Differential Equation involving mechanics

**cooltowns** · May 12th 2010, 10:10 AM

Hi

I need help with the following question. I have attached my question. I need help with part b)

this is what i have managed to do so far, but i don't think i am doing it correctly.

$m\ddot{r}=\frac{mMG}{r^{2}}$

dividing both sides my $m$

$\ddot{r}=\frac{MG}{r^{2}}$

$\int\ddot{r}dt=\int\frac{MG}{r^{2}}dt$

$\dot{r}=\frac{MG}{r^{2}}t+C<br />$

any help is appreichated

thanks

**chisigma** · May 12th 2010, 12:04 PM

This problem can be solved as in...

http://www.mathhelpforum.com/math-he...tml#post510114

Setting for semplicity sake $M G = \mu$ the DE becomes...

$r^{''} = -\frac{\mu}{r^{2}}$ (1)

Now because is...

$r^{''} = \frac{d r^{'}}{dt} = \frac{d r^{'}}{dr} \frac{d r}{dt} = r^{'} \frac{d r^{'}}{dr}$ (2)

... we obtain from (1)...

$r^{'} dr^{'} = -\frac{\mu}{r^{2}} dr$ (3)

... i.e. a separable variables DE the solution of which is...

$r^{' 2} = \frac{2 \mu}{r} + c_{1}$ (4)

... just like in Your textbook. The further step could be finding $r(t)$ ... a little less confortable step

...

Kind regards

$\chi$ $\sigma$

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