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Gravity Net Force question

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data
    The earth has a mass of 5.98*1024 Kg and the moon has a mass of 7.35*1022 kg. The distance from the centre of the moon to the centre of the earth is 3.84*108 m. A rocket with a total mass of 1200 kg is 3.0*108 m from the centre of the earth and directly in between the earth and the moon. Find the net gravity.


    2. Relevant equations
    Fnet = [itex]\frac{Gm1m2}{r^2}[/itex]


    3. The attempt at a solution
    The distance from the rocket to the moon is 3.84*108-3*108 = 8.4*107
    and the rocket to earth is 3.84*108 m

    E<--------R---->M
    let right be positive


    Fnet=Frocket to moon - Frocket to earth

    Fnet=[itex]\frac{G(1200)(7.35*10^2)}{8.4*10^7}[/itex]-[itex]\frac{G(1200)(5.98*10^2)}{3*10^8}[/itex]

    Fnet=0.83 - 5.3 N

    The answer is a negative, so the net force is pulling the rocket towards earth. Knowing this is wrong but what have i done incorrectly?
     
  2. jcsd
  3. Nov 12, 2013 #2

    CWatters

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    Perhaps check how far the Lagrange point L1 is from earth?
     
  4. Nov 12, 2013 #3

    D H

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    You dropped your units and as a result didn't realize you had a dimensionally incorrect result. Don't treat those dimensioned quantities as just numbers. Here's your error:
    G has dimensions length3·time-2·mass-1, or units of m3·s-2·kg-1. That 1200, 7.35*1022, and 5.98*1024: Those are masses, in kilograms. The denominators 8.4*107 and 3*108 are lengths, in meters. The units of your expression are thus kg·m2·s-2 rather than kg·m·s-2, the units of force.

    You didn't use this equation. You didn't square the distance in the denominator.


    It's a bit more cumbersome to do so, but always carry the units along in a computation. That extra little bit of effort will save your rear end many times over.
     
  5. Nov 12, 2013 #4

    haruspex

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    No it isn't. Read the problem statement again.
    As well as the r2→r error D H pointed out, the powers of 10 in the numerator seem to have become truncated.
     
  6. Nov 12, 2013 #5
    Fixed up the small errors that you guys mentioned

    3. The attempt at a solution
    The distance from the rocket to the moon is 3.84*108-3*108 = 8.4*107
    and the rocket to earth is 3.0*108 m

    E<--------R---->M
    let right be positive


    Fnet=Frocket to moon - Frocket to earth

    Fnet=[itex]\frac{(6.67*10^2 Nm^2/kg^2)(1200 kg)(7.35*10^{22} kg)}{(8.4*10^7 m)^2}[/itex]-[itex]\frac{(6.67*10^2 Nm^2/kg^2)(1200 kg)(5.98*10^{24} kg)}{(3*10^8 m)^2}[/itex]

    Fnet=0.83 N - 5.3 N

    Fnet= -4.47 N


    The values for the answers still the same cause I didn't write down the equation correctly here in PF (I'm sorry should have proofread more rigorously before posting). I'm still left with a negative value.

    Since I made right positive, the Frocket to moon is positive since it is the force of the rocket towards the moon. The Frocket to earth is the force of the rocket in the direction to the earth, so its a negative value. Maybe my reasoning is wrong here?​
     
    Last edited: Nov 12, 2013
  7. Nov 12, 2013 #6

    D H

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    You have close to the right number, and it is pointing in the right direction (toward the Earth is correct). Round that to the precision indicated by the values in the problem and you will have the right answer.

    In general it's a good idea to maintain a few more digits of precision in your intermediate calculations than the numbers indicate. You round to the indicated precision in your final answer.
     
  8. Nov 12, 2013 #7

    haruspex

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    To add to D H's remarks, the sequence above is not valid. 5.3 is only to one decimal place, so you cannot quote two decimal place in the next line.​
     
  9. Nov 13, 2013 #8

    CWatters

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    The L1 point is 3.23 * 10^8 meters from earth and the space craft only 3.0 * 10^8 so negative (towards the earth) appears correct.
     
  10. Nov 13, 2013 #9

    haruspex

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    I don't see any connection with Lagrange points. What's your thinking here?
     
  11. Nov 13, 2013 #10

    D H

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    The net gravitational force at the point in question must be Earthward because it's closer to Earth than is the Earth-Moon L1 point and because the net gravitational force points Earthward at L1.
     
  12. Nov 13, 2013 #11

    haruspex

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    OK, I see. (I thought it more obvious to consider the net zero point, which is about 3.46*10^8, but either will do.)
     
  13. Nov 13, 2013 #12

    CWatters

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    OK I admit it :-) I'd forgotten the zero point was a bit further from the earth than L1.
     
  14. Nov 14, 2013 #13
    Okay, since the significant digits are 1 after the decimal point the answer is -4.5 N
    . How about my calculation for Frocket to moon, i should limit it to 2 SD.​
     
  15. Nov 14, 2013 #14

    jtbell

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    Or do the whole calculation entirely in your calculator without writing down (and rounding off) intermediate results and then keying them in again.
     
  16. Nov 14, 2013 #15
    Oh okay thanks D H!
     
  17. Nov 14, 2013 #16
    Alright! thats a good tip
     
  18. Nov 14, 2013 #17

    jtbell

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    You can also avoid keying in duplicate numbers (and possibly doing them inconsistently) by recognizing things that you can factor out:

    $$F_{net} = (6.67 \times 10^{-11})(1200)
    \left( \frac{7.32 \times 10^{22}}{(8.4 \times 10^7)^2}
    - \frac{5.98 \times 10^{24}}{(3 \times 10^8)^2} \right)$$
     
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