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Homework Help: Astronaut Gravity calculation

  1. Dec 3, 2007 #1
    [SOLVED] Gravity question

    1. The problem statement, all variables and given/known data
    Calculate how far an astronaut would need to be away above the Earth in order for his weight to be 0.01 his weight on the Earth's surface.

    2. Relevant equations
    g = GM/r^2
    F = GMm/r^2

    3. The attempt at a solution
    I have no correct way to solve this. However the answer is 1.93x10^5 Km

    Would anyone please kindly show how the answer is worked out?
  2. jcsd
  3. Dec 3, 2007 #2


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    Weight on the earth surface proportional to g, where g=GM/r^2 where r is the radius of the earth and comes out to about 9.81m/sec^2. Try it. Now set 0.01*g=GM/r^2 and solve for the new r.
  4. Dec 3, 2007 #3
    I made 'r' the subject of the equation and end up with 1.99x10^5km
    Here is my working

    g = GM/r^2
    r^2 = GM/0.01
    r^2 = (6.67x10^-11 x 5.974x10^24) / 0.01
    r^2 = 3.984658x10^16
    r = [tex]\sqrt{3.984658x10^16}[/tex]
    r = 199616081.5m = 1.99x10^5km

    Please correct me.
  5. Dec 3, 2007 #4


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    i) you are paying absolutely no attention to units. ii) I think the answer you have been given is also quite wrong. What you want is 0.01*GM/(r_earth^2)=GM/r^2. That means r=10*r_earth. How far above the earth you have to be is a somewhat different question but that's 10*r_earth-r_earth=9*r_earth. I don't know where this 1.93*10^5 km is coming from. Sorry.
  6. Dec 4, 2007 #5

    D H

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    Dick is right on all points here.
    The denominator on the right-hand side should be [itex]0.01*9.80665 m/s^2[/itex] here, not just 0.01. That 0.01 is a unitless scale factor.
    If you had paid attention to units you would have been able to see that this expression is invalid. With units (but keeping that scale factor unitless), the expression becomes
    [tex]r^2 = 6.673*10^{-11} \mathrm{m}^3/\mathrm{s}^2/\mathrm{kg}
    * 5.9742*10^{24} \mathrm{kg}/0.01[/tex]
    The expression on the left-hand side has dimensions length squared. The expression on the right-hand side has units m^3/s^2, which does not jibe with length squared. The culprit is that naked scale factor. It should be paired with Earth standard gravitational acceleration:
    [tex]r^2 = 6.673*10^{-11} \mathrm{m}^3/\mathrm{s}^2/\mathrm{kg}
    * 5.9742*10^{24} \mathrm{kg}/
    (0.01*9.80665 \mathrm{m}/\mathrm{s}^2)[/tex]
    Now the right-hand side has units of square meters, matching the dimensions of the left-hand side.

    Carrying this through yields r=63760km. This is the distance from the center of the Earth, not the surface of the Earth. You need to subtract the radius of the Earth, 6378km, to get the answer to the question.

    This is a much easier way to arrive at the result. 10r=63780km to four decimal places, which differs from the more convoluted result by 20 km.

    The 1.93*10^5 km answer is simply wrong.
    Last edited: Dec 4, 2007
  7. Dec 5, 2007 #6
    Alright, go it; i got 63760km which is from the surface of the earth. Subtracting 63760km - 6378km gives me 57382km.

    AS Dick said, the answer given is quite wrong, so the answer at the back of the book must be incorrect.

    Thank you very much, much appreciated.
    Last edited: Dec 5, 2007
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