I need help on this differential equaion problem?

Question

Let equation $(1)$ be $\overrightarrow{F}= m \cdot \overrightarrow{a}$ and equation $(2)$ be $\overrightarrow{F}= \frac{-G \cdot M \cdot m}{ | \overrightarrow{r^2} |} \frac{\overrightarrow{r} }{ | \overrightarrow{r} |}$.

The problem says to equate both force equations and cancel the common factor which I get

$a = \frac{-GM}{|r|^2 }\frac{ r}{|r|}$ where $r$ is a vector.

The problem I'm having that the next question asks me to use the fact that $a(t)= r"(t)$, and $r(t)= x(t)i+y(t)j$ and convert the equation I found into an equation invovling $x(t)$ and $y(t)$.

I don't know how to do this. Also, I don't know how to deal with the absolute values and vectors. Please help.

Answer 1

Since $a(t)=r''(t)$, replacing it at the equation $a = \frac{-GM}{|r|^2 }\frac{ r}{|r|}$, we get: $$r''(t) = \frac{-GM}{|r(t)|^2 }\frac{ r(t)}{|r(t)|}$$ $$r(t)=x(t)i+y(t)j \Rightarrow |r(t)|=\sqrt{x^2(t)+y^2(t)}$$ $$r''(t)=x''(t)i+y''(t)j$$ $$\text{So } x''(t)i+y''(t)j=\frac{-GM}{\sqrt{x^2(t)+y^2(t)}^2 }\frac{ x(t)i+y(t)j}{\sqrt{x^2(t)+y^2(t)}}=\frac{-GM(x(t)i+y(t)j)}{({x^2(t)+y^2(t))}^{\frac{3}{2} }}$$

Answer 2

Use the fact that $\vec{r} = x\hat{i} + y\hat{j}$. Therefore, $\vec{r}^{\prime\prime} = x^{\prime\prime}\hat{i} + y^{\prime\prime}\hat{j}$ and $r = \sqrt{x^2 + y^2}$. Therefore, the equation \begin{equation} \vec{a} = -\frac{GM}{|r|^3}\vec{r} \end{equation} gives two scalar equations \begin{equation} x^{\prime\prime} = -\frac{GM}{(x^2 + y^2)^{3/2}}x \end{equation} and \begin{equation} y^{\prime\prime} = -\frac{GM}{(x^2 + y^2)^{3/2}}y \end{equation} In practice, it is convenient to use polar coordinates to solve this problem.