Answer:
The period of orbit = 7143.41 s = 119.1 min = 1.984 hours
Il periodo dell'orbita = 7143.41 s = 119.1 min = 1.984 hours
Explanation:
English Translation
In Apollo's missions to the moon, the command module orbited at an altitude of 110 km above the lunar surface. How long did the command module complete an orbit?
Solution
According to Kepler's laws, the square of the period of orbit is proportional to the cube of the radius of orbit.
T² ∝ r³
T² = kr³
where k = constant of proportionality. The constant of proportionality is then given as
k = (4π²/GM)
It's all obtained from equating the gravitational force on the command module by the moon and the circular motion of the command module
Let
G = Gravitational constant
M = mass of the moon
m = mass of the command module
r = radius of orbit
w = angular velocity of the command module
(GMm/r²) = mrw²
(GM/r²) = rw²
(GM/r³) = w²
But angular velocity is given as
w = (2π/T)
w² = (4π²/T²)
(GM/r³) = (4π²/T²)
We then obtain
T² = (4π²/GM)r³
T² = kr³
Mass of moon = M = (7.35 x 10²²) kg Gravitational constant = G = (6.67 x 10⁻¹¹) Nm²/kg²
radius of moon = (1.74 x 10⁶) m
Total radius of orbit = 110 km + (radius of the moon) = 110,000 + (1.74 x 10⁶)
= (1.85 × 10⁶) m
k = (4π²/GM) = (4π² ÷ [(6.67 x 10⁻¹¹) × (7.35 x 10²²)] = (8.059 × 10⁻¹²) kg/Nm²
T² = kr³
T² = (8.059 × 10⁻¹²) × (1.85 × 10⁶)³
T² = 51,028,321.74
T = √(51,028,321.74) = 7,143.41107175
T = 7143.41 s = 119.1 min = 1.984 hours
Hope this Helps!!!
