Now I have another question about the conservation of angular momentum
I just want to prove that the rotating body's linear velocity will increase as it's distance to the orbiting mass (the mass that the rotating object is rotating around it) decrease in an eliptical orbit.
I thought that if I select the constant point as the center of the orbiting mass and take two different moment on the system I can prove it.
First moment, when the rotating mass' velocity vector is perpendicular to the position vector $\vec{r}$ but magnitude of the position vector $r$ is the smallest possible value.
Second moment is the same as the first one, but the position vector's magnitude is greatest possible value.
So, if I select a cartesian coordinate system that the x-axis as horizontal, y-axis as vertical and z-axis is to perpendicular to the plane that is created by x and y axes.
So, if I select a cartesian coordinate system that the x-axis as horizontal, y-axis as vertical and z-axis is to perpendicular to the plane that is created by x and y axes.
The $\vec{r}_1=a\hat{y}$, $\vec{v}_1=v_1\hat{x}$ and $vec{r}_2=b\hat{x}$, $\vec{v}_2=v_2\hat{y}$
As the conservation of the angular momentum due to the gravitational force is a conserved force.
$\vec{J}_1=\vec{r}_1\cross{m}\vec{v}_1=m{r_1}{v_1}(\hat{y}\cross\hat{x})=m{r_1}{v_1}(-\hat{z})$
$\vec{J}_2=\vec{r}_2\cross{m}\vec{v}_2=m{r_2}{v_2}(\hat{x}\cross-\hat{y})=m{r_2}{v_2}(-\hat{z})$
$\vec{J}_1=\vec{J}_2 \Rightarrow m{r_1}{v_1}=m{r_2}{v_2}$
Here, $r_1$ is smaller than $r_2$
As, the $r_1$ decreasing $v_1$ is increasing.
So, I think I've proved it.
If It's not correct the accurate or efficiency way to prove it, please give me tips. Not the solution, I would like solve it on my own.