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Gravity on the moon Rhea

  1. Sep 8, 2011 #1
    A 3.07-kg object is thrown vertically upward from the surface of Earth, where the acceleration due to gravity is g1 = 9.81m/s2. The intial velocity is v1, and the object reaches a maximum height of y1. What is the maximum height, y2, if the object is thrown at a speed of v2 = 1.15v1 from the surface of Saturn's moon Rhea? The acceleration due to gravity on Rhea is g2 = 0.264 m/s2. Give your answer as a multiple of y1.

    Relevant equations
    I tried using the equation

    3. The attempt at a solution

    I used the equation and got
    0 = v2 + 19.62h and
    0 = 1.15v2 + .528h

    but i'm not sure what to do next. the answer is supposed to be like this:
    y2 = ____ x y1

    but i'm not too sure on where to go from where i'm at. Any hints? It would be greatly appreciated
  2. jcsd
  3. Sep 9, 2011 #2
    A few things first, that need clarification:
    You're very on target by selecting: [tex] v^2 = 2a(s-s_i) [/tex]
    But take care and have a look at your equations... How is a(the acceleration) oriented with respect to the initial velocity? i.e does it support its growth or reduce it?
    Then, if you take another look at the formulae you wrote, you'd no doubt find some arithmetical errors there that need immediate correction(hint: equation 2).
    Lastly, recall that [tex]y_2[/tex] is NOT known, so try and work around that, by firstly manipulating the first equation, and then replacing some of the variables you'll glean out of it in the second...
    Hopefully that should put you on the right track,
  4. Sep 9, 2011 #3
    So it should be like:
    0 = v2 - 19.62h and
    0 = 1.15v2 - .528h

    -v2 = -19.62h
    -1.15v2 = -.528h

    (v2 / 19.62) = h

    and use that?
  5. Sep 9, 2011 #4


    User Avatar

    Staff: Mentor

    I fixed. :!!)
    You fix. :grumpy:
    Last edited: Sep 9, 2011
  6. Sep 9, 2011 #5
    Well, yes... with a few reservations.
    Make sure all velocities in these formulae are squared; Then, observe that these are different "h"s for each planet/moon so forth.
    By plugging in the data, like you suggested though, after making all the above adjustments, you'll end up having the right answer(though your problem asks you to incorporate y_1, and y_2, so I would replace h with those first).
    But you're definitely "close to home"!
    Keep up the good work,
  7. Sep 9, 2011 #6
    *it wont take my superscript for the hearth equation*

    0 = v2 - 19.62 hearth and
    0 = 1.15v2 - .528 hrhea

    I fixed! :tongue:

    Now i have

    .051v2 = hearth
    2.178v2 = hrhea

    2.178 / .051 = 42.705

    I know i'm close it says i am! haha. but i'm just confused, if the velocity is squared on both equations, why wouldn't that be the ratio?
  8. Sep 9, 2011 #7
    That's prefectly fine, what you had written; But it seems you lost the ratio h_moon/h_earth in the process on the right-hand-side. I hate to be a stickler on these things, but if you only add that, you'd be done in a jiffy..
    Let us know what comes out of it,
  9. Sep 9, 2011 #8
    hmm. should i not rid of the 1.15 in the hrhea calculation? it seems that the ratio is when the velocity is 1 to 1 and not 1 to 1.15
  10. Sep 9, 2011 #9
    Have you verified that [tex] {v_2}^2 = (1.15 \cdot v_1)^2 [/tex], is in fact 1.15?
    after all, 1.15 squared is, something slightly different? :)
    Try this: Assuming I designate h_1 for earth and h_2 for Rhea, I would arrive at:
    {v_1}^2 = 2 \cdot h_1 \cdot g_1
    {v_2}^2 = 2 \cdot h_2 \cdot g_2
    v_2 = 1.15 v_1,[/tex] therefore by dividing both equations:
    \frac{{v_1}^2}{{v_2}^2} = \frac{h_1}{h_2} \frac{g_1}{g_2}
    Following me so far?
    Here, everything is known, but for h_2(or h_rhea, whichever), and you have the substitutes for g_2, and v_2...
    Give it a go, Daniel
  11. Sep 9, 2011 #10


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    Staff: Mentor

    Only partly. Fix fully... here, use these ( )
    Last edited: Sep 9, 2011
  12. Sep 9, 2011 #11

    so its actually 2.5047 / .051 = 49.11 ratio. I wish I didn't mess up so early in the problem lol.

    Thank you for the help!
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