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Homework Help: Gravitational Forces Problem

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data

    At what altitude above the Earth's Surface would your weight be half of what it is at the earth's surface?

    and I know that the mass of the earth is 5.974 X 10^24 and the Radius of the earth is 6.371 X 10^6

    2. Relevant equations

    I know that F = GM1M2 / R^2

    3. The attempt at a solution

    Ok so I was sitting in class and I sort of remember him changing the formula to this

    G = Me * M / R^2

    so and I think I am going this correctly I am thinking that

    G = Me * 1/2M / R^2 + h = G = Me * M / R^2

    I think that is right but I am not sure, my professor is a nice guy but he confuses me a lot. I just don't know what to do after this it has been years since I had a math class. Can anyone help?
  2. jcsd
  3. Feb 4, 2007 #2
    You want force 1 to equal half of force 2. Everything in the numerators will cancel out, which is why your professor removed the gravitational constant. What is the distance force 2 acts over? What about force 1?
  4. Feb 4, 2007 #3
    I don't know what the distance is it isn't given and I am so confused I have no idea where to even start and try and figure that out. Would that be the radius of the earth then?
  5. Feb 4, 2007 #4
    Force 2 acts over a distance of th earths radius and force 1 acts over a distance (R+h), h being the height above the earths surface.

    Remember, youre not looking for where youll have half the mass, youre looking for where theres half the weight (m*force of gravity)
  6. Feb 4, 2007 #5
    so more of what I am looking for is W1 (the weight you are on the earth's surface) / W2 (half of W1) = Re^2 / (Re + h)^2?
  7. Feb 4, 2007 #6
    Almost, both those terms appear in the denominators, so you need to flip them if you still want weight on earth / weight above earth = 2
  8. Feb 4, 2007 #7
    Um I don't think I understand. Flip them how like W2 / W1 = Re^2 / (Re + h)^2
  9. Feb 4, 2007 #8
    So if I set W1 = 80kg W2 = 40kg then cross multiplied it would sort of look like this?

    H^2 = 80kg * (6.371 X 10^6m)^2 / 40kg - (6.371 X 10^6m)
    H^2 = 4.0589 X 10^13
    H = SQRT (4.589 X 10^13)
    H = 6371000m
  10. Feb 4, 2007 #9
    What is all this?? Just use google to find the http://www.google.com/search?hl=en&...IE-SearchBox&rlz=1I7GGLR&q=mass+of+the+earth" of the earth. Then feed that back into

    F= force in newtons
    G= 6.67*10^-11
    M and m= mass is kilograms
    r= distance from center in meters

    So whatever 1/2 the weight you want(in newtons)=G*5.9742×10^24 kilograms*your mass/r^2

    This will get you the distance you need to be from earth to be half the weight.
    Last edited by a moderator: Apr 22, 2017
  11. Feb 4, 2007 #10
    Ok so I could just make up a mass and go with it from there?
  12. Feb 4, 2007 #11
    what i am saying is if you want to find half your weight on the earth, just plug in numbers so that the distance changes.
  13. Feb 4, 2007 #12
    ok so f = 6.67x10^-11 x 5.974x10^24/(6.371X10^6)^2
    F then = 9.816
  14. Feb 4, 2007 #13
    I thought you wanted to know how far you have to be for your weight to be half?? Whats 9.816 coming from??

    Say your weight is 100 pounds or 444.8 newtons. 100 pound in kilograms is equal to 45.4. Plug that in.
    From here all you have to do is some algerbra to get the distance(r), which is in meters.
  15. Feb 4, 2007 #14
    thanks man I think I got it. I am sorry that I didn't understand. my brain has been fried lookin at this thing for a couple of days now. I appreciate your help
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