3D Earth Mass Measurement
Example 8.6: Two methods to determine Earth's mass in 3D
Method 1: Using Surface Gravity
Using Newton's law of gravitation and the measured value of gravitational acceleration at Earth's surface:
\[ M_E = \frac{g R_E^2}{G} \]
Example
Question:
Weighing the Earth:
You are given the following data: \(g = 9.81\,\mathrm{ms}^{-2}\), \(R_E = 6.37 \times 10^6\,\mathrm{m}\), the distance to the moon \(R = 3.84 \times 10^8\,\mathrm{m}\) and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth \(M_E\) in two different ways.
Solution:
Method 1:
From Eq. (7.12):
\[
M_E = \frac{g R_E^2}{G}
\]
\[
= \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}}
= 5.97 \times 10^{24}\,\mathrm{kg}
\]
Method 2:
The moon is a satellite of the Earth. From Kepler’s third law,
\[
T^2 = \frac{4\pi^2 R^3}{G M_E}
\]
\[
M_E = \frac{4\pi^2 R^3}{G T^2}
\]
\[
= \frac{4 \times 3.14 \times 3.14 \times (3.84)^3 \times 10^{24}}{6.67 \times 10^{-11} \times (27.3 \times 24 \times 60 \times 60)^2}
= 6.02 \times 10^{24}\,\mathrm{kg}
\]
Both methods yield almost the same answer, the difference being less than 1%.
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 3D 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.



