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Time required to go to Mars

  1. Aug 2, 2004 #1
    A rocket moves through space at 11,000 m/s. At this rate, how much time will be required to go to mars (78.3 x 10^6 km)?

    i used t= d/v

    and got 5.05 x 10^ -4

    I'm not sure what i'm doing wrong. I cant get the correct answer
     
  2. jcsd
  3. Aug 2, 2004 #2

    Tom Mattson

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    Neither am I, because you haven't shown any of your steps.

    How did you get your answer?
     
  4. Aug 2, 2004 #3

    Doc Al

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    Why don't you show us the exact numbers you are plugging in. Don't forget to use consistent units. For example, don't mix meters and kilometers.
     
  5. Aug 2, 2004 #4
    11,000 m/s
    __________
    21,751,740 = 5.05 x 10^ -4


    I got my numerator figure by converting 78.3 x 10 ^6 km
    and got 21,751,740 m/s
     
  6. Aug 2, 2004 #5

    Tom Mattson

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    You can't convert km to m/s. The former is a distance, the latter is a speed.

    You should have converted 78.3 x 106 km to meters. The numerical value will only differ by a power of 10. Also, you're dividing in the wrong order. The speed v goes on the bottom, and the distance d goes on top.
     
  7. Aug 2, 2004 #6
    I got 71181818.18 = 71 seconds as my answer. THX :)
     
  8. Aug 2, 2004 #7

    Tom Mattson

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    The answer is t=7118181.81 seconds (which is about 82 days). You can't round that to 71 seconds (which is a minute and 11 seconds).
     
  9. Aug 2, 2004 #8

    BobG

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    Uh, I think Tom meant the they would differ by A power of 10, not that 78.3 x 10^6 km equaled 78.3 x10^10 meters. To go from km to meters, multiply by 1000 (or just change the power of ten by three - 78.3 x 10^6 to 78.3 x 10^9)

    And how did 71 plus million seconds suddenly turn into only 71 seconds. You might want to convert your 7 million+ seconds into something a little more manageable such as days, for example.
     
  10. Aug 2, 2004 #9

    NateTG

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    Tracking dimensions

    There's a really good technique that a physics teacher taught me (although it's probably common use.) For now, whenever you do a physics problem, you should keep track of the units.

    Here's an example problem which is simlar to yours:

    So, let's say we have a car that is traveling at [tex]5 \times 10^2 feet/sec[/tex] and we want to know how many days it will take to drive [tex]7 \times 10^5 miles[/tex].

    So, of course, we start with[tex]d=vt[/tex]. Since we want to solve for time, we divide both sides by [tex]v[/tex] so we have [tex]t=\frac{d}{v}[/tex]

    So, let's plug in the numbers:

    [tex]t=\frac{d}{v}=\frac{7 \times 10^5 miles}{5 \times 10^2 feet/sec}=1.4 \times 10^3 \frac{sec \times miles}{ft}[/tex]

    Of course, second miles per foot are not appropriate units, since we want days. So we can do the following:
    [tex]1 mile=5280 ft[/tex]
    [tex] 1=\frac{5280 ft }{1 mile}[/tex]
    so
    [tex]t=1.4 \times 10^3 \frac{sec \times miles}{ft} =1.4 \times 10^3 \frac{sec \times miles}{ft} \times 1 = [/tex]
    [tex]1.4 \times 10^3 \frac{sec \times miles}{ft} \times \frac{5280 ft}{1 mile}=7 \times 10^6 sec[/tex]
    Note that the miles and feet cancel nicely when you multiply out.
    Now, to get from seconds to days, you can use a similar approach:
    [tex]7 \times 10^6 sec \times \frac{1 minute}{60 seconds} \times \frac{1 hour}{60 minutes} \times \frac{1 day}{24 hours}\approx 1.2 \times 10^1 days[/tex]

    If you write up your problem in a similar fashion, you're less likely to make mistakes, and people here, and your teacher, will have an easier time helping you.

    P.S. I was playing a bit fast and loose with sig figs there. You should be more careful.
     
  11. Aug 2, 2004 #10

    HallsofIvy

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    Every physics teacher should teach about keeping track of units! That's a basic concept :).

    By the way, I am still wodering if "NateTG" is "Nate the great" who was "12 years old" for more years than I can remember! If so he has finally grown up. His responses actually make sense now!
     
    Last edited: Aug 2, 2004
  12. Aug 3, 2004 #11

    NateTG

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    Since you're in VA, probably not.
     
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