Question 7.11: Central Force Motion and Kepler's Second Law. A force direct...
Central Force Motion and Kepler’s Second Law. A force directed invariably along a line through a fixed point is called a central force. A familiar example of a central force is the tension in a pendulum string; another is the gravitational force exerted by the Earth on a satellite shown in Fig. 7.11. Derive Kepler’s second law: A particle in motion under a central force alone must move in a plane; and if its path is not a straight line through the fixed central point O, its position vector from the fixed point sweeps out equal areas in equal intervals of time.
Consider a central force F directed through the fixed origin O of an inertial frame \varphi=\left\{O ; \mathbf{e}_k\right\} in Fig. 7.11. Then the moment of F about O is zero: \mathbf{M}_O=\mathbf{x} \times \mathbf{F}=0, wherein \mathbf{x}=r e_r; and the moment of momentum principle (6.79) shows that the moment of momentum about O is a constant vector :
\mathbf{M}_O=\frac{d \mathbf{h}_O}{d t}. (6.79)
\mathbf{h}_O=\mathbf{x} \times m \mathbf{v}, a constant (7.72a)
It follows that \mathbf{h}_O=\mathbf{0} if and only if the position vector x is parallel to the velocity: \mathbf{v}=k \mathbf{x}, where k is a constant. In this instance the motion \mathbf{x}(t)=\mathbf{X}_0 e^{k t}, with \mathbf{x}_0=\mathbf{x}(0) initially, is along a straight line through O. Otherwise, when \mathbf{h}_O \neq \mathbf{0}, both x and v are always perpendicular to the constant vector h_o, and hence the particle must move in a plane whose vector equation is (7.72a). Consequently, all central force motions are plane motions.
To establish Kepler’s equal area rule, we introduce polar coordinates and write \mathbf{v}=\dot{r} \mathbf{e}_r + r \dot{\phi} \mathbf{e}_\phi, and \mathbf{x}=r \mathbf{e}_r. Then \mathbf{h}_O=h \mathbf{e}_z, is normal to the plane of motion, and (7.72a) yields m r^2 \dot{\phi} \mathbf{e}_z=h \mathbf{e}_z, that is,
r^2 \dot{\phi}=\frac{h}{m}=\gamma, a constant. (7 .72b)
Now, it may be seen in Fig. 7.11a that the element \Delta A of the plane area swept out by the position vector is \Delta A=\frac{1}{2}(r + \Delta r) r \Delta \phi. Therefore, in the \text { limit }_{\Delta t \rightarrow 0} \Delta A / \Delta t, we have
\dot{A}=\frac{1}{2} r^2 \dot{\phi} (7.72c)
This gives the rate of change of the area swept out by the position vector. Thus, with (7.72b), \dot{A}=\gamma / 2; hence A(t)=\frac{1}{2} \gamma t + A_0, where A_0=A(0) is a constant, usually taken as zero. This is Kepler’s Second Law: The radius vector sweeps out equal areas in equal intervals of time.