4.15 Planetary motion  (Page 3/4)

If “R” is the radius of curvature at a given position on the elliptical trajectory, then centripetal force equals gravitational force as given here :

$$\frac{m{v}^2}{R}=m{\omega }^{2}R=\frac{GMm}{r^2}$$

Where “M” and “m” are the mass of Sun and Earth; and “r” is the linear distance between Sun and Earth.

Except for parameters “r” and “R”, others are constant in the equation. We note that radii of curvature at perihelion and aphelion are equal. On the other hand, centripetal force is greatest at perihelion and least at aphelion. From the equation above, we can also infer that both linear and angular velocities of planet are not constant.

Angular momentum

The angular velocity of the planet about Sun is not constant. However, as there is no external torque working on the system, the angular momentum of the system is conserved. Hence, angular momentum of the system is constant unlike angular velocity.

The description of motion in angular coordinates facilitates measurement of angular momentum. In the figure below, linear momentum is shown tangential to the path in the direction of velocity. We resolve the linear momentum along the parallel and perpendicular to radial direction. By definition, the angular momentum is given by :

Angular momentum

$$L=rX{p}_{\perp }$$

$$L=rX{p}_{\perp }=rm{v}_{\perp }=rmX\omega r=m\omega r^2$$

Since mass of the planet “m” is constant, it emerges that the term “ $\omega r^2$ ” is constant. It clearly shows that angular velocity (read also linear velocity) increases as linear distance between Sun and Earth decreases and vice versa.

Maximum and minimum velocities

Maximum velocity corresponds to perihelion position and minimum to aphelion position in accordance with maximum and minimum centripetal force at these positions. We can find expressions of minimum and maximum velocities, using conservation laws.

Maximum and minimum velocities

Let “ $r_1$ ” and “ $r_2$ ” be the minimum and maximum distances, then :

$$r_1=a\left(1-e\right)$$

$$r_2=a\left(1+e\right)$$

We see that velocities at these positions are perpendicular to semi major axis. Applying conservation of angular momentum,

$$L=r_{1}m{v}_1=r_{2}m{v}_2$$

$$r_1{v}_1=r_2{v}_2$$

Applying conservation of energy, we have :

$$\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}$$

Substituting for “ $v_2$ ”, “ $r_1$ ” and “ $r_2$ ”, we have :

$$v_1=v_{\max }=\sqrt{\{\frac{GM}{a}\left(\frac{1+e}{1-e}\right)}$$

$$v_2=v_{\min }=\sqrt{\{\frac{GM}{a}\left(\frac{1-e}{1+e}\right)}$$

Energy of sun-planet system

As no external force is working on the system and there is no non-conservative force, the mechanical energy of the system is conserved. We have derived expression of linear velocities at perihelion and aphelion positions in the previous section. We can, therefore, find out energy of “Sun-planet” system by determining the same at either of these positions.

Let us consider mechanical energy at perihelion position. Here,

$$E=\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}$$

Substituting for velocity and minimum distance, we have :

$$\Rightarrow E=\frac{mGM\left(1+e\right)}{2a\left(1+e\right)}-\frac{GMm}{a\left(1-e\right)}$$

$$\Rightarrow E=\frac{mGM}{a\left(1-e\right)}\left(\frac{1+e}{2}-1\right)$$

$$\Rightarrow E=\frac{mGM}{a\left(1-e\right)}\left(\frac{e-1}{2}\right)$$

$$\Rightarrow E=-\frac{GMm}{2a}$$

We see that expression of energy is similar to that of circular trajectory about a center with the exception that semi major axis “a” replaces the radius of circle.