91Ó°ÊÓ

Newton's law of gravitation states that every point mass attracts every other point mass with a force along the line intersecting both points. This force is proportional to the product of their masses and inversely proportional to the square of the distance between them. In formula form, it is written as: \( F = G \frac{m_1 m_2}{r^2} \) where:

• \( F \) is the gravitational force between the masses
• \( G \) is the gravitational constant \( (6.67430 \times 10^{-11} \; \text{N} \; \text{m}^2 / \text{kg}^2) \)
• \( m_1 \) and \( m_2 \) are the masses of the two objects
• \( r \) is the distance between the centers of the two masses

In the context of our exercise, we consider two masses: the core mass \( M \) and a particle of small mass. The distance \( r \) is equal to the radius \( R \). Using Newton's law, we find the gravitational force and then derive the gravitational acceleration by dividing the force by the mass of the particle. The simplified version for gravitational acceleration \( a \) at the surface becomes: \( a = \frac{GM}{R^2} \).
This formula allows us to calculate the gravitational acceleration acting on any object at the surface of the core.

The shell theorem is an important concept in gravitational physics. It states two key points about a spherical shell of uniform density:

• Outside the shell, the gravitational force at any point can be calculated as though all the mass of the shell were concentrated at its center.
• Inside the shell, the force is zero as long as the point is not in the mass of the shell.

In our exercise, the planet has an outer shell with inner radius \( R \) and outer radius \( 2R \), and we consider points both at the surface of the core \( ( R ) \) and at a point \( ( 3R ) \) from the center.
At \( R \), we are at the surface of the core, so only the core's mass contributes to the gravitational force. At \( 3R \), we consider the total mass of the core plus the entire shell mass as the particle is outside the shell. Hence, we compute the gravitational force using the combined mass: \( M_{total} = M + 4M = 5M \). Then using Newton's law of gravitation, the gravitational acceleration at distance \( 3R \) can be determined by: \( a = \frac{GM_{total}}{(3R)^2} \).

Gravitational force is the attractive force acting between any two masses. It is one of the four fundamental forces of nature. The magnitude of this force is calculated by Newton's law of gravitation. At the macroscopic level, although gravity is the weakest force, it is essential for maintaining the structure of the universe.
In the context of our practice problem, we have two points of interest to find the gravitational force and subsequently the gravitational acceleration:

• At radius \( R \): The particle feels a force due to the core's mass \( M \). The acceleration \( a \) can be derived using the formula: \( a = \frac{GM}{R^2} \).
• At radius \( 3R \): The particle feels a combined force due to both the core and the outer shell. According to the shell theorem, the mass of the shell can be considered concentrated at the center, giving a total mass of \( 5M \). The resulting acceleration \( a \) can be determined using: \( a = \frac{GM_{total}}{(3R)^2} \).

Calculating these allows us to understand how the gravitational force and acceleration differ at varying distances from the center of the planet.