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Math Help - can any body help me

  1. #1
    Junior Member sweet's Avatar
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    can any body help me

    hi
    how can i solved this problem ...
    by using integrating of area proof
    the ratio between Aphelion velocity and Perihelion velocity equal the ratio between both of them radius
    if The orbit is an ellipse???
    I need your help as soon as you can
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sweet
    hi
    how can i solved this problem ...
    by using integrating of area proof
    the ratio between Aphelion velocity and Perihelion velocity equal the ratio between both of them radius
    if The orbit is an ellipse???
    I need your help as soon as you can
    If you mean:

    Using Kepler's second law - that the area swept out in unit time by a planet
    is a constant - then yes.

    At Aphelion and Perihelion the velocity of the planet is at right angles to the
    vector from the Sun to the Planet, so the area swept out in unit time in
    each case is:

    <br />
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2<br />

    and:

    <br />
\frac{d}{dt}A_{Peri}=v_{Peri}\ r_{Peri}/2<br />

    at Aphelion and Perihelion respectivly. Kepler's second law tells us these are
    equal so:

    <br />
\frac{d}{dt}A_{Ap}=\frac{d}{dt}A_{Peri}<br />
,

    so:

    <br />
v_{Ap}\ r_{Ap}/2=v_{Peri}\ r_{Peri}/2<br />

    Hence:

    <br />
\frac{v_{Ap}}{v_{Peri}}=\frac{r_{Peri}}{v_{Ap}}<br />

    RonL
    Last edited by CaptainBlack; May 22nd 2006 at 01:54 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sweet
    hi
    how can i solved this problem ...
    by using integrating of area proof
    the ratio between Aphelion velocity and Perihelion velocity equal the ratio between both of them radius
    if The orbit is an ellipse???
    I need your help as soon as you can
    If I understand correctly you need to show that:
    \frac{v_{Aph}}{v_{Per}}=\frac{r_{Aph}}{r_{Per}}

    Unfortunately it isn't true. So I hope I misunderstood your question!

    This is an interesting problem, since my book does not give a formula for the speed during an orbit.

    For starters, I'll assume you are referring to an inverse-square force law (ie. the &quot;Kepler Problem&quot. Note that it is possible to produce elliptical orbits from other force laws, so this is necessary information.

    The position in terms of the eccentricity is:
    r = \frac{a(1-e^2)}{1+e \, cos \theta}
    where a is the semi-major axis a = \frac{r_{Aph}+r_{Per}}{2}=-\frac{GM}{2E} where E is the energy of the body in orbit, M is the central mass, and G is the Universal Gravitation constant.

    For perihelion we set \theta = 0 \,rad and for aphelion we set \theta = \pi \, rad. This gives a ratio of the radii as:

    \frac{r_{Aph}}{r_{Per}} = \frac{\frac{a(1-e^2)}{1-e}}{\frac{a(1-e^2)}{1+e}} = \frac{1+e}{1-e}

    Now we need the apsidal speeds. As with all velocities, the direction is tangent to the path traced by the object over time. Hence we are looking for r \frac{d \theta}{dt}.

    We don't know d \theta /dt and we aren't going to get it. However, we do know a relation involving both it and the (constant) angular momentum:
    \frac{d \theta}{dt} = \frac{L}{mr^2} where m is the reduced mass of the system, and L is the angular momentum.

    So at aphelion r \frac{d \theta}{dt} = \frac{L(1-e)}{ma(1-e^2)}.

    and at perihelion r \frac{d \theta}{dt} = \frac{L(1+e)}{ma(1-e^2)}.

    Thus:

    \frac{v_{Aph}}{v_{Per}} = \frac{\frac{L(1-e)}{ma(1-e^2)}}{\frac{L(1+e)}{ma(1-e^2)}} = \frac{1-e}{1+e}

    which is not the same as the expression we want. It is, however, the reciprocal of the quotient of the radii, so I'm hoping I understood your question incorrectly.

    -Dan
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  4. #4
    Junior Member sweet's Avatar
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    Cool

    CaptainBlack

    thank u soooo much

    but what u mean about A_{Ap}

    if u have some paper about Kepler's problem and two body problem

    plese give it to me i will grateful for u
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  5. #5
    Junior Member sweet's Avatar
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    thank u topsquark
    u r method very long but i will think about it

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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by sweet
    CaptainBlack

    thank u soooo much

    but what u mean about A_{Ap}

    if u have some paper about Kepler's problem and two body problem

    plese give it to me i will grateful for u
    \frac{d}{dt}A_{Ap} is the rate that area is swept out by a line segment conecting the Sun to Planet
    when the Planet is at Aphelion. Not a strictly correct notation, perhaps \left \frac{dA}{dt} \right| _{Ap} would be better.

    RonL
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  7. #7
    Junior Member sweet's Avatar
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    ok ...but i know

    <br />
\frac{d}{dt}A=r^2 \frac{d}{dt}[thet] /2<br /> <br />

    and u writ it in the form
    <br />
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2<br />
    ??????
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by sweet
    ok ...but i know

    <br />
\frac{d}{dt}A=r^2 \frac{d}{dt}[thet] /2<br /> <br />

    and u writ it in the form
    <br />
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2<br />
    ??????

    At aphelion:

    \frac{d}{dt}\theta=v_{Ap}/r_{Ap}

    because the Planet is moving at rightangles to the segment connecting the
    Sun to the Planet.

    RonL
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  9. #9
    Junior Member sweet's Avatar
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    i get it

    CaptainBlack u r the best
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