Respuesta :
Answer:
R=4.22*10⁴km
Explanation:
The tangential speed [tex]v[/tex] of the geosynchronous satellite is given by:
[tex]v=\frac{2\pi R}{T}[/tex]
Because [tex]2\pi R[/tex] is the circumference length (the distance traveled) and T is the period (the interval of time).
Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:
[tex]F_c=\frac{mv^{2} }{R}[/tex]
If we substitute the expression for [tex]v[/tex] in this formula, we get:
[tex]F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}[/tex]
Since the centripetal force is the gravitational force [tex]F_g[/tex] between the satellite and the Earth, we know that:
[tex]F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }[/tex]
Where G is the gravitational constant ([tex]G=6.67*10^{-11} Nm^{2}/kg^{2}[/tex]) and M is the mass of the Earth ([tex]M=5.97*10^{24}kg[/tex]). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:
[tex]R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km[/tex]
This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.
