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Gravitational Fields question

  1. Jun 9, 2008 #1
    [SOLVED] Gravitational Fields question

    1. The problem statement, all variables and given/known data
    1. Mount Everest is approximately 10 km high. how much less would a mountaineer of mass 100kg (including backpack) weigh at its summit, compared to her weight at sea level? Would this difference be measureable with bathroom scales?

    Mass of Earth = 6.0 * 10^24
    Radius (km) = 6400

    2. Relevant equations
    g = GM/r²

    where G = 6.67 * 10^-11

    3. The attempt at a solution
    Every time I try, I either get numbers far too big or far too small so I think I'm having difficulty reading what the question is actually asking me to do.

    g = 9.8

    (9.8 * r²) / G = M

    (9.8 * 10,000²) / 6.67 * 10^-11 = M

    M = 1.5 * 10^19 which is far too large.

    So I think I'm just doing something completely wrong.

  2. jcsd
  3. Jun 9, 2008 #2

    D H

    Staff: Mentor

    You have a units problem here. Carry your units throughout, including physical constants such as G.
  4. Jun 9, 2008 #3
    thanks for the reply. But actually, I think you've lost me even more :)
  5. Jun 9, 2008 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're looking for the weight of the mountaineer at 2 different heights. Since weight is:

    [tex]W = mg[/tex]

    where m is the mass of the mountaineer and you already have the equation for g, can you solve it now?
  6. Jun 9, 2008 #5

    D H

    Staff: Mentor

    There are several things wrong here.

    First, you are using the wrong units. What are the units of G? (Hint: G is not 6.67*10-11. It has a numerical value of 6.67*10-11 in one particular system of units. What are those units?) You must use consistent units when solving a problem. If you don't you will get the wrong answer.

    Second, you are using the wrong r. You are using the height of Mount Everest. Newton's law of gravity pertains to the distance from the center of the Earth. How far is the peak of Mount Everest from the center of the Earth?

    And third, you are looking for weight, not mass. The mass is given as 100 kg.
  7. Jun 9, 2008 #6
    Ok, so at sea level, the lady weighs 100*9.8 = 980N

    so I need to find the value for g at the summit of the mountain.

    I'm still confused with this constant business though, on wikipedia, it says the units of G are N(m/kg)². So do i need to square the mass?

    g = GM/r²

    g = 6.67*10^-11 * 100² / (6400,000 + 10,000)²

    g = 1.6*10^-20

    g = 1.62*10^-22 (if the mass isn't squared).

    W = 100 * 1.62*10^-20 = 1.62*10^-18

    another incredibly small number :(
  8. Jun 9, 2008 #7

    D H

    Staff: Mentor

    You don't square the mass.

    What is the gravitational force exerted on the backpacker by the Earth at sea level and at the top of Mt. Everest?
  9. Jun 9, 2008 #8
    g = GM/r²

    gr²/M = G

    9.8 * 6400,000² / 100 = G

    G = 4.01408*10^12

    gr² / G = M

    9.8 * (6400,000 + 10,000)² / 4.01408*10^12 = M

    M = 100.313

    Weight = Mg

    983.06 - 980 = 3.06N

    So the difference in weight is 3.06N?

    The only problem is that she gained weight at the summit, she didn't lose it. So what did I do wrong there?
  10. Jun 9, 2008 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're trying to find g. You know what G is since its a constant. Also M is the mass of the Earth not the mountaineer's. The mountaineer has a mass of m.
  11. Jun 9, 2008 #10
    god damn it. Ok.

    Weight at sea level = 980N

    weight at summit = mg

    g = 6.67*10^-11 * 6.0 * 10^24 / 6410,000²

    g = 9.74

    100 * 9.74 = 974

    980 - 974 = 6N

    so the difference in weight is 6N?

    Is that correct?

    Can't believe my entire mistake was using the wrong mass.
  12. Jun 9, 2008 #11


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah thats the correct method. I'd calculate the sea level one using the radius of the Earth you've been given, then you should be fine.
  13. Jun 9, 2008 #12
    ah cool, ok, thanks for the help everyone.
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