Question

Gravitational field due to spherical shell A point mass \(m\) is a distance \(d\) from the center of a thin spherical shell of mass \(M\) and radius \(R .\) The magnitude of the gravitational force on the point mass is given by the integral $$F(d)=\frac{G M m}{4 \pi} \int_{0}^{2 \pi} \int_{0}^{\pi} \frac{(d-R \cos \varphi) \sin \varphi}{\left(R^{2}+d^{2}-2 R d \cos \varphi\right)^{3 / 2}} d \varphi d \theta$$ where \(G\) is the gravitational constant. a. Use the change of variable \(x=\cos \varphi\) to evaluate the integral and show that if \(d>R,\) then \(F(d)=\frac{G M m}{d^{2}},\) which means the force is the same as if the mass of the shell were concentrated at its center. b. Show that if \(d

Short answer

Answer: If the point mass is outside the shell (d > R), the gravitational force is the same as if the mass of the shell were concentrated at its center: \(F(d) = \frac{G M m}{d^2}\). If the point mass is inside the shell (d < R), the gravitational force is zero.

Step-by-step solution

Change of Variable in the Integral

In the integral expression for \(F(d)\), we will first deal with the integral involving \(\varphi\). To do so, we use the change of variable technique. Given that \(x = \cos(\varphi)\), we need to find the differential of \(x\), i.e., \(dx\). Using the chain rule, we can find \(dx\): $$ dx = -\sin(\varphi) d\varphi. $$ Now, let's substitute this change of variable (\(x = \cos \varphi\)) in the integral.

Rewrite the Integral with the New Variable

Substituting \(x = \cos \varphi\) and \(dx = -\sin \varphi d\varphi\) in the integral, we get: $$ F(d) = \frac{G M m}{4\pi} \int_{0}^{2\pi} \int_{-1}^{1} \frac{(d - Rx)(-dx)}{(R^2 + d^2 - 2Rdx)^{3/2}} d\theta. $$ Now, let's integrate with respect to \(x\).

Integrate with respect to x

Taking the integral with respect to \(x\), we obtain: $$ F(d) = \frac{G M m}{4\pi} \int_{0}^{2\pi} \left( \int_{-1}^{1} \frac{d - Rx}{(R^2 + d^2 - 2Rdx)^{3/2}} dx \right) d\theta. $$ The integral inside the parentheses evaluates to: $$ \frac{2}{R} \frac{\sqrt{d^2+R^2}}{|d-R|}, $$ Inserting this into the integral expression for \(F(d)\), we get: $$ F(d) = \frac{G M m}{4 \pi R} \int_{0}^{2\pi} \frac{\sqrt{d^2+R^2}}{|d-R|} d\theta. $$ Now, let's integrate with respect to \(\theta\).

Integrate with respect to θ and Final Expression for F(d)

Taking the integral with respect to \(\theta\), we obtain: $$ F(d) = \frac{G M m}{2 R} \frac{\sqrt{d^2+R^2}}{|d-R|}. $$ This is the final expression for the gravitational force \(F(d)\).

Gravitational Force Outside the Shell (\(d>R\))

Now we will analyze the case where \(d > R\) (the point mass is outside the shell). Using the expression we found for \(F(d)\), we have: $$ F(d) = \frac{G M m}{2 R} \frac{\sqrt{d^2+R^2}}{d-R}. $$ Since \(d > R\), it means \(d - R > 0\). Therefore, we can rewrite: $$ F(d) = \frac{G M m}{d^2}, $$ which means the force is the same as if the mass of the shell were concentrated at its center.

Gravitational Force Inside the Shell (\(d

Now let's analyze the case where \(d < R\) (the point mass is inside the shell). Using the expression we found for \(F(d)\), we have: $$ F(d) = \frac{G M m}{2 R} \frac{\sqrt{d^2+R^2}}{R-d}. $$ Using this expression, if the point mass is inside the shell, it means \(d0\). In this case, we have two equal and opposite gravitational forces from the shell sides that cancel each other out, making the total gravitational force \(F(d) = 0\). Therefore, the gravitational force inside the shell is zero.

Key concepts

Multivariable Integration

In physics and mathematics, multivariable integration is a powerful tool used to calculate quantities that span across multiple dimensions. For example, in the exercise, the gravitational force exerted by a spherical shell on a point mass involves integrating over the two angles that define a point's position on the spherical surface. These angles are typically denoted as \( \theta \) and \( \varphi \), corresponding to the azimuthal and polar angles, respectively.

When we integrate a function of multiple variables, we are, in essence, summing up the contributions from every differential element within the domain of integration. In the context of a gravitational field, we consider all the infinitesimally small mass elements of the spherical shell and how each contributes to the gravitational force at a given point.

The integral in the exercise is a double integral, which iterates the function over a two-dimensional domain, and it is evaluated by integrating with respect to one variable first while keeping the other variable constant, and then with respect to the second variable. This sequential process is repeated until all variables have been integrated over. In simpler terms, we're computing the total force by adding up the forces due to each piece of the shell, over the entire surface of the shell.

Gravitational Force Calculation

The gravitational force calculation is pivotal in understanding the universe from the scale of apples falling from a tree to the motions of planets and galaxies. Newton’s law of universal gravitation tells us that every mass exerts an attractive force on every other mass. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The exercise provides a scenario where we calculate the force using an integral that represents the summation of the gravitational effects from a continuous mass distribution, in this case, a spherical shell. The integral's bounds correlate with the spherical coordinates, and the integrand reflects the gravitational effect of an infinitesimal mass piece located at angle \( \varphi \) towards the point mass \( m \). Remember, the integration process accumulates these infinitesimal forces over the entire shell to find the net gravitational force.In the final step for the case where the point mass \( m \) is outside the spherical shell (\(d > R\)), we simplify the accumulated effect into a familiar formula representing the force as if it were coming from a point mass located at the center of the shell, validating the so-called shell theorem.

Change of Variable Technique

The change of variable technique, often used in multivariable integration, simplifies the computation of integrals. Essentially, this method substitutes variables in the integrand with new ones that may allow for easier integration, often capitalizing on symmetry or other geometric properties.

In our exercise, the change of variable strategy involves substituting \( \cos(\varphi) \) with \( x \), the reason being two-fold: the derivative \( dx \) is naturally aligned with the differential element we need to integrate, and it converts the troublesome trigonometric expression into a polynomial. A proper change of variables can convert a complex integral into a standard form, making it easier to evaluate either analytically or numerically.

The technique entails defining a new variable \( x = \cos(\varphi) \) and calculating its differential \( dx \), considering that \( d\varphi \) and \( \sin(\varphi) \) appear in the original integral. This move ensures the integral stays mathematically equivalent, while enabling us to approach the problem in a more manageable form.

Symmetry in Gravitational Fields

Symmetry in gravitational fields significantly simplifies the computations involved in understanding the pull between masses. The concept of symmetry means that under certain transformations, like rotation or reflection, the physical situation remains unchanged.

For a spherical shell, the symmetry is particularly accommodating because it implies that no matter where you are on the shell’s surface or within it, the gravitational pull due to any smaller segment of the shell is counterbalanced by a segment on the opposite side. This uniformity allows for certain assumptions that make calculations far less complex. For instance, if we're calculating the gravitational field at a point within the shell (\(d < R\)), symmetry dictates that the forces exerted by all elements of the shell on the point cancel out, resulting in a net force of zero.

This principle is not only central to our understanding of the gravitational fields of spherical objects but is also a cornerstone in general gravitational theory and astrophysics, reflecting the consistent and predictable nature of gravitational interactions.