# Differential Equation involving mechanics

#### cooltowns

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)

#### Attachments

• 21.8 KB Views: 9
Last edited:

#### chisigma

MHF Hall of Honor
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

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