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Ch. 1 Solutions

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Basic

Medium

Basic

B-1. Estimating the Mass of Neptune

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Problem: Express the mass of Neptune \(M_N\) in terms of the orbital radius of Uranus \(r_U\), the orbital radius of Neptune \(r_N\), and the observed deviation in acceleration \(\delta a\).

Solution:

The gravitational acceleration exerted by Neptune on Uranus, from Newton's law of universal gravitation, is:

\(\delta a = \frac{GM_N}{(r_N - r_U)^2}\)

Here, for simplicity, we approximate that Uranus and Neptune are on the same side, with a separation distance of \(|r_N - r_U|\). Solving for \(M_N\):

\(\boxed{M_N = \frac{\delta a \cdot (r_N - r_U)^2}{G}}\)

Note: The actual calculations by Le Verrier and Adams were far more complex than this, taking into account the ellipticity of the orbits and the time variation of the relative positions of the planets. This problem is a simplification of the essential idea.

B-2. Why T-V and Not T+V

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(a) The case \(L = T - V\) (correct Lagrangian):

\(L = \frac{1}{2}m\dot{y}^2 - mgy\)

The partial derivatives:

\(\frac{\partial L}{\partial \dot{y}} = m\dot{y}, \qquad \frac{\partial L}{\partial y} = -mg\)

Substituting into the Euler-Lagrange equation \(\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} - \frac{\partial L}{\partial y} = 0\):

\(\frac{d}{dt}(m\dot{y}) - (-mg) = 0 \quad\Rightarrow\quad m\ddot{y} + mg = 0\)

\(\boxed{m\ddot{y} = -mg}\)

This is Newton's equation of motion, correctly expressing that "gravity acts downward (in the \(-y\) direction) and the ball accelerates downward."

(b) The case \(L' = T + V\) (incorrect Lagrangian):

\(L' = \frac{1}{2}m\dot{y}^2 + mgy\)

The partial derivatives:

\(\frac{\partial L'}{\partial \dot{y}} = m\dot{y}, \qquad \frac{\partial L'}{\partial y} = +mg\)

Substituting into the Euler-Lagrange equation:

\(\frac{d}{dt}(m\dot{y}) - mg = 0 \quad\Rightarrow\quad m\ddot{y} - mg = 0\)

\(\boxed{m\ddot{y} = +mg \quad(\text{unphysical})}\)

This means "gravity acts upward and the ball accelerates upward," which contradicts experiment.

Physical meaning: The form of the Lagrangian is not a "law of nature" but a "hypothesis." The form \(L = T - V\) is supported only by the post hoc justification that "if we tentatively adopt it and apply the Euler-Lagrange equation, Newton's equation of motion emerges." In other words, even in the Lagrangian formalism, the correct form is ultimately chosen by consistency with experiment. This is consistent with the prologue's stance that "all physical models are hypotheses."

Verification: For a ball thrown upward with initial velocity \(\dot{y}(0) = v_0\), case (a) gives \(y(t) = v_0 t - \frac{1}{2}gt^2\) (parabolic motion that rises to a maximum height and then falls), while case (b) gives \(y(t) = v_0 t + \frac{1}{2}gt^2\) (an unphysical solution that accelerates upward indefinitely). Only (a) agrees with experiment.

Medium

M-1. Derivation of Kepler's Third Law

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Problem: Using gravitational force \(F = GMm/r^2\) and the condition for circular motion \(F = mv^2/r\), derive \(T^2 \propto r^3\).

Solution:

For a planet (mass \(m\)) in a circular orbit, gravitational force provides the centripetal force:

\(G\frac{Mm}{r^2} = \frac{mv^2}{r}\)

Canceling \(m\) from both sides and solving for \(v\):

\(v^2 = \frac{GM}{r}\)

Since the period of circular motion is \(T = 2\pi r / v\):

\(T = \frac{2\pi r}{v} = 2\pi r \cdot \frac{1}{\sqrt{GM/r}} = 2\pi \sqrt{\frac{r^3}{GM}}\)

Squaring both sides:

\(\boxed{T^2 = \frac{4\pi^2}{GM} r^3}\)

Since \(G\) and \(M\) (the mass of the Sun) are constants, \(T^2 \propto r^3\) is demonstrated. This is Kepler's Third Law.

Key Point: From Newton's model, the proportionality constant \(4\pi^2/(GM)\) is specifically determined. Kepler's law merely identified the relationship \(T^2 \propto r^3\) empirically, but Newton's model predicts the value of the proportionality constant as well. This is the difference between "description" and "explanation."

M-2. Calculation of the Gravitational Potential

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Problem: Solve \(\nabla^2 \Phi = 0\) in the region \(r \neq 0\) under the assumption of spherical symmetry, and derive \(\Phi = -GM/r\).

Solution:

Since we assume spherical symmetry, let \(\Phi = \Phi(r)\). The Laplacian in spherical coordinates is:

\(\nabla^2 \Phi = \frac{1}{r^2}\frac{d}{dr}\left(r^2 \frac{d\Phi}{dr}\right)\)

In the region \(r \neq 0\), we have \(\rho = 0\), so \(\nabla^2 \Phi = 0\):

\(\frac{1}{r^2}\frac{d}{dr}\left(r^2 \frac{d\Phi}{dr}\right) = 0\)

Since \(r^2 \neq 0\), the expression inside the parentheses must be a constant:

\(r^2 \frac{d\Phi}{dr} = C_1\)

\(\frac{d\Phi}{dr} = \frac{C_1}{r^2}\)

Integrating:

\(\Phi = -\frac{C_1}{r} + C_2\)

Imposing the boundary condition \(\Phi \to 0\) as \(r \to \infty\) (the potential vanishes at infinity) gives \(C_2 = 0\).

Next, we determine \(C_1\). Since a mass \(M\) is located at the origin, we integrate Poisson's equation over a sphere enclosing the origin (using Gauss's theorem):

\(\int \nabla^2 \Phi \, dV = 4\pi G \int \rho \, dV = 4\pi G M\)

Applying Gauss's theorem to the left-hand side:

\(\oint \nabla\Phi \cdot d\mathbf{S} = \oint \frac{d\Phi}{dr} \cdot r^2 \sin\theta \, d\theta \, d\phi = \frac{C_1}{r^2} \cdot 4\pi r^2 = 4\pi C_1\)

Therefore \(4\pi C_1 = 4\pi GM\), i.e., \(C_1 = GM\).

\(\boxed{\Phi = -\frac{GM}{r}}\)

Verification: Computing the force from this \(\Phi\) gives \(F = -m \frac{d\Phi}{dr} = -m \cdot \frac{GM}{r^2}\) (negative indicating the attractive direction). The magnitude is \(|F| = GMm/r^2\), which is consistent with Newton's law of universal gravitation.