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'True' Weight.

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data
    An object is located a distance d above the surface of a large planet of radius r. At this position, its true weight is one percent (1.000 %) less than its true weight on the surface. What is the ratio of d/r?

    2. Relevant equations


    3. The attempt at a solution
    so if something weights 100 kg on the earth, at 'that' distance, it should weigh 99k.

    so d should be extremely small compared to R.

    D/R should be a big number then. But I have no idea how to put this into calculations.
     
  2. jcsd
  3. Mar 6, 2015 #2

    billy_joule

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    Based on what?
    What is Newton's law of gravitation?
     
  4. Mar 6, 2015 #3
    D is just the distance from the surface to the point in space, it is not the distance from the center of the earth to the point in space.
    So if a person would only get 1% lighter in terms of weight at that point, it shouldn't be very far away from the surface of the earth.

    newton's law of gravitation is g = GMe / R^2e

    Me = mass of earth

    Re = radius of earth
     
  5. Mar 6, 2015 #4

    billy_joule

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  6. Mar 6, 2015 #5
  7. Mar 6, 2015 #6

    jbriggs444

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    That's half of it. The right side of your equation is the force of gravity a distance D above the surface of the unknown planet. The left hand side is 99 percent of the force of gravity at the surface of the unknown planet. You are correctly equating the two numbers.

    Now can you express the surface gravity of the unknown planet in terms of Newton's universal law of gravitation?

    On a cosmetic note... it's not the earth. So the mass should probably be symbolized as ##M_p## (mass of the unknown planet) rather than ##M_e##.
     
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