Respuesta :
Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y
Explanation:
According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
[tex]T^{2}=\frac{4\pi^{2}}{GM}r^{3}[/tex] (1)
Where;
[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 semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
So for satellite X, the orbital period [tex]T_{X}[/tex] is:
[tex]T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}[/tex] (2)
Where [tex]r_{X}=1.2(10)^{6}m[/tex]
[tex]T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}[/tex] (3)
[tex]T_{X}=413.712 s[/tex] (4)
For satellite Y, the orbital period [tex]T_{Y}[/tex] is:
[tex]T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}[/tex] (5)
Where [tex]r_{Y}=1.9(10)^{5}m[/tex]
[tex]T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}[/tex] (6)
[tex]T_{Y}=26.064 s[/tex] (7)
This means [tex]T_{X}>T_{Y}[/tex]
Now let's calculate the tangential speed for both satellites:
For Satellite X:
[tex]V_{X}=\sqrt{\frac{GM}{r_{X}}}[/tex] (8)
[tex]V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}[/tex]
[tex]V_{X}=18224.783 m/s[/tex] (9)
For Satellite Y:
[tex]V_{Y}=\sqrt{\frac{GM}{r_{Y}}}[/tex] (10)
[tex]V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}[/tex]
[tex]V_{Y}= 45801.13 m/s[/tex] (11)
This means [tex]V_{Y}>V_{X}[/tex]
Therefore:
Satellite X has a greater period and a slower tangential speed than Satellite Y
