Measuring Earth's Mass
Example 8.6: Two methods to determine Earth's mass
Method 1: Using Surface Gravity
Using Newton's law of gravitation and the measured value of gravitational acceleration at Earth's surface:
\[ M_E = \frac{g R_E^2}{G} \]
Earth's mass calculation:
\[ M_E = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} \]
Calculated mass: 5.97 × 1024 kg
Method 2: Using Moon's Orbit
Using Kepler's third law and the Moon's orbital characteristics:
\[ T^2 = \frac{4\pi^2 R^3}{G M_E} \]
\[ M_E = \frac{4\pi^2 R^3}{G T^2} \]
Earth's mass calculation:
\[ M_E = \frac{4π^2 (3.84 × 10^8)^3}{6.67 × 10^{-11} × (27.3 × 24 × 3600)^2} \]
Calculated mass: 6.02 × 1024 kg
Both methods yield nearly identical results (<1% difference):
5.97 × 1024 kg (Surface) vs 6.02 × 1024 kg (Orbit)
Physics Explanation:
This simulation demonstrates two independent methods to calculate Earth's mass, showing the consistency of physical laws.
Method 1: Surface Gravity
From Newton's law of gravitation:
\[ g = \frac{G M_E}{R_E^2} \]
Rearranged to solve for Earth's mass:
\[ M_E = \frac{g R_E^2}{G} \]
Using measured values of g (9.81 m/s²), Earth's radius (6.37×10⁶ m), and gravitational constant G (6.67×10⁻¹¹ Nm²/kg²).
Method 2: Moon's Orbit
From Kepler's third law (for circular orbits):
\[ T^2 = \frac{4π^2 R^3}{G M_E} \]
Rearranged to solve for Earth's mass:
\[ M_E = \frac{4π^2 R^3}{G T^2} \]
Using Moon's orbital radius (3.84×10⁸ m) and period (27.3 days).
Significance
The close agreement between these independent methods provides strong evidence for the validity of Newton's law of universal gravitation.
