Chapter 11.2, Problem 62E

(a)

To determine

To Show: The value of $\theta =\frac{vt}{r_0}$ and then $r\left(t\right)=\langle r_0\cos \frac{vt}{r_0},r_0\sin \frac{vt}{r_0}\rangle$ .

Explanation of Solution

Given Data:

The satellite of the mass $m$ that moves at the constant speed of the $v$ around the planet of mass M in the circular orbit of the radius $r_0$ is measured from the planets centre of mass. Then, determine the satellite’s orbital period T .

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the circular arc from the point where $t=0$ to the point where $m$ is. The length of the arc is $r_0\theta$ and it is equal to $vt$ then,

  $\begin{array}{l}{r}_0\theta =vt\\ \theta =\frac{vt}{r_0}\end{array}$

Then, the magnitude of the position vector is $r_0$ as the position vector is $r\left(t\right)$ then,

  $r\left(t\right)=\langle r_0\cos \frac{vt}{r_0},r_0\sin \frac{vt}{r_0}\rangle$

(b)

To determine

To find: The acceleration of the satellite.

Answer to Problem 62E

The acceleration vector is $\frac{-v^2}{r_0}\langle \cos \frac{vt}{r_0},\sin \frac{vt}{r_0}\rangle$ .

Explanation of Solution

Consider the velocity vector of the satellite is,

  $\begin{array}{c}a\left(t\right)=\frac{d\left(v\left(t\right)\right)}{dt}\\ =\frac{d}{dt}\langle r_0\cos \frac{vt}{r_0},r_0\sin \frac{vt}{r_0}\rangle \\ =\frac{d}{dt}\langle -v\sin \frac{vt}{r_0},v\cos \frac{vt}{r_0}\rangle \end{array}$

Then, the acceleration vector of the satellite is,

  $\begin{array}{c}a\left(t\right)=\frac{d\left(v\left(t\right)\right)}{dt}\\ =\frac{d}{dt}\langle -v\cos \frac{vt}{r_0},v\cos \frac{vt}{r_0}\rangle \\ =\frac{-v^2}{r_0}\langle \cos \frac{vt}{r_0},\sin \frac{vt}{r_0}\rangle \end{array}$

(c)

To determine

To Show: The expression $v^2=\frac{GM}{r_0}$ .

Explanation of Solution

Consider from part (b) the acceleration vector is,

  $a\left(t\right)=-{\left(\frac{v}{r_0}\right)}^{2}r\left(t\right)$

Then, by the second law of the gravitation,

  $\begin{array}{c}F=\left(\frac{-GMm}{r_0^2}\right)\frac{r}{r_0}\\ =-m{\left(\frac{v}{r_0}\right)}^{2}r\\ =\frac{GM}{r_0}\end{array}$

Hence, porved.

(d)

To determine

To Show: The orbital period T satisfies $vT=2\pi r_0$ .

Explanation of Solution

Consider that the satellite completes the circle that means $\theta =2\pi$ then the time take is known as the orbital period and is presented by $T$ then,

  $\begin{array}{l}\theta =2\pi \\ \frac{vT}{r_0}=2\pi \\ vT=2\pi r_0\end{array}$

Hence, proved.

(e)

To determine

To Show: The expression for the period of satellite is $T^2=\frac{4{\pi }^2}{GM}{r}_0^3$ .

Explanation of Solution

Substitute the values $v=\frac{2\pi r_0}{T}$ and $v^2=\frac{GM}{r_0}$ .

  $\begin{array}{l}{\left(\frac{2\pi r_0}{T}\right)}^2=\frac{GM}{r_0}\\ \frac{1}{T^2}=\frac{GM}{4{\pi }^2{r}_0^3}\\ T^2=\frac{4{\pi }^2}{GM}{r}_0^3\end{array}$

Hence, proved.