Answer
(a) No
(b) $7.76\frac{Km}{s}$
(c) $1.49h$
Work Step by Step
(a) The orbital speed is given as $v=\sqrt{\frac{GM_E}{R_E+h}}$. This equation shows that the orbital speed of the shuttle does not depend on its mass.
(b) We know that
$v=\sqrt{\frac{GM_E}{R_E+h}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{6.67\times 10^{-11}(5.97\times 10^{24})}{6.37\times 10^6+250\times 10^3}}$
$v=7760\frac{m}{s}=7.76\frac{Km}{s}$
(c) The required time can be calculated as
$T=\frac{2\pi r}{v}$
We plug in the known values to obtain:
$T=\frac{2\pi(6.37\times 10^6+250\times 10^3)}{7.76\times 10^3}$
$T=5360s=1.49h$
