Chapter 10.3, Problem 80E

To determine

To show: For a satellite following a curve that is a conic section and sweeping out equal area in equal time.

The acceleration vector is given by:

  $〈\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}〉=-\frac{4K^2}{c{r}^2}\cdot \langle \cos t,\sin t\rangle$

Also, to show that the satellite is directed toward the origin and has magnitude inversely proportional to the square of the distance.

Answer to Problem 80E

It is shown that the acceleration vector is $〈\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}〉=-\frac{4K^2}{c{r}^2}\cdot \langle \cos t,\sin t\rangle$ and the satellite is directed toward the origin and has magnitude inversely proportional to the square of the distance.

Explanation of Solution

Given information:

The position vector of the satellite is given by:

  

Concept used:

For polar coordinates of a point $\left(r,t\right)$ , the rectangular coordinates of that point are equal to $\left(x,y\right)=\left(r\cos t,r\sin t\right)$ .

Here,

  $\begin{array}{c}x(t)=\frac{c\cos t}{1+e\cos t}\\ =\frac{c}{1+e\cos t}\cdot \cos t\end{array}$

and

  $\begin{array}{c}y(t)=\frac{c\sin t}{1+e\cos t}\\ =\frac{c}{1+e\cos t}\cdot \sin t\end{array}$

Where, $r=\frac{c}{1+e\cos t}$

Differentiate x with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{c\cdot (-\sin t)\cdot (1+e\cos t)-c\cos t\cdot e\cdot (-\sin t)}{{\left(1+e\cos t\right)}^2}\\ =\frac{-c\sin t\cdot (1+e\cos t)-c\cos t\cdot (-e\sin t)}{{\left(1+e\cos t\right)}^2}\\ =\frac{-c\sin t-ce\sin t\cos t+ce\sin t\cos t}{{\left(1+e\cos t\right)}^2}\\ =\frac{-c\sin t}{{\left(1+e\cos t\right)}^2}\\ ={\left(\frac{c}{1+e\cos t}\right)}^2\frac{-\sin t}{c}\\ =\frac{-r^2\sin t}{c}\end{array}$

Differentiate y with respect to t .

  $\begin{array}{c}\frac{dy}{dt}=\frac{c\cdot \cos t\cdot \left(1+e\cos t\right)-c\sin t\cdot e\cdot \left(-\sin t\right)}{{\left(1+e\cos t\right)}^2}\\ =\frac{c\cos t+ce{\cos }^{2}t+ce{\sin }^{2}t}{{\left(1+e\cos t\right)}^2}\\ =\frac{c\cos t+ce\cdot \left({\cos }^{2}t+{\sin }^{2}t\right)}{{\left(1+e\cos t\right)}^2}\\ =\frac{c\cos t+ce\cdot 1}{{\left(1+e\cos t\right)}^2}\left[\text{Trigonometric identity}:{\cos }^{2}t+{\sin }^{2}t=1\right]\\ =\frac{c\cos t+ce}{{\left(1+e\cos t\right)}^2}\\ =\frac{c\cdot \left(\cos t+e\right)}{{\left(1+e\cos t\right)}^2}\\ ={\left(\frac{c}{1+e\cos t}\right)}^2\cdot \left(\frac{\cos t+e}{c}\right)\\ =r^2\left(\frac{\cos t+e}{c}\right)\\ \end{array}$

Now differentiate $\frac{dx}{dt}$ with respect to t .

  $\begin{array}{c}\frac{d^{2}x}{d{t}^2}=\frac{d}{dt}\left(\frac{-r^2\sin t}{c}\right)\\ =\frac{-r^2\cos t}{c}\end{array}$

  $\begin{array}{c}\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left(r^2\left(\frac{\cos t+e}{c}\right)\right)\\ =\frac{-r^2\sin t}{c}\end{array}$

Thus, the acceleration vector will be :

  $\begin{array}{l}〈\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}〉=〈\frac{-r^2\cos t}{c},\frac{-r^2\sin t}{c}〉\\ =\frac{-r^2}{c}〈\cos t,\sin t〉\end{array}$

Using the fact $2K=r^2\frac{d\theta }{dt}$

Here $\theta =t$

Thus,

  $2K=r^2$

It implies

  $\begin{array}{c}〈\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}〉=\frac{-r^2\cdot r^2}{c\cdot r^2}〈\cos t,\sin t〉\\ =\frac{-{\left(r^2\right)}^2}{c\cdot r^2}〈\cos t,\sin t〉\\ =\frac{-{\left(2K\right)}^2}{c\cdot r^2}〈\cos t,\sin t〉\\ =\frac{-4K^2}{c\cdot r^2}〈\cos t,\sin t〉\end{array}$

Since the position vector is directed away from the origin and the vector is directed toward the origin.

  

It implies that the acceleration vector of the satellite is directed toward the origin.

The magnitude of the acceleration vector of the satellite is given by:

  $\begin{array}{c}\left|a\left(t\right)\right|=\sqrt{{\left(\frac{-4K}{c{r}^2}\cos t\right)}^2+{\left(\frac{-4K}{c{r}^2}\sin t\right)}^2}\\ =\sqrt{{\left(\frac{4K^2}{c{r}^2}\right)}^2\left({\cos }^{2}t+{\sin }^{2}t\right)}\\ =\sqrt{{\left(\frac{4K^2}{c{r}^2}\right)}^2\left(1\right)}\\ =\sqrt{{\left(\frac{4K^2}{c{r}^2}\right)}^2}\\ =\left|\frac{4K^2}{c{r}^2}\right|\end{array}$

Here, $a\left(t\right)\propto \frac{1}{r^2}$

Thus, the magnitude of the acceleration vector of the satellite is inversely proportional to the square of the distance of the satellite from the origin.