Answer: [tex]V=\sqrt{\frac{1.32(10)^{14} m^{3}/s^{2}}{R}}[/tex]
Explanation:
Approaching the orbit of the satellite around the Earth to a circular orbit, we can use the equation of velocity in the case of uniform circular motion:
[tex]V=\sqrt{G\frac{M}{r}}[/tex] (1)
Where:
[tex]V[/tex] is the velocity of the satellite
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant
[tex]M=5.972(10)^{24}kg[/tex] is the mass of the Earth
[tex]r=3R[/tex] is the radius of the orbit
Rewriting (1):
[tex]V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24}kg}{3R}}[/tex] (2)
Then:
[tex]V=\sqrt{\frac{1.32(10)^{14} m^{3}/s^{2}}{R}}[/tex] (3)
