Talha's Physics Academy
Gravitation - Derivation for Mass of Earth
Q. Using Newton’s Law of gravitation, derive an expression for Mass of Earth and hence find its value.
Mass of Earth
Consider a body of mass '$m$' placed on or near the surface of the Earth.
- Let the mass of the Earth be '$M_e$'
- Let the radius of the Earth be '$R_e$'
Figure: Gravitational interaction between a body on the Earth's surface and the Earth.
Mathematical Derivation
Step 1: Gravitational Force
According to Newton’s Law of Universal Gravitation, the gravitational force of attraction ($F$) between the Earth and the body is given by:
$F = G \frac{M_e m}{R_e^2}$ --- (i)
Step 2: Weight of the Body
We know that the gravitational force of attraction exerted by the Earth on a body is equal to the weight ($W$) of that body:
$W = m g$ --- (ii)
Step 3: Comparing Forces
Equating equation (i) and equation (ii) (since $F = W$):
$m g = G \frac{M_e m}{R_e^2}$
Canceling the mass of the body '$m$' from both sides:
$g = G \frac{M_e}{R_e^2}$
Rearranging the equation to solve for the mass of the Earth ($M_e$):
$M_e = \frac{g R_e^2}{G}$
Calculation of Earth's Mass
From standard astronomical data:
- Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$
- Radius of the Earth, $R_e = 6.38 \times 10^6 \, \text{m}$
- Universal gravitational constant, $G = 6.67 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2$
Substituting these values into the derived formula:
$M_e = \frac{(9.8 \, \text{m/s}^2) \times (6.38 \times 10^6 \, \text{m})^2}{6.67 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2}$
$M_e = \frac{9.8 \times (40.7044 \times 10^{12})}{6.67 \times 10^{-11}}$
$M_e \approx 6.0 \times 10^{24} \, \text{kg}$
Conclusion: Thus, the calculated mass of the Earth is approximately $6.0 \times 10^{24} \, \text{kg}$.
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