# Thread: Differential Equation involving mechanics

1. ## Differential Equation involving mechanics

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.

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

dividing both sides my $\displaystyle m$

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

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

$\displaystyle \dot{r}=\frac{MG}{r^{2}}t+C$

any help is appreichated

thanks

2. This problem can be solved as in...

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

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

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

Now because is...

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$