Answer: 3071.79 m/s
Explanation:
Approaching satellite's orbit 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[/tex] is the radius of the orbit
Now, if we want the satellite to move in a geosynchronous orbit, it needs to travel at a specific orbiting radius and period to fulfill this condition. This means the satellite's orbital period [tex]T[/tex] must match Earth's rotation on its axis, which takes one sidereal day ( approximately 24 h).
So, if the satellite travels one complete circle [tex]C=2\pi r[/tex] in a period [tex]T[/tex], its velocity is also expressed as:
[tex]V=\frac{2\pi r}{T}[/tex] (2)
Where [tex]T=24 h \frac{3600 s}{1 h}=86400 s[/tex]
Combining (1) and (2):
[tex]\sqrt{G\frac{M}{r}}=\frac{2\pi r}{T}[/tex] (3)
Isolating [tex]r[/tex]:
[tex]r=\sqrt[3]{x\frac{GMT^{2}}{4 \pi^{2}}}[/tex] (4)
[tex]r=42,240,234.3 m[/tex] (5)
Substituting (5) in (1):
[tex]V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{42,240,234.3 m}}[/tex] (6)
Finally:
[tex]V=3071.79 m/s[/tex]
