Mass and Density of Earth

1. Concept Overview

The mass and density of Earth are fundamental parameters in understanding gravitational attraction, planetary motion, and satellite dynamics.
By combining Newton’s Law of Gravitation with the concept of gravitational acceleration, we can derive the Earth’s mass and determine its average density.

2. Explanation and Mathematical Derivation

Let:

• Mass of Earth = [M]
• Radius of Earth = [R]
• Acceleration due to gravity on the surface = [g]
• Gravitational constant = [G]

According to Newton’s Law of Gravitation, the gravitational force exerted by the Earth on an object of mass [m] at its surface is:

[ F = \dfrac{G M m}{R^2} ]

But, the same force can also be expressed as:

[ F = m g ]

Equating both expressions of force:

[
m g = \dfrac{G M m}{R^2}
]

Canceling [m] from both sides:

[
g = \dfrac{G M}{R^2}
]

From this, we can find the mass of the Earth as:

[
M = \dfrac{g R^2}{G}
]

$ \displaystyle \because G=6.67\times {{10}^{{-11}}}\text{ N}{{\text{m}}^{\text{2}}}\text{k}{{\text{g}}^{{\text{-2}}}}$ and $\displaystyle g=9.8\text{ m}{{\text{s}}^{{\text{-2}}}}$ and $\displaystyle R=6.4\times {{10}^{5}}\text{ m}$

$\displaystyle \begin{array}{l}\therefore M=\dfrac{{9.8\times {{{\left( {6.4\times {{{10}}^{5}}} \right)}}^{2}}}}{{6.67\times {{{10}}^{{-11}}}}}\\\Rightarrow M=6.018\times {{10}^{{24}}}\text{ kg}\end{array}$

Now, the density of Earth ([\rho]) is defined as:

[
\rho = \dfrac{\text{Mass}}{\text{Volume}} = \dfrac{M}{\dfrac{4}{3} \pi R^3}
]

Substituting the value of [M]:

[
\rho = \dfrac{\dfrac{g R^2}{G}}{\dfrac{4}{3} \pi R^3} = \dfrac{3 g}{4 \pi G R}
]

Thus, the average density of Earth is:

[
\rho = \dfrac{3 g}{4 \pi G R}
]

3. Dimensions and Units

4. Key Features

1. The mass of the Earth is determined indirectly using [g], [R], and [G].
2. The value of [g] varies slightly across the surface due to rotation and shape of Earth.
3. Earth is not a perfect sphere — its density is not uniform. It increases toward the core.
4. Average density gives an idea of Earth’s internal structure (mantle, core, crust).
5. Density is independent of the object’s mass but depends on the Earth’s radius and gravity.

5. Important Formulas to Remember

6. Conceptual Questions with Solutions

7. FAQ / Common Misconceptions

8. Practice Questions (with Step-by-Step Solutions)

Q1. Find the mass of the Earth given [g = 9.8 m/s^2], [R = 6.37 \times 10^6 m], and [G = 6.67 \times 10^{-11} N·m^2/kg^2].

Solution:
[M = \dfrac{g R^2}{G}] [= \dfrac{9.8 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}}] [= 5.97 \times 10^{24} kg]

Q2. Calculate the mean density of the Earth using the same data.

Solution:
[\rho = \dfrac{3 g}{4 \pi G R}] [= \dfrac{3 \times 9.8}{4 \pi \times 6.67 \times 10^{-11} \times 6.37 \times 10^6}] [= 5.5 g/cm^3]

Q3. If Earth’s radius were doubled but mass remained the same, what would happen to density?

Solution:
[\rho’ = \dfrac{M}{\dfrac{4}{3} \pi (2R)^3}] [= \dfrac{\rho}{8}]
Hence, density becomes one-eighth.

Q4. Derive the expression for [\rho] in terms of [M] and [R].

Solution:
[\rho = \dfrac{M}{\dfrac{4}{3} \pi R^3}] [= \dfrac{3M}{4 \pi R^3}]

Q5. If [R = 6.4 \times 10^6 m], [M = 5.98 \times 10^{24} kg], find [g] on the surface.

Solution:
[g = \dfrac{G M}{R^2}] [= \dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(6.4 \times 10^6)^2}] [= 9.8 m/s^2]