Answer
$(a)\space 20130\space km$
$(b)\space 3.875\space km/s$
Work Step by Step
(a) Here we use the equation $T=\sqrt {\frac{4\pi^{2}r^{3}}{GM_{E}}}$ to find the orbital period. where, T - Orbital period, V - speed of the satellite, $M_{E}$ - the mass of the earth, G - universal gravitational constant, r - radius of the orbit.
$T=\sqrt {\frac{4\pi^{2}r^{3}}{GM_{E}}}$ ; Let's plug known values into this equation.
$T= 2\pi \sqrt {\frac{r^{3}}{GM_{E}}}$
$11.97\space h(\frac{3600\space s}{h})=2\pi \sqrt {\frac{r^{3}}{6.67\times10^{-11}\space Nm^{2}/kg^{2}\times5.97\times10^{24}\space kg}}$
$47032164\times39.8\times10^{13}\space m^{3}= r^{3}$
$26551721\space m= r$
The altitude of the satellite (h) = r - radius of the earth
$h=26.5\times10^{6}\space m-6.37\times10^{6}\space m$
$h=20.13\times10^{6}\space m=20130\space km$
(a) Here we use the equation $V=\sqrt {\frac{GM_{E}}{r}}$ to find the speed of the satellite
$V=\sqrt {\frac{GM_{E}}{r}}$; Let's plug known values into this equation
$V=\sqrt {\frac{6.67\times10^{-11}\space Nm^{2}/kg^{2}\times5.97\times10^{24}\space kg}{26.5\times10^{6}\space m}}$
$V=\sqrt {\frac{39.8\times10^{13}}{26.5\times10^{6}}}\space m/s= 3875.4\space m/s=3.875\space km/s$
