Answer:
T = 5,06 10³ s
Explanation:
For this exercise, let's use Newton's second law, where force is the force of gravitational attraction
F = m a
The acceleration is centripetal
a = v² / r
Let's replace
G m [tex]M_{e}[/tex] / [tex]R_{e}[/tex]² = m v² / [tex]R_{e}[/tex]
G [tex]M_{e}[/tex] / [tex]R_{e}[/tex] = v²
The velocity module is constant, so we can use the equation of uniform motion
v = d / t
Where the distance is the length of the circle and in this case the time is called the period
d = 2π [tex]R_{e}[/tex]
We replace
(2π[tex]R_{e}[/tex] / T)² = G [tex]M_{e}[/tex] / [tex]R_{e}[/tex]
T² = 4π² [tex]R_{e}[/tex]³ / G [tex]M_{e}[/tex]
T = √ 4π² [tex]R_{e}[/tex]³ / G [tex]M_{e}[/tex]
Let's calculate
T = √ (4π² (6.37 10⁶)³ / 6.67 10⁻¹¹ 5.98 10²⁴)
T = √ (255.83 10⁵) = √ (25.583 10⁶)
T = 5,05796 10³ s
