it's 12900 km above earths surface come into play here??
     
  4. Oct 18, 2009 #3

    D H

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    No. That would be wrong.

    Hint: Newton's law of gravitation.
     
  5. Oct 18, 2009 #4
    Hmm I know the basics behind that law but don't see how I can numerically use that principal to solve this?
     
  6. Oct 18, 2009 #5
    I honestly thought that law came into effect in the form of an equation when you're working with two different masses and whatnot...how would I apply it in this situation?
     
  7. Oct 18, 2009 #6
    Unless...does the mass of the entire Earth come into play here? That wouldn't really make much sense, because that isn't really known is it? Again what are the two unique masses I'm supposed to use here?
     
  8. Oct 18, 2009 #7

    D H

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    Correct.

    Where did you ever get this idea?
     
  9. Oct 18, 2009 #8

    Monocerotis

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    Fg = G(m1)(m2)/d2

    G = 6.67x10-11
    m1 = mass of ship
    m2 = mass of earth
    d = distance (note: squared)
     
  10. Oct 18, 2009 #9
    Then what is the known mass of the earth? Because I just searched it and got 4 different (similar) answers...
     
  11. Oct 18, 2009 #10
    5.9742 × 10^24 seems to be the most popular, yes?
     
  12. Oct 18, 2009 #11
    I got 3380000000 and my units would be newtons, is this correct?
     
  13. Oct 18, 2009 #12

    D H

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    Monocerotis,

    1. Please don't give too much help here. Our job is to help students do their own homework, not do it for them.

    2. G=6.67x10-11 what? What units, to be specific.
    G is not a unitless constant. Solar system dynamists are just as likely to use G=1.4879×10-34 AU3/day^2/kg as they are to use G=6.674×10-11 m3/s2/kg.
     
  14. Oct 18, 2009 #13

    D H

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    Two problems here.
    1. Units.
    2. The distance is to the center of the Earth, not the Earth's surface.
     
  15. Oct 18, 2009 #14
    So my final units wouldn't be newtons? I know the G constant is in m^3 kg^-1 s^-2 but since it asks for the final answer in newtons I'm assuming that's what the end product will be?
     
  16. Oct 18, 2009 #15
    Well how the heck am I supposed to know the distance to the center of the earth when it would surely vary at different poles/locations on earth?

    And how can the units be converted to newtons if what I figured out isn't in newtons? Because the original problem says calculate the force in newtons...
     
  17. Oct 18, 2009 #16

    D H

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    Ignoring the 21 km difference between the equatorial and polar radius is a heck of a lot better than ignoring the radius of the Earth altogether.

    What are the units for G? Is this compatible with the other units you are using?
     
  18. Oct 18, 2009 #17
    The units for force is newtons normally, and gravity is a force that's being calculated here, I still don't see what you're getting at. I appreciate the help, I hope I don't sound ungrateful, it just seems you're answering my questions with more complex questions that I can't answer.
     
  19. Oct 18, 2009 #18

    D H

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    You used G=6.67x10-11. What are the units associated with that number?
     
  20. Oct 18, 2009 #19

    D H

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    Suppose you had been told the satellite was 8016 miles above the surface of the Earth instead of 12900 km. How would you have solved the problem given this information?
     
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