Differential Equation involving mechanics

Mar 2009
66
0
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

(Happy)
 

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chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
This problem can be solved as in...

http://www.mathhelpforum.com/math-help/differential-equations/143954-finding-general-solution-difficult-differential-equation.html#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 (Thinking) ...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)