Respuesta :
Answer:
Orbital period, T = 2.02 hours
Explanation:
It is given that, an artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s². We have to find the orbital period (T) of the satellite.
Firstly, calculating the distance between Earth and satellite. The acceleration due to gravity is given by :
[tex]a=\dfrac{GM}{r^2}[/tex]
G = universal gravitational constant
M = mass of earth
[tex]r=\sqrt{\dfrac{GM}{a}}[/tex]
[tex]r=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{6.03\ m/s^2}}[/tex]
r = 8126273.3 m..........(1)
Now, according to Kepler's third law :
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
Putting the value of r from equation (1) in above equation as :
[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.97\times 10^{24}}\times (8126273.3)^3[/tex]
[tex]T^2=53202721.01\ s[/tex]
T = 7294.01 seconds
Since, 1 hour = 3600 seconds
Converting seconds to hour we get :
So, T = 2.02 hour
So, the orbital period of the satellite is 2.02 hours.
