Problem 45
Question
A sphere of density \(\rho\) and radius \(a\) has a concentric cavity of radius
\(b\) as shown in the Fig. \(7.13\).
Gravitational potential energy as a function of \(r\), where \(r\) is the distance
from the centre of the sphere. When \(0
Step-by-Step Solution
Verified Answer
The correct answer is (B). With the considerations and calculations taking into account a sphere of radius \(a\) and a concentric cavity of radius \(b\), the gravitational potential energy as a function of distance from the center of the sphere when \(0
1Step 1: Understand gravitational potential energy
The gravitational potential energy \(U\) is the work done against the gravitational force to bring a mass \(m\) from infinity to a point \(r\). In this case, the potential energy assuming an uniform density sphere without a cavity would be \(U = -\frac{4\pi G\rho m r^2}{3}\).
2Step 2: Consider the Sphere with a Cavity
When considering the sphere with the cavity, we can conceptualize this as the difference in potential energy between two spheres: a large sphere of radius \(a\) and a smaller sphere of radius \(b\) which represents the cavity. So the new potential energy function will look like this: \(U(r) = -\frac{4\pi G\rho m a^2}{3} + \frac{4\pi G\rho m b^2}{3}\).
3Step 3: Apply Given Conditions
The above expression is applicable if \(b
Key Concepts
Uniform Density SphereGravitational ForceSpherical Cavity
Uniform Density Sphere
Imagine a ball perfectly shaped with every portion having exactly the same amount of material—this is what we call a Uniform Density Sphere. In mathematical terms, if a sphere has a density \rho, it means that for every cubic centimeter, gram, or any unit of volume, we will find the same amount of mass throughout the sphere.
Why is this important in physics, especially in gravitational studies? Because a sphere with uniform density generates a gravitational field that can be predicted and calculated with known equations. A real-life example might be certain planets or moons which can be approximately considered to have uniform density for some calculations. The notion simplifies complex problems, such as finding the gravitational potential energy anywhere in or outside the sphere.
Gravitational Potential Energy in Uniform Density Spheres
The gravitational potential energy in a uniform density sphere is affected by its radius and the mass of the object being influenced by gravity. Without any cavities or empty spaces, a uniform sphere's potential energy can be expressed with the formula \( U = -\frac{4\pi G\rho m r^2}{3} \) where \( m \) is the mass of the object, \( G \) is the gravitational constant, \( r \) is the distance from the center of the sphere, and \( \rho \) is the uniform density of the sphere. This formula makes certain gravitational calculations much more straightforward.
Why is this important in physics, especially in gravitational studies? Because a sphere with uniform density generates a gravitational field that can be predicted and calculated with known equations. A real-life example might be certain planets or moons which can be approximately considered to have uniform density for some calculations. The notion simplifies complex problems, such as finding the gravitational potential energy anywhere in or outside the sphere.
Gravitational Potential Energy in Uniform Density Spheres
The gravitational potential energy in a uniform density sphere is affected by its radius and the mass of the object being influenced by gravity. Without any cavities or empty spaces, a uniform sphere's potential energy can be expressed with the formula \( U = -\frac{4\pi G\rho m r^2}{3} \) where \( m \) is the mass of the object, \( G \) is the gravitational constant, \( r \) is the distance from the center of the sphere, and \( \rho \) is the uniform density of the sphere. This formula makes certain gravitational calculations much more straightforward.
Gravitational Force
The Gravitational Force is a natural phenomenon by which all things with mass or energy are brought toward one another. On Earth, it gives weight to physical objects and causes the ocean tides. The gravity of Earth is the force that holds us to the ground.
In outer space, the gravitational force is what keeps the planets in orbit around the Sun and moons in orbit around their planets. Sir Isaac Newton quantified this force in his law of universal gravitation, stating that every particle attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Calculating Gravitational Force Inside a Sphere
In the context of our uniform density sphere, if we wish to calculate the gravitational force on an object inside the sphere, it turns intriguing. While outside the sphere, all its mass contributes to the gravitational attraction, when an object is inside, only the mass in the spherical shell outside the object's distance from the center exerts a gravitational pull on the object. Interestingly, this means that the gravitational force at any point inside a uniform density sphere (with no cavity) is proportional to that point's distance from the center. This is a peculiar consequence of the symmetry of the sphere and the nature of gravitational forces.
In outer space, the gravitational force is what keeps the planets in orbit around the Sun and moons in orbit around their planets. Sir Isaac Newton quantified this force in his law of universal gravitation, stating that every particle attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Calculating Gravitational Force Inside a Sphere
In the context of our uniform density sphere, if we wish to calculate the gravitational force on an object inside the sphere, it turns intriguing. While outside the sphere, all its mass contributes to the gravitational attraction, when an object is inside, only the mass in the spherical shell outside the object's distance from the center exerts a gravitational pull on the object. Interestingly, this means that the gravitational force at any point inside a uniform density sphere (with no cavity) is proportional to that point's distance from the center. This is a peculiar consequence of the symmetry of the sphere and the nature of gravitational forces.
Spherical Cavity
A Spherical Cavity within a sphere introduces an interesting twist to gravitational calculations. If you carve out a spherical section from the center or at any place within a uniform density sphere, you get what we call a spherical cavity. The key feature of this cavity is that it contains no mass.
Effect on Gravitational Potential Energy
When we think about a sphere with a cavity inside and its gravitational potential energy, we cannot ignore the absence of mass in that cavity. The gravitational effects from the sphere's mass no longer apply within the volume of the cavity, rendering the influence of gravity inside the cavity different from that of a full sphere.
For instance, if an object were placed within the cavity, it would feel no gravitational pull from the mass that would've been present if the cavity weren't there. According to Newton's Shell Theorem, an object within a hollow shell experiences zero net gravitational force from the shell. So when considering gravitational potential energy inside the cavity, the expression changes to reflect the missing mass. This is why the solution in our exercise for the gravitational potential energy when \( 0
Effect on Gravitational Potential Energy
When we think about a sphere with a cavity inside and its gravitational potential energy, we cannot ignore the absence of mass in that cavity. The gravitational effects from the sphere's mass no longer apply within the volume of the cavity, rendering the influence of gravity inside the cavity different from that of a full sphere.
For instance, if an object were placed within the cavity, it would feel no gravitational pull from the mass that would've been present if the cavity weren't there. According to Newton's Shell Theorem, an object within a hollow shell experiences zero net gravitational force from the shell. So when considering gravitational potential energy inside the cavity, the expression changes to reflect the missing mass. This is why the solution in our exercise for the gravitational potential energy when \( 0
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