Question

Nelle missioni di Apollo sulla Luna,il modulo di comando orbitava a un altitudine di 110 km al di sopra della superficie lunare.Quanto tempo impiegava il modulo di comando a completare un'orbita

Answer 1

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!!!