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Gravitation Continued

  1. Aug 25, 2004 #1
    Thanks for you help on the first post. I just have two follow up questions. So I understand now that the moons acceleration is part of its circular orbit, but what about the acceleration of the earth. How can I picture that? If I consider the earth stationary, then the moon orbits the earth, and it would seem that only the moon uses its acceleration vector as centripital. How does the earth use its acceleration? Also, I was wondering about planetary collisions. Say a body was moving through space and then the earth got in its way. I would guess that if it was moving straight for the earth, it would smash right into it, speeding up according to newtons gravitation equation in the process. But what if it is skewed, so that it does not head right for the planet, but perhaps is a little bit off. Is there any easy way to determine if it would become a satelite, or would it be deflected off into space, following a new curve?

    Thanks again,

    Cyrus
     
  2. jcsd
  3. Aug 25, 2004 #2
    Nothing is stationary on the universe.
     
  4. Aug 25, 2004 #3

    Doc Al

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    The moon pulls the earth

    The Earth accelerates as well, though not as much. (The gravitational pull on each is the same, but the Earth is 81 times as massive as the Moon.) Considering the Earth-Moon as an isolated system, they both orbit around the center of mass of the system, which is located inside the Earth. And that Earth-Moon center of mass is what orbits around the Sun. (Of course this is a simplified model: there are many other factors, as you can imagine.)
     
  5. Aug 25, 2004 #4

    BobG

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    Yes. Assuming it's small, and the Earth is the only other object in the universe, making things a little simpler, you measure the position of the body and the velocity of the body, relative to the Earth. The specific energy of object (energy per unit of mass) is just:

    [tex]\frac{v^2}{2}-\frac{\mu}{r}=\varepsilon[/tex]
    where v is velocity, r is position and [tex]\mu=3.986x10^{5}km^3/sec^2[/tex]
    If the specific energy is less than 0, the object will orbit the Earth's center. If equal to or greater than 0, the object will be deflected into a new curve.

    To find out if the object will actually collide with the Earth, or just orbit around it, you need to find the perigee radius. If the object is orbiting the Earth and you just need the distance rather than the exact location, you first find the specific angular momentum using the cross product of the position and velocity:

    [tex]\vec{h}=\vec{r} \times \vec{v}[/tex]

    Find the magnitude of the angular momentum using the Pythagorean theorem.

    Then find the semi-major axis (average distance of the object from the Earth):

    [tex]a=-\frac{\mu}{2 \varepsilon}[/tex]

    The eccentricity of the orbit is:

    [tex]e=\sqrt{1+\frac{2 \varepsilon h^2}{\mu^2}}[/tex]

    After all that, it gets easy. The radius of perigee is:

    [tex]r_p=a(1-e)[/tex]

    Edit: And, finally, if your perigee radius is less than the radius of the Earth, it means the object is going to collide with the Earth.

    Toss in the effect of the planets and the Sun, and the problem gets a little more complicated.
     
    Last edited: Aug 25, 2004
  6. Aug 25, 2004 #5
    Thanks a bunch BobG your a real pal! Hemmiiiiiiiii ;-) (I love that one where the guy tells the neighbor, yeah so I told the little lady its got a hemmi, done deal. Meanwhile, hes grillin some food and catches his glove on fire, dunking it into the ice chest, and trys to play it off by saying, thirsty? HA!)
     
  7. Aug 26, 2004 #6

    BobG

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    Actually, I think he says it has a hemi, not a hemmi. Most people who own hemmi's aren't dumb enough to stick their hand in a fire. :rofl:

    Edit:

    This is a hemi: http://www.thehemi.com/

    This is a hemmi: http://weblab.research.att.com/~davek/slide/hemmi/postversalogr.jpg
     
    Last edited: Aug 26, 2004
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