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How High Can You Jump on asteroid 243 Ida?

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    The asteroid 243 Ida has a mass of about 4.0×1016kg and an average radius of about 16 km (it’s not spherical, but you can assume it is). If you can jump 61 cm straight up on earth, how high could you jump on 243 Ida? (Assume the asteroid’s gravity doesn’t weaken significantly over the distance of your jump.).

    I got the correct answer, I just want to verify my method.
    2. Relevant equations
    Kinematic Equations, Newton Laws, Newtons Gravitational, haven't covered work/energy yet.

    3. The attempt at a solution
    Initially I tried using the kinematic equations, I could get them to workout because my initial velocity was unknown, and I figured it would vary due to my acceleration due to gravity. Is this correct? Or can you use kinematic equations to solve this?

    My next guess was to express Earth's gravity in a ratio with Ida 243's gravity. That is 9.8/(1.0428*10-2) This gave me 9.39766 * ida's g. I figured then my distance to be 9.39766*.62 because my acceleration downwards would be less, which means the height of my jump should be proportionally related to the decrease/increase of gravity. This gave me 573m, which is correct. However, I want to be sure I went about solving this correctly.

    Thanks.
     
  2. jcsd
  3. Mar 16, 2016 #2

    mfb

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    Staff: Mentor

    The fraction is much larger than 9.39766. And "something*g" is not a ratio.
    Why should distance be proportional to gravitational acceleration? It is, but what is the argument? Why not proportional to the square root or square of it?


    573 meter (I get 573.8m, so 574m would be better) is "correct" within the incorrect assumptions the problem statement asks you to do. In reality you could jump higher, as you lift of faster (gravity slows you during the acceleration phase as well, and it does so more on Earth than on Ida).

    Edit: typo
     
    Last edited: Mar 16, 2016
  4. Mar 16, 2016 #3
    Sorry, I didn't carry the decimals correctly in my post. 939.78 would be correct, however, the approximation gives a round off error of 573 meters. When using exacts I get 574m when rounded. I am also aware that multiplication is not a ration, I refer back to my original post:

    A ratio.

    Because a = Δv/Δt where Δv = Δx/Δt thus follows a=Δx/t2 so as x increases/decreases, acceleration does.

    I'm assuming this is 574m? Checking a couple times this is what I got.

    This is relatively why I was confused since no initial velocity was given. Yet if I started with the assumption that my initial and final were both 0 and I got the same answer. I see WHY I did, but I'm confused if this is just coincidence based on my assumption. Using v2 = v02 + 2aΔx. set
    v0e2 + 2aeΔxe = v0i2 + 2aiΔxi based on my assumption then

    aeΔxe = aiΔxi
    ae(xefinal-xeinitial) = ai(xifinal-xiinitial)
    xifinal = ((ae(xefinal-xeinitial))/ai) + xiinitial

    which gives me:
    754 ~ .62(9.8/1.0428*10-2) + 0


    Essentially exactly what I did to solve the problem. Was this just coincidence? That is, I shouldn't have assumed my initial velocity was 0 since it my initial velocity would be the instantaneous velocity when my feet leave the ground, correct?
     
  5. Mar 16, 2016 #4

    mfb

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    Not a very strict way of showing it, but in principle a dimensional analysis is sufficient, yes.
    Oops, typo. 574 of course.
    You can calculate the velocity directly after losing contact to the ground, but you don't need it.

    Everything that works out in terms of units will have the right answer (unless you apply the g_Earth/g_Ida factor in the wrong way, but that is easy to spot).
     
  6. Mar 16, 2016 #5
    Cool, thanks. Beyond dimensional analysis how would you attempt to prove this? It seemed to almost come intuitively to me because of our work in 1D problems, and even 2D when examining components.
     
  7. Mar 16, 2016 #6

    mfb

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    There is a standard formula for the distance needed to change the velocity from 0 to v (or the opposite direction) at a constant acceleration g.
     
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