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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) \(5.01 \mathrm{m},(\mathrm{ii}) 4.99 \mathrm{m},\) (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) \(5.32 \times 10^{-9} \ \mathrm{N}\), (ii) and (iii) \(0 \ \mathrm{N} \). Graph force versus distance equals zero at and within radius, decreases after.

Step by step solution

01

Understanding the Problem

We need to calculate the gravitational force a spherical shell exerts on a point mass at different distances. The key concept here is that the gravitational force of a spherical shell on an object outside of it can be calculated as if all the shell's mass were concentrated at its center, but inside it, the force is zero.
02

Calculate Force for Distance Greater than Radius

For the distance of 5.01 m (outside the shell), the force can be calculated using Newton's law of gravitation: \[ F = G \frac{M \cdot m}{r^2} \]where \( G = 6.674 \times 10^{-11} \ \mathrm{N}\, \mathrm{m}^2\, \mathrm{kg}^{-2} \), \( M = 1000.0 \ \mathrm{kg}\), \( m = 2.00 \ \mathrm{kg} \), and \( r = 5.01 \ \mathrm{m} \). Thus, \[ F = 6.674 \times 10^{-11} \frac{1000 \times 2}{(5.01)^2} \].
03

Evaluate the Formula for Outside Shell

Calculate the force: \[ F = 6.674 \times 10^{-11} \frac{2000}{25.1001} = 5.32 \times 10^{-9} \ \mathrm{N} \]. This is the gravitational force for distance 5.01 m.
04

Calculate Force for Distance Less than or Equal to Radius

For distances less than or equal to the radius (4.99 m and 2.72 m), the gravitational force is zero because all the mass is outside the point mass. According to the shell theorem, the gravitational force inside a hollow spherical shell is zero.
05

Graph Setup

For the graph, the y-axis will represent the gravitational force and the x-axis will represent the distance from the center of the sphere. The key points will be:- At \( r = 5.00 \ \mathrm{m} \) (the shell's surface), the force begins to drop from a max value. - For \( r < 5.00 \ \mathrm{m} \), the force is zero. - For \( r > 5.00 \ \mathrm{m} \), it decreases with \(1/r^2\).
06

Conclusion

Summarizing the calculations: 1. At 5.01 m: \( F = 5.32 \times 10^{-9} \ \mathrm{N} \).2. At 4.99 m and 2.72 m: \( F = 0 \ \mathrm{N} \).The graph will show zero force from \( r = 0 \) to \( r = 5.00 \ \mathrm{m} \) and then decrease inversely with the square of the distance beyond the surface.

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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 was introduced by Isaac Newton and refers to the gravitational force exerted by a spherical object on another object. An important aspect is how gravity works inside and outside a spherical shell. Although this may seem complex, the theorem provides a very straightforward solution.

According to the Shell Theorem:
  • The gravitational force exerted by a uniform spherical shell on a point mass located outside the shell is as if all the mass of the shell were concentrated at the shell's center. This means that if you're outside the shell, the entire mass acts as though it's at a single point right at the center, simplifying calculations greatly.
  • Inside the shell, surprisingly, the gravitational force is zero. This happens because the mass of the shell pulls equally in all directions. The opposing gravitational forces cancel each other out, resulting in a net force of zero.
This theorem is incredibly useful in physics, not just for theoretical purposes but also for practical calculations in gravitational fields of spherical objects like planets or stars. It helps us calculate forces easily without needing to account for each tiny particle within the sphere.
Newton's Law of Gravitation
Newton's Law of Gravitation is fundamental in understanding how objects interact with each other via gravity. This law explains the force that attracts two bodies towards each other.

The formula for the gravitational force between two masses is given by: \[ F = G \frac{M \cdot m}{r^2} \]
  • Here, \(F\) is the gravitational force between two objects.
  • \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \, \mathrm{N}\, \mathrm{m}^2\, \mathrm{kg}^{-2}\).
  • \(M\) and \(m\) are the masses of the two objects, and \(r\) is the distance between their centers.
Newton’s law reveals that the force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.

This means that as the distance between the objects increases, the gravitational force decreases rapidly. Conversely, larger masses exert stronger gravitational forces. It’s a beautifully simple yet powerful law that describes why and how everything in the universe is drawn towards each other.
Spherical Shell
A spherical shell is a common model used in physics to simplify gravitational problems. It represents a thin layer of mass distributed evenly over the surface of a sphere.

This model is notably used in problems where we need to calculate the force of gravity around hollow objects. Thanks to the Shell Theorem, calculations involving spherical shells become a lot easier.
  • Outside the shell: The effects of the shell's gravity act as if the entire mass were concentrated at a single point in the center. This allows us to treat the spherical shell just like a point mass when calculating the gravitational force at distances greater than the radius of the shell.
  • Inside the shell: No matter where you are within this hollow shell, the gravitational pull from all sides cancels out and results in zero net force at any point inside. This might seem counterintuitive at first, but it's a direct consequence of the symmetrical distribution of mass.
Understanding spherical shells is crucial for any student dealing with astronomical calculations or studying the basics of gravitational physics.

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Most popular questions from this chapter

At a certain instant, the earth, the moon, and a stationary \(1250-\mathrm{kg}\) spacecraft lie at the vertices of an equilateral triangle whose sides are \(3.84 \times 10^{5} \mathrm{km}\) in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{100}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice. \()\)

(a) Calculate how much work is required to launch a spacecraft of mass \(m\) from the surface of the earth (mass \(m_{\mathrm{E}},\) radius \(R_{\mathrm{E}} )\) and place it in a circular low earth orbit - that is, an orbit whose altitude above the earth's surface is much less than \(R_{\mathrm{E}}\) . (As an example, the International Space Station is in low earth orbit at an altitude of about \(400 \mathrm{km},\) much less than \(R_{\mathrm{E}}=6380 \mathrm{km} .\) You can ignore the kimetic energy that the spacecraft has on the ground due to the earth's rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement: "In terms of energy, low earth orbit is halfway to the edge of the universe."

Your starship, the Aimless Wanderer, lands on the mysteri- ous planet Mongo. As chief scientist-engineer, you make the following measurements: A \(2.50-\mathrm{kg}\) stone thrown upward from the ground at 12.0 \(\mathrm{m} / \mathrm{s}\) returns to the ground in \(8.00 \mathrm{s} ;\) the circumference of Mongo at the equator is \(2.00 \times 10^{5} \mathrm{km}\) ; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the Aimless Wanderer goes into a circular orbit \(30,000 \mathrm{km}\) above the surface of Mongo, how many hours will it take the ship to complete one orbit?

A particle of mass 3\(m\) is located 1.00 \(\mathrm{m}\) from a particle of mass \(m .\) (a) Where should you put a third mass \(M\) so that the net gravitational force on \(M\) due to the two masses is exactly zero? (b) Is the equilibrium of \(M\) at this point stable or unstable (i) for points along the line connecting \(m\) and \(3 m,\) and (ii) for points along the line passing through \(M\) and perpendicular to the line connecting \(m\) and 3\(m ?\)

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