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Mass of the Earth - Please Help

  1. May 31, 2008 #1

    1. The problem statement, all variables and given/known data

    I'm trying to figure out the mass of the earth.

    2. Relevant equations
    I know what it actually is, 5.97*10^24 kg, but it's just how to do it that's got me stumped.

    I know the F = Gm1m2/d squared (or r squared depending what you're taught) formula.

    3. The attempt at a solution

    I have the mass of my object, which is 0.012kg, i know my d, that's 6.3781*10^3 squared (which is 4.06801586*10^7), I know G (6.67*10^-11), but I'm just confused about the F, in my book we worked it was 931 N = ...., but I think that was just 9.8 N (gravitational acceleration) multiplied by my teacher's weight, as an example.

    If I do 0.012*9.8 I get 0.1176 N, so does that mean my equation should look like:

    0.1176 N = (6.67*10^-11) * 0.012 * Me (mass of earth)

    Then I rearrange the formula to get

    But the answer I keep getting, no matter how many different ways I try, even when I do get the right answer, I'm about 1 000 000 off, i.e. I get 5.97*10^18. WHY oh why is this happening? I have to show working for my assignment. Now I'm lost when I even follow my own book cuz we used different masses i.e. 60 kg instead of 12 GRAMS, (the 12 grams is the weight of my sinker).

    Please, please, pleaase any help is REALLY appreciated. o:):approve:
  2. jcsd
  3. May 31, 2008 #2
    Units are your downfall. You're really close. Always watch units.

    (Think..what could be off to give you a factor of 10^6 off in terms of order of magnitude?)
  4. May 31, 2008 #3
    Given the radius of the Earth and G constant [tex]G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}[/tex], then by the second formula
    You get
    [tex]F=\frac{Gm_{earth}m_{1}}{R^2}=m_{1}g[/tex], where g is acceleration due to gravity

    Now, divide both side by [tex]m_{1}[/tex] to cancel out the mass m1. We don't need it.
    Then, we get [tex]m=gR_{earth}^2/G[/tex]. Plug and chug.
    Last edited: May 31, 2008
  5. May 31, 2008 #4

    That I haven't converted something properly? I thought about how I converted g to kg, but that all seems right...

    Maybe I have the radius of the earth wrong.. is it meant to ... nah its not meant to be in meters is it?

    (Thank you by the way :smile:)
  6. May 31, 2008 #5
    Check the units of the newton (kg*m)/sec^2
    and the units of G : m^3 kg^-1 s^-2

    and you will have learned your first (and hopefully forever memorable) lesson that you should ALWAYS include units.
  7. May 31, 2008 #6
    Konthelion, yes that works! But... I'm still getting to the power of 18, instead of 24. I'm still that million off, and I cannot figure what I'm doing wrong...

    (Thank you :smile:)
  8. May 31, 2008 #7
    Check your distance d i.e. radius of the Earth, it is [tex]6.37 × 10^6[/tex] not [tex]6.37x10^3 [/tex]
  9. May 31, 2008 #8
    Seconds squared is acceleration right? Um..

    Conversion.. Umm... 1000 squared is one million.. so... umm, yeah the

    WAIT I think I've got it...

    I converted the radius from km to metres.... and it gave me 4.068015*10^13, instead of 10*7, which is the 6 more decimals I need..

    Oooh thank you thank you thank you I'm going to see if it works now, it should!!:biggrin::biggrin::blushing::rofl:
  10. May 31, 2008 #9
    Ooh, So does that mean...
    Yes thats it... I'm sure...

    Thank you, both of you. Soo much :smile: :smile:

    I just have to find my paper with it written haha I have so much working out here...

    Let's see if I do this right... :biggrin:
  11. May 31, 2008 #10

    It worked!!

    Aww thank you enormously! THANK YOU! :biggrin: :smile: :biggrin: :smile: YAY!

    I know it seems ridiculously small that I shouldn't get something like this, but thank you.

    Thats all it was, the units! Thank you so much! :smile:
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