Answer:
T = 1733.16 s = 28.88 min
Explanation:
The orbital velocity of a satellite about Earth is given as follows:
[tex]v = \sqrt{\frac{GM}{R}}[/tex]
where,
v = orbital speed = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²
M = Mass of Earth = 5.982 x 10²⁴ kg
R = Orbit Radius = 3.117 x 10⁶ m
Therefore,
[tex]v = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(5.982\ x\ 10^{24}\ kg)}{(3.117\ x\ 10^{6}\ m)}}\\\\v = 11.3\ x\ 10^{3}\ m/s[/tex]
but the velocity is given as:
[tex]v = \frac{distance}{time}[/tex]
for distance = circumference = 2πR
time = time period = T = ?
Therefore,
[tex]11.3\ x\ 10^{3}\ m/s = \frac{2\pi(3.117\ x\ 10^{6}\ m)}{T}\\\\T = \frac{2\pi(3.117\ x\ 10^{6}\ m)}{11.3\ x\ 10^{3}\ m/s}\\\\[/tex]
T = 1733.16 s = 28.88 min
