Problem 30 - University Physics with Modern Physics

Question

A uniform, spherical, \(1000.0-\mathrm{kg}\) shell has a radius of 5.00 \(\mathrm{m} .\) (a) Find the gravitational force this shell exerts on a \(2.00-\mathrm{kg}\) point mass placed at the following distances from the center of the shell: (i) \(5.01 \mathrm{m},\) (ii) \(4.99 \mathrm{m},\) (iii) 2.72 \(\mathrm{m}\) . (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty.\)

Step-by-Step Solution

Answer

(i) 5.312 × 10^-8 N, (ii) 0 N, (iii) 0 N. Force is zero inside, decreases outside.

Step 1: Understanding the Shell's Gravitational Field

According to the shell theorem, a spherically symmetric shell exerts no net gravitational force on a point mass inside the shell. For points outside the shell, the gravitational force is the same as if all the shell's mass were concentrated at its center.

Step 2: Apply Gravitational Force Formula (Outside the Shell)

The gravitational force for a point mass outside the shell is given by Newton's Law of Universal Gravitation: \( F = \frac{G M m}{r^2} \) where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( M \) is the mass of the shell \( 1000 \, \text{kg} \), \( m \) is the mass of the point \( 2 \, \text{kg} \), and \( r \) is the distance from the center of the shell.

Step 3: Calculate Force at r = 5.01 m

Since 5.01 m is outside the shell, use the formula:\[ F = \frac{6.674 \times 10^{-11} \times 1000 \times 2}{(5.01)^2} \]Calculating gives:\[ F \approx 5.312 \times 10^{-8} \, \text{N} \]

Step 4: Calculate Force at r = 4.99 m

Since 4.99 m is inside the shell, the shell theorem tells us that the gravitational force is zero inside the shell, so:\[ F = 0 \, \text{N} \]

Step 5: Calculate Force at r = 2.72 m

Similarly, because 2.72 m is also inside the shell, the gravitational force is zero. Thus:\[ F = 0 \, \text{N} \]

Step 6: Sketch Force vs Distance Graph

For \( r < 5.00 \, \text{m} \), the force is zero. For \( r \geq 5.00 \, \text{m} \), the force follows the inverse square law, \( F \propto \frac{1}{r^2} \). The graph is a horizontal line at \( F = 0 \, \text{N} \) for \( r < 5.00 \, \text{m} \) and decreases following \( \frac{1}{r^2} \) for \( r > 5.00 \, \text{m} \).

Key Concepts

Shell Theorem

The Shell Theorem is a crucial principle in gravitational physics. It provides insights into how spherical shells of matter exert gravitational forces. Here's the simple break down:
- **Outside the shell:** According to the Shell Theorem, if you are outside a spherical shell with uniform mass distribution, the shell behaves as if all its mass is concentrated at its center. This means the gravitational field outside is similar to that of a point mass of the same total mass located at the center of the sphere.
- **Inside the shell:** Quite fascinatingly, if you are inside the shell, the theorem states that the gravitational forces cancel out, resulting in a net gravitational force of zero. This happens because the gravitational pull from the mass distributed around you is evenly balanced in all directions, thereby nullifying any net force.
This concept helps in simplifying complex gravitational calculations when dealing with large spherical bodies like planets or stars. For any point mass located inside a spherical shell, the gravitational force it experiences is precisely zero. This fundamental insight shapes our understanding of celestial mechanics.

Gravitational Force Calculation

Calculating gravitational force is central to understanding interactions between masses. To find this force, we use Newton's Law of Universal Gravitation. The formula is very straightforward: \[ F = \frac{G M m}{r^2} \]- **F** is the gravitational force between two masses.- **G** is the gravitational constant: approximately \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\).- **M** and **m** are the masses of the two objects interacting.- **r** is the distance between the centers of the two masses.
For example, in our exercise where a 2 kg point mass is influenced by a 1000 kg shell, if the point mass is outside the shell, as at 5.01 m, we substitute these values into the formula to find the gravitational force. This calculation includes squaring the distance \((r^2)\), which is crucial as it follows the inverse square law. Hence, gravitational force decreases with the square of the distance, making distance a vital factor in gravitational interactions.

Inverse Square Law

The inverse square law is a principle that articulates how certain physical quantities diminish with distance. In the context of gravity, this law states that the gravitational force between two objects is inversely proportional to the square of the distance between their centers. Simply put: - If you double the distance between two masses, the gravitational force becomes one-fourth as strong.- Tripling the distance reduces the force to one-ninth.
This inverse relationship is expressed in the formula used to calculate gravitational force:\[ F = \frac{G M m}{r^2} \]The term \(r^2\) in the denominator manifests the inverse square dependence. This concept is vital in astronomy and physics, helping us understand how celestial bodies interact over vast distances. The inverse square law not only applies to gravity but also to other fields like electromagnetism and light propagation, forming a foundational element of physical science.