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Gravitational Potential Energy problem-Physics1-help

  1. Dec 22, 2008 #1
    1. The problem statement, all variables and given/known data

    An asteriod hits the Moon and ejects a large rock from its surface. The rock has enough speed to travel to a point between the earth and the moon where the gravitational forces on it from the earth and the moon are equal and opposite in direction. At that point the rock has a very small velocity toward Eart. What is the speed of the rock when it encounters Earth's atmosphere at an altitude of 700 Km above the surface?


    2. Relevant equations

    Gravitational potential energy: PE= -Gm1m2/r
    Conservation of Mechanical Energy: KE(intial)+PE(intial)=KE(final)+PE(final)


    3. The attempt at a solution
    well firstly ididnt get what they mean that its moving towrds earth, I mean if the forces between the moon and earth are equal and opposite in direction shouldn't the velocity of the rock be zero, Ijust need calrification here becuase that was confusing me a bit. And also I tired finding the velocity that it moves with toward earth but I dont know anything about the distance that it travelled. I just basically need someone to clarify this problem for me thats all and thankyou:).
     
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  3. Dec 22, 2008 #2

    Hootenanny

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    Okay, I'll try and clarify the problem for you. Imagine that the rock is traveling form the moon towards the earth. Now, when it reaches the point where there is no net force acting on the rock it still moving towards the earth some non-zero velocity. Recall Newton's First Law, if there are no external forces acting on a body, then the body remains at rest or _________
     
  4. Dec 22, 2008 #3
    ohh yea!! I see and one more thing when they say enough speed to travel between a point between the earth and the moon where the gravitational forces are equal and opposite in direction, could they possibly be hinting that the rock would travel initially at the escape speed from the moon or I am I analyzing it the wrong way. thats all and thank you once more:)
     
  5. Dec 22, 2008 #4

    Hootenanny

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    I wouldn't say that the rock was launched at the escape velocity. Instead, I would say that at the point between the earth and the moon where the gravitational forces are equal and opposite in direction, that the rock has a very small velocity - so small in fact that you could consider it to be zero. :wink:
     
  6. Dec 22, 2008 #5
    wait, but if its zero how does the rock keep moving towards earth untill it reaches an altitude of 700 km :confused:?
     
  7. Dec 22, 2008 #6

    Hootenanny

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    I didn't say it was zero, I said you could consider it to be zero:
    If it had a non-zero velocity at this point, then it would simply there forever, never moving. It does have some small velocity at this point to ensure that it doesn't just sit there but moves towards earth instead. However, this velocity is so small that you can neglect it and assume that the rock starts moving from rest at this point.

    Do you follow?
     
  8. Dec 23, 2008 #7
    oh ok yea I see where your coming from,, alright thanks alot your help is appreciated:smile:,, and oh yea one more thing I've been trying to figure out the distance where this rock was undergoing zero net force,, and for some reason I am not getting any where I dont know why. I mean this is how Iam thinking of it I figure out the intial distance between the rock and the earth then since I am going to assume that it starts from rest I use this equation
    v= √2(-GMearth/7078+GMearth/Rre).

    Where 7078 is the distance from the earth center to the point of 700km altitude in km and Rre is the initial distance between the earth and the rock. Doing that I could figure out the final velocity, but I tired finding the initial distance between rock and earth but i dont seem to be getting any where.
     
  9. Dec 23, 2008 #8

    Hootenanny

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    I think it's best if we start from scratch.

    Using Newton's Law of gravitation, can you write down an equation the represents the net force felt by the rock (i.e. the sum of the forces exerted by the moon and earth)?
     
  10. Dec 23, 2008 #9
    umm alright yea sure, so it would be something like this

    Fnet=GMeMr/Rre^2-GMoMr/Rro^2=0

    where Me=mass of earth
    Mr=mass of rock
    Rre=distance between earth and rock
    Mo=Mass of moon
    Rro=distance between moon and rock
     
  11. Dec 23, 2008 #10

    Hootenanny

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    Looks good. So we want to figure out where abouts this Fnet is zero. At the moment we have two reference points (the earth and the moon) and we're measuring the distance from both points of reference re and ro.

    To make our life easier, let's pick just one reference point say the centre of the earth. Can you now re-write ro in terms of re and the distance from the centre of the earth to the centre of the moon deo?

    Notice that you can also cancel off the mass of the rock.
     
  12. Dec 23, 2008 #11
    ok so Ro=0.2725Re and deo=3.85x10^8 m

    so our equation would be somethin like GMe/Deo^2-GMo/0.2725Rer^2=0 ?
     
  13. Dec 23, 2008 #12

    Hootenanny

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    Hmmm, not quite. Let's try it like this:

    [tex]R_o + R_e = D_{eo}[/tex]

    Hence

    [tex]R_o = D_{eo} - R_e[/tex]

    The using your equation

    [tex]\frac{G M_o}{r_o^2} = \frac{G M_e}{r_e^2}[/tex]

    We obtain

    [tex]\frac{M_o}{\left(D_{eo} - R_e\right)^2} = \frac{M_e}{R_e^2}[/tex]

    Do you follow?
     
  14. Dec 23, 2008 #13
    alright so you mean it that way ok so far so good...
     
  15. Dec 23, 2008 #14

    Hootenanny

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    Good. So to answer your question you need to know where the velocity of this rock can be considered zero. Therefore, you need to determine Re.

    Once you have Re the problem is a straight forward application of conservation of energy.
     
  16. Dec 23, 2008 #15
    alright thank you very much , i don't know what i would've done without your help:approve::shy:
     
  17. Dec 23, 2008 #16

    Hootenanny

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    No problem, it was a pleasure. Feel free to come back if you get stuck later on.

    Merry Christmas!
     
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