3D Orbital Period Calculation
This 3D simulation demonstrates the conversion of the constant k and calculation of the Moon's orbital period around Earth.
Given constant k:
10⁻¹³ s² m⁻³
Moon's distance from Earth:
3.84 × 10⁵ km
Conversion of Constant k
We need to express k in days and kilometres:
k = 10⁻¹³ s² m⁻³
= 10⁻¹³ [1/(24 × 60 × 60)² d²] [1/(1/1000)³ km⁻³]
= 1.33 × 10⁻¹⁴ d² km⁻³
= 10⁻¹³ [1/(24 × 60 × 60)² d²] [1/(1/1000)³ km⁻³]
= 1.33 × 10⁻¹⁴ d² km⁻³
1 day = 24 × 60 × 60 = 86400 seconds
1 km = 1000 meters
Therefore:
1 s = 1/86400 d
1 m = 1/1000 km
Calculation of Moon's Orbital Period
Using the relationship T² = kR³ (Kepler's Third Law):
T² = (1.33 × 10⁻¹⁴ d² km⁻³) × (3.84 × 10⁵ km)³
T² = (1.33 × 10⁻¹⁴) × (5.66 × 10¹⁶)
T² ≈ 753.33 d²
T ≈ √753.33 ≈ 27.3 days
T² = (1.33 × 10⁻¹⁴) × (5.66 × 10¹⁶)
T² ≈ 753.33 d²
T ≈ √753.33 ≈ 27.3 days
Converted constant k:
1.33 × 10⁻¹⁴ d² km⁻³
Moon's orbital period (T):
27.3 days
Key Notes
1. The same relationship holds for elliptical orbits if we replace the distance with the semi-major axis of the ellipse.
2. For elliptical orbits, Earth would be at one of the foci of the ellipse.
3. The animation above shows the Moon's orbit (simplified as circular) with the correct 27.3 day period.
