Problem 33 - University Physics with Modern Physics

Question

A uniform, solid, 1000.0-kg sphere has a radius of 5.00 m. (a) Find the gravitational force this sphere exerts on a 2.00-kg point mass placed at the following distances from the center of the sphere: (i) 5.01 m, (ii) 2.50 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) 5.299 x 10^-9 N, (ii) 1.335 x 10^-8 N.

Step 1: Apply Gravitational Force Formula

The gravitational force between two masses is given by the formula:\[F = \frac{G \cdot M \cdot m}{r^2}\]where \(F\) is the gravitational force, \(G\) is the universal gravitational constant \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\), \(M\) is the mass of the sphere, \(m\) is the mass of the point, and \(r\) is the distance between the center of the sphere and the point mass.

Step 2: Calculate Force when r=5.01 m

Since 5.01 m is greater than the radius of the sphere (5.00 m), treat the sphere as a point source with mass concentrated at its center. Apply the formula:\[F = \frac{(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2) \times (1000.0 \text{ kg}) \times (2.00 \text{ kg})}{(5.01 \text{ m})^2}\]Compute:\[F \approx 5.299 \times 10^{-9} \text{ N}\]

Step 3: Calculate Force within the Sphere, r=2.50 m

When the mass is inside the sphere (2.50 m), only the mass within the sphere up to that radius contributes to the gravitational force. Use the formula:\[F = \frac{G \cdot M' \cdot m}{r^2}\]where \(M' = \frac{4}{3}\pi r^3 \rho\) and \(\rho = \frac{M}{\frac{4}{3}\pi R^3}\). Solving:\[M' = \frac{(2.50)^3}{5.00^3} \cdot 1000 \approx 125\text{ kg}\]Compute the force with:\[F \approx \frac{(6.674 \times 10^{-11}) \times 125 \times 2.00}{2.50^2} \approx 1.335 \times 10^{-8} \text{ N}\]

Step 4: Sketch Gravitational Force vs Distance

The gravitational force inside the sphere increases linearly with distance from the center up to the surface, due to the increasing mass. Beyond the surface, it decreases inversely proportional to the square of the distance. The plot has a slope for \(r < R\) and follows an inverse-square law for \(r \geq R\).

Key Concepts

Universal Gravitational Constant

The concept of the universal gravitational constant, denoted as \(G\), is fundamental to describing the gravitational force between two masses. This constant acts as the proportionality factor in Newton's Law of Universal Gravitation, which mathematically is expressed as \[F = \frac{G \cdot M \cdot m}{r^2}\]. Here:

• \(F\) represents the gravitational force.
• \(M\) and \(m\) are the masses of two objects interacting gravitationally.
• \(r\) is the distance between the centers of the two masses.
• \(G\) is approximately \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\).

This constant is remarkably small, indicating that gravitational forces are weak compared to other fundamental forces, such as electromagnetic forces. Despite its weakness, gravity's effect becomes significant due to the large masses of celestial bodies, like planets and stars. Its presence ensures that gravitational calculations remain consistent, reliable, and universally applicable.
The use of \(G\) allows for precise predictions of gravitational interactions whether on Earth or in space, emphasizing its universal nature.

Inverse-Square Law

The inverse-square law is a critical concept underpinning many physical phenomena, including gravity. It describes how a particular quantity, such as gravitational force, diminishes with the square of the distance from the source of that quantity. In simpler terms, according to this law, if you double the distance from a mass, the gravitational force you experience is quartered, not halved.For gravitational force, this law is written as \[F \propto \frac{1}{r^2}\], which shows how force and distance \(r\) are related inversely through the square of \(r\). This relationship means that as you move further away from a massive object, the gravitational pull it exerts on you decreases rapidly.
The inverse-square law plays a vital role in determining how gravitational interactions weaken over distance. For students, understanding this concept helps in visualizing and calculating how forces change when objects are moved closer or further apart in space, simplifying complex concepts like orbital mechanics and satellite paths.

Mass Distribution Inside a Sphere

Understanding how mass distribution affects gravitational force is essential, especially when dealing with problems like calculating the gravitational pull within and outside a sphere. For a sphere with uniform mass distribution, the following principles apply:

• When outside the sphere, the entire mass can be considered concentrated at its center. This simplification allows the use of the formula \(F = \frac{G \cdot M \cdot m}{r^2}\).
• Inside the sphere, only the mass within the radius you're considering affects the gravitational force. This is due to the shell theorem, which suggests that the net gravitational force inside a hollow sphere is zero.

The mass contributing to the force inside is given by \[M' = \rho \cdot \frac{4}{3}\pi r^3\], where \(\rho\) is the density. The force experienced can be calculated using the effective mass up to the point of interest.
This concept is key to understanding variations in gravitational strength from the center to the edge of a sphere. When working with gravitational problems, knowing how to adjust calculations based on whether your point of interest is inside or outside a massive object like a planet can greatly aid in accurate predictions.