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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 \(r \rightarrow \infty.\)

Short Answer

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

Step by step solution

01

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.
02

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.
03

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:\[ F \approx 5.312 \times 10^{-8} \, \text{N} \]
04

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:\[ F = 0 \, \text{N} \]
05

Calculate Force at r = 2.72 m

Similarly, because 2.72 m is also inside the shell, the gravitational force is zero. Thus:\[ F = 0 \, \text{N} \]
06

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 \( r > 5.00 \, \text{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Kirkwood Gaps. Hundreds of thousands of asteroids orbit the sun within the asteroid belt, which extends from about \(3 \times 10^{8} \mathrm{km}\) to about \(5 \times 10^{8} \mathrm{km}\) from the sun. (a) Find the orbital period (in years) of (i) an asteroid at the inside of the belt and (ii) an asteroid at the outside of the belt. Assume circular orbits. (b) In 1867 the American astronomer Daniel Kirkwood pointed out that several gaps exist in the asteroid belt where relatively few asteroids are found. It is now understood that these Kirkwood gaps are caused by the gravitational attraction of Jupiter, the largest planet, which orbits the sun once every 11.86 years. As an example, if an asteroid has an orbital period half that of Jupiter, or 5.93 years, on every other orbit this asteroid would be at its closest to Jupiter and feel a strong attraction toward the planet. This attraction, acting over and over on successive orbits, could sweep asteroids out of the Kirkwood gap. Use this hypothesis to determine the orbital radius for this Kirkwood gap. (c) One of several other Kirkwood gaps appears at a distance from the sun where the orbital period is 0.400 that of Jupiter. Explain why this happens, and find the orbital radius for this Kirkwood gap.

What is the escape speed from a 300-km-diameter asteroid with a density of 2500 \(\mathrm{kg} / \mathrm{m}^{3} ?\)

The acceleration due to gravity at the north pole of Neptune is approximately 10.7 \(\mathrm{m} / \mathrm{s}^{2} .\) Neptune has mass \(1.0 \times 10^{26} \mathrm{kg}\) and radius \(2.5 \times 10^{4} \mathrm{km}\) and rotates once around its axis in about 16 \(\mathrm{h}\) . (a) What is the gravitational force on a 5.0 -kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

The star Rho \(^{1}\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho \(^{1}\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho \(^{1}\) Cancri?

At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8 ). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 \(\mathrm{km} / \mathrm{s} .\) (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzchild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

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