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Homework Help: Earth's Gravitational Field Strength

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data

    a) Find the altitude above the Earth's surface where Earth's gravitational field strength would be two-thirds of its value at the surface. Assume re = 6.371 103 km.
    __________ km
    (b) Find the altitude above the Earth's surface where Earth's gravitational field strength would be one-third of its value at the surface.

    [Hint: First find the radius for each situation; then recall that the altitude is the distance from the surface to a point above the surface. Use proportional reasoning.]

    2. Relevant equations

    This is what I am having trouble with. What equation do I use?

    3. The attempt at a solution

    I am having trouble with the equation. Which equation do I use for gravitational strength that involves height...or distance from the surface of the earth??
  2. jcsd
  3. Oct 9, 2007 #2


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    Homework Helper

    Gravitation field strength = GM/r^2

    Find the distance from the center of the earth... then subtract the earth's radius... that gives altitude.
  4. Oct 9, 2007 #3


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    Homework Helper

    First find the distance from the earth's center r when:

    GM/r^2 = (2/3)GM/re^2

    solve for r in terms of re. Then plug in the number for re to get r... then subtract the earth's radius from r... that gives the altitude...

    or another way to look at it:

    GM/(re + a)^2 = (2/3)GM/re^2

    solving for a will give you the answer you need.

    that's for 2/3... then do the same thing for 1/3...
  5. Sep 25, 2011 #4
    If I use that formula I solve for "a" but the coefficient in from of Re is negative... what am I doing wrong?
  6. Sep 25, 2012 #5
    I've had an attempt at the first question you posted, and I shall try to tell you how I got my answer.

    Using the equation A = GM/r^2 where 'A' here is my Gravitational Field Strength, we can say that if 'R' is the radius at which the Gravitational Field Strength is two-thirds that of the field strength on Earth's surface, then GM/R^2 = (2/3) GM/6371.1^2 = (2/3) GM/40590915.21.

    'G', the Gravitational Constant is 6.67 x 10^-11 and 'M', the mass of the Earth in this case, is 5.97219 x 10^24 kg.

    But we don't need to know that because we can simply the equation above, diving all sides by 'G' and 'M' to get 1/R^2 = 2/(3(40590915.21)) so 1/R^2 = 2/121772745.63. Diving both the numerator and denominator on the right side, we get 1/60886372.815 and since that is equal to 1/R^2 we can deduce that 60886372.815 = R^2.

    R = sqrt(60886372.815) = 7802.97205012km. That's the radius from the centre of mass.

    In order to find the altitude from the surface, we take that value and subtract the radius at the surface from it.

    7802.97205012 - 6371.1 = 1431.87205012km. And that's your answer.

    Hope this helped, and I'll probably cover the second question a bit later.
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