Answer
$ (a) 500 mi $, $1716 mi $
$ (b) 1354 mi; 27.7 mi/min $
Work Step by Step
$r=\frac{4995}{1+0.12 \cos\theta}$ ................ eq (1)
Reference to the figure
(a) At perigee $\theta =0^\circ$.
Putting $\theta =0^\circ$. in equation (1)
$r=\frac{4995}{1+0.12 \cos0^\circ} = 4460$(correct to 4 significant figures
Let R be the radius of the earth in miles' . Since $ R= 3960 mi$, $\Longrightarrow$
altitude of the satellite at perigee $=r-R=4460 -3960=500 $ mi
at apogee $\theta =180^\circ$.
Putting $\theta =180^\circ$. in equation (1)
$r=\frac{4995}{1+0.12 \cos180^\circ} = 5676$(correct to 4 significant figures)
The altitude of the satellite at apogee$ =r-R=5676-3960 =1716$ mi
(b)
Equation (1) may be written as
$r=4995 (1+0.12\cos\theta)^{-1}$
Taking derivative with respect to t
$\frac{dr}{dt}= \frac{ d[4995 (1+0.12\cos\theta)^{-1}] }{dt}$
$\frac{dr}{dt}= \frac{ d[4995 (1+0.12\cos\theta)^{-1}] }{d\theta} \frac{d\theta}{dt}$
$\frac{dr}{dt}=4995 \frac{ d[ (1+0.12\cos\theta)^{-1}] }{d\theta} \frac{d\theta}{dt}$
$\frac{dr}{dt}=-4995 (1+0.12\cos\theta)^{-1-1} [ \frac{ d(1+0.12\cos\theta)}{d\theta} ]\frac{d\theta}{dt}$
$\frac{dr}{dt}=-4995 (1+0.12\cos\theta)^{-2} [ \frac{d(1)}{d\theta} +\frac{d (0.12\cos\theta)}{d\theta}]\frac{d\theta}{dt}$
$\frac{dr}{dt}=-4995 (1+0.12\cos\theta)^{-2} [ 0 +0.12\frac{d (\cos\theta)}{d\theta}]\frac{d\theta}{dt}$
$\frac{dr}{dt}=-4995 (1+0.12\cos\theta)^{-2} [ 0.12(-\sin\theta)]\frac{d\theta}{dt}$
$\frac{dr}{dt}=4995 (1+0.12\cos\theta)^{-2} [ 0.12(\sin\theta)]\frac{d\theta}{dt}$
$\frac{dr}{dt}= \frac{4995\times 0.12 \sin\theta}{ (1+0.12\cos\theta)^2}\frac{d\theta}{dt}$ ........................ eq (2)
Given
$ \frac{d\theta}{dt}=2.7^\circ/min$
Since
$ 180^\circ=\pi$ radian
$1^\circ =\frac{\pi}{180}$ radian $\Longrightarrow$
$2.7^\circ= \frac{\pi}{180} \times 2.7$ radians
Putting $ \frac{d\theta}{dt}= \frac{\pi}{180} \times 2.7$ radians and $ \theta =120^\circ$
in equation (2)
$\frac{dr}{dt}= \frac{4995\times 0.12 \sin120^\circ}{ (1+0.12\cos120^\circ)^2} \times \frac{\pi}{180} \times 2.7 $
$\frac{dr}{dt}=\frac{4995\times 0.12\times \frac{\sqrt 3}{2}}{[(1+0.12\times (-\frac{1}{2})]^2} \times\frac{2.7\pi}{180}$
$\frac{dr}{dt}=\frac{519.1}{ \frac{2209}{2500}} \times\frac{2.7\pi}{180}$
$\frac{dr}{dt}=\frac{519.1 \times2500}{ 2209} \times\frac{2.7\pi}{180}= 27.7$ mi/min (correct to three signigiciant figure)