Answer: [tex]V=5839.051m/s[/tex]
Explanation:
According to the Third Kepler’s Law of Planetary motion:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where;:
[tex]T=1.26(10)^{4}s[/tex] is the period of the satellite
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M=5.98(10)^{24}kg[/tex] is the mass of the Earth
[tex]a[/tex] is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity [tex]V[/tex] is given by:
[tex]V=\sqrt{\frac{GM}{a}}[/tex] (2)
Now, from (1) we can find [tex]a[/tex], in order to substitute this value in (2):
[tex]a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}[/tex] (3)
[tex]a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}[/tex] (4)
[tex]a=11705845.57m[/tex] (5)
Substituting (5) in (2):
[tex]V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}[/tex] (6)
[tex]V=5839.051m/s[/tex] (7) This is the speed at which the satellite travels
