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Math Help - Newtons' Universal Law of Gravity and polar co-ordinates

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    Newtons' Universal Law of Gravity and polar co-ordinates

    planet of mass
    m moves in a circular orbit of radius R around a Sun (of mass M),

    under the influence of the Sun’s gravitational force, given by

    F=\frac{-GMm \hat{\rho}}{R^2}

    where
    G is Newton’s gravitational constant and \rho is radial basis vector.


    (i) Write down Newton's second law.

    My answer: m\left(R \ddot{\theta} \hat{\theta} -R \dot{\theta}^2 \hat{\rho}\right) = \frac{-GMm \hat{\rho}}{R^2}


    ii) Show \hat{\theta} component of second law leads to \ddot{\theta} = 0 and the \hat{\rho} leads to \dot{\theta}^2 = \frac{GM}{R^3}


    There's more I'll add later
    Last edited by mr fantastic; May 27th 2011 at 04:11 AM. Reason: Significantly cleaned up the latex and formatting, corrected a quoted equation, fixed spelling etc.
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    Quote Originally Posted by Duke View Post
    planet of mass
    m moves in a circular orbit of radius R around a Sun (of mass M),

    under the influence of the Sun’s gravitational force, given by

    F=\frac{-GMm\rho}{R^2}

    where G is Newton’s gravitational constant and \rho is radial basis vector.


    (i) Write down Newton's second law.

    My answer: m\left(R \ddot{\theta} \hat{\theta} -R \dot{\theta}^2 \hat{\rho}\right) = \frac{-GMm \hat{\rho}}{R^2}

    ii) Show \hat{\theta} component of second law leads to \ddot{\theta} = 0 and the \hat{\rho} leads to \dot{\theta}^2 = \frac{GM}{R^3}


    There's more I'll add later
    (i) can be found in most textbooks. (ii) is simply equating components.
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    Thanks for that useful post Mr fantastic
    Can anyone help?
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    Quote Originally Posted by Duke View Post
    Thanks for that useful post Mr fantastic
    Can anyone help?
    I spent 20 minutes fixing your incomprehensible post (numerous latex errors, formatting problems etc.) and then posted a reply that told you exactly how to do (ii). And, as already said, (i) can be found in most textbooks that cover this material, not to mention your class notes.

    In response to all this you decide to be sarcastic. You could not be bothered taking the trouble to fix your post but apparently expect people to come along, decipher it and then type out a complete solution.

    I suggest you take the time to actually think about what I posted.
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    I said your post was useful. It was not intended to be sarcastic. Anyway, I don't have a textbook so that advice isn't helpful.
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    Quote Originally Posted by Duke View Post
    I said your post was useful. It was not intended to be sarcastic.
    Even giving you the benefit of the doubt here, you should have known that your post # 3 would come across as sarcastic (I would certainly read it that way, if I was not given any further information), since immediately after calling mr f's post "useful", you ask for more help, generally. The implication here is that mr f's post was useless, and so you're casting about for some "real" help. If you had genuinely found mr f's post useful, it would have made more sense for you to ask clarifying questions specific to mr f's post.

    It is not enough to have the truth - you must present it in a winsome way. (Just look at the Challenger disaster as an example of not presenting the truth winsomely enough!) To be able to do this is a crucial life skill. I speak feelingly, because I have long had problems with this, and the lack of this skill has cost me a very great deal of time and money.

    Anyway, I don't have a textbook so that advice isn't helpful.
    For (i), the LHS is the mass times the acceleration, written in tangential and normal components. The RHS is the sole force of gravitation.
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    So I am right with (i)?

    With part (ii), its an identity so can I just equate co-efficents
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    Quote Originally Posted by Duke View Post
    So I am right with (i)?

    With part (ii), its an identity so can I just equate co-efficents
    Yes, all correct. Actually, a better link for you is here, remembering that

    \dot{\theta}=\omega.
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    the next part asked me to solve the equation, assuming {\theta }(0)=0 I got {\theta }(t)=t{/sqrt}(GM/R^3) and then I had to find the period around the planets sun. so period= 2pi/w and w= {\theta }(t) but I'm getting a factor of 1/t that shouldn't be there (it's a show that question). Thanks
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    Quote Originally Posted by Duke View Post
    the next part asked me to solve the equation, assuming {\theta }(0)=0 I got {\theta }(t)=t{/sqrt}(GM/R^3)
    Did you mean {\theta }(t)=t{\sqrt}(GM/R^3)?

    That would be the correct answer. But, because you had the LaTeX square root symbol incorrect (have to use backslashes, not forward-slashes), it looked like your square root was in the denominator.

    and then I had to find the period around the planets sun. so period= 2pi/w and w= {\theta }(t) but I'm getting a factor of 1/t that shouldn't be there (it's a show that question). Thanks
    \omega = \dot{\theta}

    not

    \omega = \theta.

    You could either take the derivative of your answer, or simply go back to the original differential equation. Since it's a linear function you're dealing with, the t should go away upon differentiation.
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