Answer:
[tex]v = 3869 m/s[/tex]
Explanation:
As we know that the orbital speed of the satellite is given as
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
also we know that
time period of the revolution is given as
[tex]T = \frac{2\pi r}{v}[/tex]
now from above equation we know that
[tex]T = \frac{2\pi (\frac{GM}{v^2})}{v}[/tex]
[tex]T = \frac{2\pi GM}{v^3}[/tex]
so we will have
[tex]v = (\frac{2\pi GM}{T})^{1/3}[/tex]
now plug in all data in this equation
[tex]v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}[/tex]
[tex]v = 3869 m/s[/tex]
