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Attractive force

  1. Jul 29, 2013 #1

    Zondrina

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

    Find the attractive force between the Sun and Venus. The distance between their centers is 1.08x1011m.

    2. Relevant equations

    ##F = \frac{Gm_1m_2}{d^2}##
    ##m_{sun} = 1.98 * 10^{30}##
    ##m_{venus} = 4.83 * 10^{24}##
    ##G = 6.67 * 10^{-11}##

    3. The attempt at a solution

    ##F = \frac{Gm_1m_2}{d^2}##

    ##= \frac{(6.67)(1.98)(4.83)(10^{43})}{(1.12)(10^8)}##
    ##= \frac{(63.8)(10^{35})}{1.12}##
    ##= 57 * 10^{35} N##

    Does this look okay?
     
  2. jcsd
  3. Jul 29, 2013 #2

    Mark44

    Staff: Mentor

    No. It looks like you aren't using the right value of d2. How did you get 1.12 X 108?
     
  4. Jul 29, 2013 #3

    Zondrina

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    I did (1.08)(1.08)(104)(104) and I rounded a few places off. I see it was 1011 though, my mistake. My 11 looked like a 4 on paper by accident.

    So I would get (1.12)(1022).

    Thanks for noticing that.

    EDIT : So I get ##57*10^{21}N##. I also see this is an intro phys question, but I was accidentally inside the calc & beyond section, sorry about that.
     
  5. Jul 29, 2013 #4

    SteamKing

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    Very sloppy work. Almost no units shown except for the final result. Missing exponents and some exponents dropped altogether. Incorrect arithmetic calculations buried within expressions.
     
  6. Jul 29, 2013 #5

    Mark44

    Staff: Mentor

    I noticed that, as well, but didn't mention it.

    By not including units, Zondrina, you made extra work for the people checking what you did.

    Also, if you square 1.08, you get 1.1664. I still don't see how you got 1.12 out of that. The general rule is to not round off until all your calculations are done. If you round before then, it will affect your final answer.
     
  7. Jul 29, 2013 #6

    Zondrina

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    I usually define my variables with their magnitudes and directions before I do calculations and keep the units out of the calculations to avoid clutter ( I know this is a bit lazy, but I find it makes the actual arithmetic stand out more to avoid mistakes ).

    These questions are usually cooked so that the unit for force is Newtons. So I took that for granted when I did the calculation. I'll try to be more symbolic next time.

    EDIT : Indeed, it's closer to 1.17.
     
    Last edited: Jul 29, 2013
  8. Jul 29, 2013 #7

    Mark44

    Staff: Mentor

    But you shouldn't round that number. The calculation should be done like this (omitting the powers of 10)
    $$\frac{(6.67)(1.98)(4.83)}{1.1664}$$
     
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