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average acceleration

Discussion in 'MCAT Study Question Q&A' started by ssiding, Dec 18, 2008.

  1. ssiding

    2+ Year Member

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    The sun is approximately 150 million kilometers from the earth. A year, the time for one revolution ofthe earth around the sun is roughly ​
    π × 107 seconds. Assuming a circular orbit, what bestapproximates the average acceleration of the earth due to the sun?
    A. 6 × 10^-6​
    m/s2

    B. 6 × 10^-3​
    m/s2

    C. 6 × 10^2​
    m/s2

    D. 4 × 10^19​
    m/s2

     
  2. ssiding

    2+ Year Member

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    First of I figure out the circumference of the circle on which the earth travels. C=(pi X diameter). This gives me the distance travels, we know the time so I use the equation (pi X 150,000,000,000meters X 2)/(pi X 10 e 7). I cross out the pi and get 300,000,000,000/10,000,000 with is 30,000m/s. Then I divide 30K by pi X 10 e 6 which ends up being 1 e -3 m/s2. This is not a choice.
     
  3. Vihsadas

    Vihsadas No summer
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    Here's a hint.

    Think about the difference between linear acceleration and centripetal acceleration. Which do you think should apply in this case and why?

    So you should understand that in an orbiting system, the linear acceleration is zero. the centripetal acceleration, however, causes the orbiting body to continually change direction so it's path is a curve. So, you should be using the equation for centripetal accel: a = v^2/r.

    Now, in this case, what is v? It's the linear velocity.
    Then r is the distance from the sun to the earth.

    I think the most difficult thing about this problem is just thinking about the differences between the types of accelerations, and then really it's an issue of making sure you carry over your units properly. GL.
     

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