Q.Weighing the Earth: You are given the following data: $g = 9.81\text{ m s}^{-2}$, $R_E = 6.37 \times 10^{6}\text{ m}$, the distance to the moon $R = 3.84 \times 10^{8}\text{ m}$ 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.

To obtain the mass of the Earth ($M_E$), we apply Newton's Law of Universal Gravitation in two distinct scenarios. We use the universal gravitational constant $G = 6.674 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$.

Method 1: Using acceleration due to gravity ($g$) at Earth's surface

Concept: Newton's Law of Universal Gravitation and the definition of acceleration due to gravity.

1. The gravitational force on a mass $m$ at Earth's surface is $F = \frac{G M_E m}{R_E^2}$.
2. This force is also expressed as $F = mg$.
3. Equating these gives $g = \frac{G M_E}{R_E^2}$.
4. Solving for $M_E$: $M_E = \frac{g R_E^2}{G} = \frac{9.81\text{ m s}^{-2} \times (6.37 \times 10^{6}\text{ m})^2}{6.674 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}} = 5.96 \times 10^{24}\text{ kg}$.

Method 2: Using the Moon's orbital period ($T$)

Concept: Newton's Law of Universal Gravitation applied to orbital motion (Kepler's Third Law).

1. The gravitational force between Earth and Moon provides the centripetal force for the Moon's orbit: $\frac{G M_E M_M}{R^2} = \frac{M_M v^2}{R}$.
2. The Moon's orbital speed $v$ is related to its period $T$ by $v = \frac{2 \pi R}{T}$. …

To determine the mass of the Earth, we can leverage Newton's Law of Universal Gravitation in two distinct scenarios. The core idea is that the gravitational force exerted by the Earth is responsible for both the acceleration of objects near its surface and the centripetal force that keeps the Moon in orbit. By quantifying these effects and knowing other relevant parameters, we can isolate and calculate the Earth's mass.

We will need the universal gravitational constant, $G = 6.674 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$.

Method 1: Using the acceleration due to gravity ($g$) at Earth's surface

Concept and Intuition:

Any object near the Earth's surface experiences an acceleration due to gravity, $g$. This acceleration is a direct consequence of the gravitational force exerted by the Earth. According to Newton's Law of Universal Gravitation, the force between the Earth (mass $M_E$) and an object (mass $m$) at its surface (distance $R_E$ from the center) is $F_g = \frac{G M_E m}{R_E^2}$. From Newton's second law, this force also equals $mg$. By equating these two expressions for the gravitational force, the mass of the object $m$ cancels out, allowing us to find $M_E$.

1. Relate gravitational force to acceleration due to gravity: The gravitational force $F_g$ exerted by the Earth on an object of mass $m$ at its surface is given by Newton's Law of Universal Gravitation:

This force also causes the object to accelerate downwards with acceleration $g$. By Newton's second law, $F_g = mg$.

2. Equate the expressions for force and solve for $M_E$:

Setting the two expressions for $F_g$ equal:

Notice that the mass of the object, $m$, cancels out from both sides. This shows that $g$ is independent of the object's mass.

Rearranging this equation to solve for $M_E$:

> [!FORMULA]
> The mass of the Earth can be found using the acceleration due to gravity:
> $$M_E = \frac{g R_E^2}{G}$$

3. Substitute the given values and calculate:

Given:

* $g = 9.81\text{ m s}^{-2}$

* $R_E = 6.37 \times 10^{6}\text{ m}$

* $G = 6.674 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$

Method 2: Using the Moon's orbital data

Concept and Intuition:

The Moon orbits the Earth because of the gravitational force between them. This gravitational force acts as the centripetal force required to keep the Moon in its nearly circular orbit. By equating the gravitational force between the Earth and Moon to the centripetal force needed for the Moon's orbit, we can derive an expression for the Earth's mass. This approach is essentially a direct application of Kepler's Third Law, which can be derived from Newton's laws.

1. Identify the forces involved: The gravitational force $F_g$ between the Earth (mass $M_E$) and the Moon (mass $m_M$) is given by:

where $R$ is the distance between the Earth and the Moon.

This gravitational force provides the centripetal force $F_c$ required to keep the Moon in its orbit. For an object moving in a circle, the centripetal force is:

where $v$ is the orbital speed of the Moon.

2. Express orbital speed in terms of orbital period:

The Moon completes one revolution (a distance of $2\pi R$) in a time period $T$. So, its orbital speed $v$ is:

Substitute this into the centripetal force equation:

3. Equate gravitational and centripetal forces and solve for $M_E$: Setting $F_g = F_c$: …

Step 1 (Method 1): At Earth's surface, the gravitational force on a test mass $m$ equals $mg$, and by Newton's law of gravitation it also equals $\dfrac{GM_Em}{R_E^2}$. Equate and cancel $m$: $g=\dfrac{GM_E}{R_E^2}\Rightarrow M_E=\dfrac{gR_E^2}{G}$.

Step 2: Substitute $g=9.81$, $R_E=6.37\times10^6$ m, $G=6.674\times10^{-11}$ to get $M_E\approx5.96\times10^{24}$ kg.

Step 3 (Method 2): The Moon orbits under Earth's gravity, so the gravitational force provides the centripetal force: $\dfrac{GM_Em_M}{R^2}=\dfrac{m_Mv^2}{R}$, with $v=\dfrac{2\pi R}{T}$.

Step 4: Substituting and cancelling $m_M$ gives $M_E=\dfrac{4\pi^2R^3}{GT^2}$. …