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Law of universal gravitation

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/10 its value at the Earth's surface?


    2. Relevant equations
    FG= GmM/r^2
    g= Gm/r^2

    G=6.67x10^-11 (Nm^2)/kg^2
    m(earth)=5.97x10^24 kg
    g=9.8m/s^2

    3. The attempt at a solution
    g=1/10g (earth)
    g= Gm/r^2

    r^2= (Gm/g)1/10
    r^2= (Gm/g)10

    Answer: r=2.02x10^7 m^2

    Using dimensional analysis I somehow got m^2, but I know the units should be in meters, not meters squared.
    My other concern is where I change r^2= (Gm/g)1/10 to r^2= (Gm/g)10

    I'm not even sure if it was correct to do that. By changing the multiplication from 1/10 to 10 was the only way I got the right answer. I could maybe post an attachment with a picture of my work if it would help you understand what I did.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Dec 2, 2012 #2

    TSny

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    Homework Helper
    Gold Member

    Hi mosque. Welcome to PF!

    The acceleration is g/10. So you have ##r^2= \frac{GM}{(g/10)}##

    Try simplifying this by multiplying the numerator and denominator by 10.

    Note that you had to take a square root to get the answer. What does that do to the units?
     
  4. Dec 2, 2012 #3
    Wow that's so simple. Now I understand what I was doing wrong. Now that I think about it dividing the gravity by 10 is the same as multiplying it by 10, but instead I worked out Gm/g first and then tried to multiply by 1/10.

    Also, once I plugged in the number I did take the square root to get the final answer, but I did the dimensional analysis after so I completely forgot to take the square of the meter.

    Thank you so much! :)
     
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