Question

A satellite in circular orbit 1150 kilometers above Earth makes one complete revolution every 140 minutes. Assuming that Earth is a sphere of radius 6400 kilometers, what is the linear speed (in kilometers per minute) of the satellite? (Round your answer to one decimal place.) km/min

Answer 1

Step-by-step explanation:

The radius of the orbit is r = 1150 km + 6400 km = 7550 km or

$r = 7.55×10^6\:\text{m}$

and it takes 140 minutes to complete 1 revolution. This is the same aa

$140\:\text{min}×\dfrac{60\:\text{s}}{1\:\text{min}} = 8400\:\text{s}$

The linear speed of the satellite then is simply equal to the orbital circumference divided by the orbital period T or

$v = \dfrac{2\pi r}{T}$

$\:\;\:\:=\dfrac{2\pi(7.55×10^6\:\text{m})}{8.4×10^3\:\text{s}}$

$\:\:\:\:=5647.4\:\text{m/s}$

In km/min, this is

$v = \dfrac{2\pi(7550\:\text{km})}{140\:\text{min}} = 338.8\:\text{km/min}$