Central force problem with relation to Kepler's law The problem is:
Consider a particle moving in $\mathbb R^3$ with location as a function of time described by the vector function $\mathbf r$(t). Assume that the motion is driven by a central force of unspecified magnitude in the direction $\mathbf -r $(t) from the particle back to the spatial origin $\mathbf 0$. Use Newton's law of motion $\mathbf F$=m$\mathbf a$ to show that the path $\mathbf r$(t) of the particle:
1) lies in a plane through the origin, and
2)the radius vector $\bf{r}$ sweeps out equal areas in equal times.
How would you begin this central force problem?
I think we would start by using the definition of a central force:
$$ {\bf F(r}(t)) = \lvert F(r)\rvert \times {{\bf r}(t) \over |r(t)|} $$
And then equate with F = ma , and try to solve for ${\bf r}(t)$?
Any pointer in the right direction would be appreciated, thanks!
 A: 
PART ONE
Denote the velocity of point $M$ with $\vec v$. In a general case that vector and force vector $\vec F$ define a plane. Denote that plane with $xOy$ with O being the center of attraction. Note that the axis $z$ (not shown in the picture) is perpendicular to the plane $xOy$. 
If you apply Newton's law of motion in the $z$ direction you get:
$$ma_{z}=0$$
...which basically means that the acceleration in the $z$ direction is equal to zero. The point will be leaving the plane $xOy$ only if it has non-zero component of velocity along the $z$ axis. But $v_{z}=0$ so the point will continue to move in the plane defined by vectors $\vec F$ and $\vec v$. Despite the fact that vectors $\vec F$ and $\vec v$ change, the plane $xOy$ remains fixed.
PART TWO
Denote with $\vec r_0$ the unit vector along the radial axis OM. Denote with $\vec p_0$ the unit circular vector (vector obtained by rotatining $\vec r_0$ for $90^\circ$ counterclockwise).
One can easily show that:
$$\vec r_0 = \vec{i}\cos\varphi+\vec{j}\sin\varphi$$
$$\vec p_0 = -\vec{i}\sin\varphi+\vec{j}\cos\varphi$$
$$\frac{d\vec r_0}{dt} = (-\vec{i}\sin\varphi+\vec{j}\cos\varphi)\frac{d\varphi}{dt}=\vec p_0 \frac{d\varphi}{dt}\tag{1}$$
$$\frac{d\vec p_0}{dt} = (-\vec{i}\cos\varphi-\vec{j}\sin\varphi)\frac{d\varphi}{dt}=-\vec r_0 \frac{d\varphi}{dt}\tag{2}$$
Let's start with something obvious:
$$\vec r = r \vec r_0$$
$$\vec v = \frac{d\vec r}{dt}=\frac{dr}{dt}\vec r_0+r\frac{d\vec r_0}{dt}\tag{3}$$
Replace (1) into (3) and you get:
$$\vec v = \frac{dr}{dt}\vec r_0+r\frac{d\varphi}{dt}\vec p_0\tag{4}$$
You have basicaly obtained the radial and circular component of velocity:
$$v_r=\frac{dr}{dt}$$
$$v_p=r\frac{d\varphi}{dt}\tag{4'}$$
Ok, let's calculate the acceleration, starting from (4):
$$\vec a = \frac{d^2r}{dt^2}\vec r_0+\frac{dr}{dt}\frac{d\vec r_0}{dt}+\frac{dr}{dt}\frac{d\varphi}{dt}\vec p_0+r\frac{d^2\varphi}{dt^2}\vec p_0+r\frac{d\varphi}{dt}\frac{d\vec p_0}{dt}\tag{5}$$
Replace (1) and (2) into (5), regroup and you get:
$$\vec a=a_r\vec r_0+a_p\vec p_0$$
with radial and circular component of acceleration defined with:
$$a_r=\frac{d^2r}{dt^2}-r\left( \frac{d\varphi}{dt}\right)^2$$
$$a_p=r\frac{d^2\varphi}{dt^2}+2\frac{dr}{dt}\frac{d\varphi}{dt}\tag{6}$$
Back to Newton's law:
$$m\vec a=\vec F$$
In scalar form:
$$m a_r=F_r=-F$$
$$m a_p=F_p=0\tag{7}$$
From (6) and (7) we get:
$$a_p=0$$
$$r\frac{d^2\varphi}{dt^2}+2\frac{dr}{dt}\frac{d\varphi}{dt}=0$$
$$r\frac{d^2\varphi}{dt^2}+2\frac{dr}{dt}\frac{d\varphi}{dt}=0$$
$$r^2\frac{d^2\varphi}{dt^2}+2r\frac{dr}{dt}\frac{d\varphi}{dt}=0$$
$$\frac{d}{dt}(r^2\frac{d\varphi}{dt})=0$$
$$r^2\frac{d\varphi}{dt}=\text{const}\tag{8}$$
Now look at the elementary area (shaded triangle) swept by vector $\vec r$:
$$dA=\frac12 r (v\space dt) \sin\alpha=\frac12 r (v \sin\alpha)dt$$
Notice that $v\sin\alpha$ is actually the circular component of velocity defined in (4'):
$$\frac{dA}{dt}=\frac12 rv_p=\frac12 r\cdot r\frac{d\varphi}{dt}=\frac 12 r^2\frac{d\varphi}{dt}$$
Taking (8) into account we finally get:
$$\frac{dA}{dt}=\text{const}$$
...which simply translates to the fact that the radius vector $\vec r$ sweeps out equal areas in equal times.
With gravity being a central force, we have just proved the second Kepler's law. Now go ahead and prove the remaining two :)
