Question

A uniform, spherical, 1000.0-kg shell has a radius of 5.00 m. (a) Find the gravitational force this shell exerts on a 2.00-kg point mass placed at the following distances from the center of the shell: (i) 5.01 m, (ii) 4.99 m, (iii) 2.72 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 \(r\) \(\rightarrow\) \(\infty\).

Short answer

(i) \(5.30 \times 10^{-9}\) N, (ii) 0 N, (iii) 0 N. The force is zero inside the shell and decreases as 1/\(r^2\) outside.

Step-by-step solution

Understanding the Problem

We need to calculate the gravitational force between a uniform spherical shell and a point mass at varying distances. The shell's mass \(M\) is 1000.0 kg, and the point mass \(m\) is 2.00 kg. The shell's radius is 5.00 m. According to the shell theorem, if a point mass is outside a spherical shell, the shell can be treated as a point mass at its center. If it is inside, the gravitational force is zero.

Applying Newton's Gravity Law for Case (i)

For the point mass at a distance \(r = 5.01\) m, which is outside the shell, use the formula:\[F = \frac{G M m}{r^2}\]where \(G = 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2\). Substitute \(M = 1000.0\) kg, \(m = 2.00\) kg, and \(r = 5.01\) m.

Calculation for 5.01 m

Substitute the values into the gravitational force formula:\[F = \frac{(6.674 \times 10^{-11})(1000)(2.00)}{(5.01)^2}\]Calculate to find \(F \approx 5.30 \times 10^{-9}\) N.

Applying Shell Theorem for Case (ii)

For the point mass at 4.99 m, which is inside the shell, according to the shell theorem, the gravitational force is zero because the mass of the shell does not exert a net gravitational force within itself.

Applying Shell Theorem for Case (iii)

For the point mass at 2.72 m, which is also inside the shell, the gravitational force is again zero, per the shell theorem, as there is no net gravitational force exerted within the shell.

Sketching the Graph

Plot a graph where the x-axis represents the distance \(r\) from the center of the sphere, and the y-axis represents gravitational force \(F\). From \(r = 0\) to \(r = 5.00\) m, \(F = 0\). From \(r > 5.00\) m, \(F\) follows the curve \(F = \frac{G M m}{r^2}\), decreasing as \(r\) increases.

Key concepts

Gravitational Force

Gravitational force is a fundamental force that acts between two masses, attracting them towards each other. This force is described by Newton's law of universal gravitation, which we will discuss in detail later on. The key idea is that the gravitational force is always attractive and depends on both the masses involved and the distance between them.

In our original exercise, the gravitational force is calculated between a spherical shell and a point mass. The formula used to find this force is given by:

• Force (\( F \)) is proportional to the product of the two masses, the shell (\( M \)) and the point mass (\( m \)).
• This force is inversely proportional to the square of the distance (\( r \)) between the center of the spherical shell and the point mass.

It's important to note that gravitational force diminishes rapidly as the distance between the two masses increases, illustrating the inverse square law nature of this force.

Spherical Shell

A spherical shell is a three-dimensional shape that is hollow and defined by two radii, one outer and one possibly inner if it's a thick shell. In physics, particularly in relation to the gravitational force, spherical shells hold special significance due to the shell theorem derived from Newtonian gravity.

The shell theorem states:

• Outside the shell: When a point mass is located outside a spherical shell, the shell’s gravitational force acts as though all its mass were concentrated at its center.
• Inside the shell: If the point mass is inside the spherical shell, there is no net gravitational force exerted by the shell, regardless of the internal structure of the shell.

This property greatly simplifies calculations as it allows us to treat the shell as a point mass for external forces, providing that neat simplification evident in the given exercise solutions.

Newton's Law of Universal Gravitation

Newton's law of universal gravitation is a pivotal theory in physics that describes the attractive force between two bodies. It states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of this force is given by the formula:\[F = \frac{G M m}{r^2}\]Here:

• \( F \) is the gravitational force between the two masses.
• \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)), a constant of nature that quantifies the strength of gravity.
• \( M \) and \( m \) are the masses of the two objects.
• \( r \) is the distance between the center of the two masses.

This groundbreaking formula shows how gravitational force is universal across the cosmos, applying not just to celestial bodies like the moon and Earth but also to the interactions between everyday objects. It illustrates the direct proportionality to the product of the masses and the inverse square relationship with distance, a feature fundamental to understanding gravitational interactions.