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I'm so sick of this. (gravitation problem)

  1. Oct 17, 2005 #1
    Astronomers detect a distant meteoroid moving along a straight line that, if extended, would pass at a distance 3RE from the center of the Earth, where RE is the radius of the Earth. What minimum speed must the meteoroid have if the Earth’s gravitation is not to deflect the meteoroid to make it strike the Earth?
    Ans:
    sqrt.(GMearth/4RE)

    MY homework! SO i got it in the right category.

    I tried, and tried. Not getting anywhere. I WOULD paste what I've done, but what I've done is wrong. Cause I interpreted the Q wrongly. I had my asteroid start to move towards the earth only when it was 3RE distance away; whatmore, I tried doing it via dynamics/kinematics; which was stupid cause the direction of the gravitational force is constantly changing. (it is right?)
    Please help. Advice or even just small, little hints. Thanks!
     
  2. jcsd
  3. Oct 17, 2005 #2

    Diane_

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    Small hint? OK. Look at it as gravity providing a centripetal force. If the speed were low enough, the "meteoroid" would be captured into a circular orbit. The speed is clearly less than that.
     
  4. Oct 17, 2005 #3
    Thanks but uh, I dun really understand. If the speed was low enough, then it would be captured into orbit... okay i can accept that. But why should the speed be lower? If the speed was any lower, then it'd also be captured into orbit wouldn't it? Wouldn't it have to be faster, so that it doesn't get deflected and explode into earth?
    Yea... I'm lost.
     
    Last edited: Oct 17, 2005
  5. Oct 17, 2005 #4

    Diane_

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    Yes and no. It's going to be in an orbit of some sort no matter what it's speed. If the speed is high enough, then the orbit will be hyperbolic (i.e. not closed) and it'll escape. If the speed is lower, it'll be in an orbit more or less elliptical. If the orbit happens to be SO elliptical that it intersects with the surface of the Earth, then it impacts.

    When you throw an object into the air, you're putting it in orbit. It's just that the orbit has such a small perigee that it hits the surface. If, however, the Earth were permeable, then the rock or whatever that you throw would orbit the Earth's center of gravity forever.

    I'm not sure that helps. Do you know how to find the centripetal force for a particular curve? Do you know Newton's Law of Gravitation? If you put those two together (with the appropriate distances), you'll end up with an expression for the necessary speed for an orbit at some particular height, which will look very much like the solution you've been given. (Note: It won't be the same, but very similar.)
     
  6. Oct 18, 2005 #5
    Ok, thanks vry much! I think i should be able to at least stumble through the question now. Before, I was just staring.
     
  7. Oct 18, 2005 #6

    lightgrav

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    Try writing angular momentum conservation and a
    2nd equation of energy conservation (with grav.PE).
    At closest approach (1RE) the velocity is tangential.
     
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