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Magazine Chapter 13: Problem 84
A uniform solid sphere of radius \(R\) produces a gravitational acceleration of \(a_{g}\) on its surface. At what distance from the sphere's center are there points (a) inside and (b) outside the sphere where the gravitational acceleration is \(a_{g} / 3\) ?
Short Answer
Expert verified
Inside: \(r = \frac{R}{3}\) and outside: \(r = \sqrt{3} R\).
Step by step solution
01
Understanding the Problem
We are given a uniform solid sphere that has a gravitational acceleration of \(a_{g}\) on its surface. We need to find two distances, one inside the sphere and one outside, where the gravitational acceleration is \(a_{g}/3\).
02
Gravitational field inside a sphere
For a point inside a sphere at a distance \(r\) from its center, the gravitational acceleration is given by \(g(r) = \frac{G M(r)}{r^2}\). The mass \(M(r)\) is the mass inside a radius \(r\) and is \(\frac{M}{R^3}r^3\), where \(M\) is the total mass of the sphere and \(R\) is its radius.
03
Determine the acceleration inside
Substitute the expression for \(M(r)\) into the formula for \(g(r)\) to get: \[g(r) = \frac{G}{r^2}\bigg(\frac{M}{R^3}r^3\bigg) = \frac{G M r}{R^3}\]. Set this equal to \(\frac{a_g}{3}\) to find the distance from the center where the acceleration is \(\frac{a_g}{3}\).
04
Solve for distance inside
We know that at the surface, \(G M / R^2 = a_{g}\). Hence, inside the sphere, the equation \(\frac{G M r}{R^3} = \frac{a_g}{3}\) simplifies to \( r = \frac{R}{3} \). This means the acceleration is \(\frac{a_g}{3}\) at \(r = \frac{R}{3}\) from the center.
05
Gravitational field outside a sphere
At a distance \(r\) outside the sphere, the acceleration is \(g(r) = \frac{G M}{r^2}\). Since the gravitational acceleration on the surface is \(a_g = \frac{G M}{R^2}\), set \(\frac{G M}{r^2} = \frac{a_g}{3}\) to find the external distance.
06
Solve for distance outside
From the equation \(\frac{G M}{r^2} = \frac{G M}{3 R^2}\), simplify to \(r^2 = 3 R^2\). Taking the square root, you find \(r = \sqrt{3} R\). This is the distance at which the gravitational acceleration is \(a_g / 3\) outside the sphere.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Uniform Solid Sphere
A uniform solid sphere is an object that has the same density throughout its volume. Each small part has the same amount of mass per unit volume. This uniformity means the mass is evenly spread across the sphere.
When discussing gravitational fields, we often assume spheres to be uniform to simplify calculations. This model helps in predicting how gravity behaves in various points inside and outside the sphere.
In the context of gravitational studies:
When discussing gravitational fields, we often assume spheres to be uniform to simplify calculations. This model helps in predicting how gravity behaves in various points inside and outside the sphere.
In the context of gravitational studies:
- Each part of the sphere pulls gravitationally.
- Gravity at any point is due to the combined effect of the entire mass.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object accelerates due to the gravitational force exerted on it by a mass, like a planet or star. For Earth, we often refer to it as "g," approximately 9.81 m/s虏.
It measures how much velocity an object gains every second as it falls freely. For a sphere, gravitational acceleration depends on its mass and how far you are from its center.
Key aspects to remember include:
It measures how much velocity an object gains every second as it falls freely. For a sphere, gravitational acceleration depends on its mass and how far you are from its center.
Key aspects to remember include:
- Gravitational acceleration on the surface of a sphere is constant when the sphere is uniform.
- Inside the sphere, it varies depending on your distance from the center.
- Outside the sphere, the decrease of gravitational acceleration follows the square of the distance from its center.
Distance from Center
The concept of distance from the center refers to how far a point is located from the center of a sphere, like a planet or moon. This distance affects how strongly you feel the gravitational pull.
Within the sphere, as you travel further inside from the center, gravitational acceleration begins at zero and increases as you move towards the surface. However, beyond the sphere, as the distance increases, the gravitational acceleration decreases.
For a uniform solid sphere:
Within the sphere, as you travel further inside from the center, gravitational acceleration begins at zero and increases as you move towards the surface. However, beyond the sphere, as the distance increases, the gravitational acceleration decreases.
For a uniform solid sphere:
- Inside, gravitational acceleration is proportional to distance from the center.
- Outside, gravitational acceleration decreases with the inverse square of the distance.
- Understanding the impact of distance helps calculate gravitational effects at various points near or within massive bodies.
Inside and Outside a Sphere
When discussing gravitational acceleration, it's important to differentiate between being inside or outside a sphere.
Inside the sphere, gravitational force behaves differently compared to outside because all mass contributes to the force in a unique way:
For points inside:
Inside the sphere, gravitational force behaves differently compared to outside because all mass contributes to the force in a unique way:
For points inside:
- Gravitational effects come only from the mass enclosed by your position.
- Acceleration increases linearly with distance from the center.
- All mass of the sphere contributes to gravitational force as if concentrated at the center.
- Acceleration decreases with the square of the distance.
