Advanced Gravitation: Calculus & Theory

1. Foundations: The Law and The Field

The study of gravitation in an advanced context utilizes vector calculus to describe the interaction between masses, transitioning from scalar magnitudes to field concepts.

1.1. Newton's Law of Universal Gravitation (Vector Form)

The gravitational force $\mathbf{F}_{12}$ exerted on a mass $m_1$ by a mass $m_2$ is fundamentally attractive, described by:

$$ \mathbf{F}_{12} = - G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}}_{12} \tag{1.1} $$

Here, $G$ is the gravitational constant, $r$ is the magnitude of the distance between the masses, and $\hat{\mathbf{r}}_{12}$ is the unit vector pointing from $m_1$ to $m_2$. The negative sign rigorously enforces the attractive nature of the force.

2. Gravitational Field and Potential Energy

2.1. The Gravitational Field $\mathbf{g}$

The gravitational field $\mathbf{g}$ represents the force per unit mass exerted by a massive object $M$ on any point in space. It is mathematically defined as:

$$ \mathbf{g} = \frac{\mathbf{F}}{m_t} \tag{2.1} $$

For a point mass $M$, substituting Equation (1.1) into (2.1) (with $m_t$ replacing $m_1$ and $M$ replacing $m_2$):

$$ \mathbf{g}(\mathbf{r}) = - G \frac{M}{r^2} \hat{\mathbf{r}} \tag{2.2} $$

2.2. Derivation of Gravitational Potential $V$

Since the gravitational force is conservative, it can be derived from a scalar potential field $V$, often called the gravitational potential, which is the potential energy per unit mass. The relationship between the field and the potential is defined by the gradient operator $\nabla$:

$$ \mathbf{g} = - \nabla V \tag{2.3} $$

Proof (Radial Derivation):

  1. In spherical coordinates, for a spherically symmetric field (like that of a point mass), the gradient simplifies to only the radial component: $$ \nabla V = \left( \frac{\partial V}{\partial r} \right) \hat{\mathbf{r}} $$
  2. Substitute this and Equation (2.2) into Equation (2.3): $$ - G \frac{M}{r^2} \hat{\mathbf{r}} = - \left( \frac{\partial V}{\partial r} \right) \hat{\mathbf{r}} $$
  3. Equating the magnitudes gives the differential equation: $$ \frac{dV}{dr} = G \frac{M}{r^2} $$
  4. Integrate with respect to $r$: $$ V(r) = \int G \frac{M}{r^2} dr = - G \frac{M}{r} + C $$
  5. Apply the boundary condition that the potential is zero at infinity ($V(\infty) = 0$): $$ 0 = - G \frac{M}{\infty} + C \implies C = 0 $$

Thus, the gravitational potential due to a point mass $M$ is:

$$ V(r) = - G \frac{M}{r} \tag{2.4} $$

The Gravitational Potential Energy $U$ of a mass $m$ in this field is $U = m V$:

$$ U(r) = - G \frac{M m}{r} \tag{2.5} $$

3. Gauss's Law for Gravitation (Integral and Differential Forms)

3.1. Integral Form (Flux Theorem)

Gauss's Law relates the flux $\Phi_g$ of the gravitational field $\mathbf{g}$ through a closed surface $S$ to the mass enclosed $M_{\text{encl}}$ by that surface:

$$ \Phi_g = \oint_S \mathbf{g} \cdot d\mathbf{A} = - 4 \pi G M_{\text{encl}} \tag{3.1} $$

The negative sign reflects that the attractive field $\mathbf{g}$ is directed inward, opposite to the outward-pointing normal of the surface area element $d\mathbf{A}$.

3.2. Differential Form (Poisson's Equation)

We derive the differential form by applying the Divergence Theorem to Equation (3.1), where the enclosed mass $M_{\text{encl}}$ is expressed as an integral of the mass density $\rho$ over the volume $V$:

Proof (Differential Form):

  1. Apply the Divergence Theorem to the left side of Equation (3.1): $$ \oint_S \mathbf{g} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{g}) dV $$
  2. Express the enclosed mass $M_{\text{encl}}$ as a volume integral of the mass density $\rho$: $$ - 4 \pi G M_{\text{encl}} = - 4 \pi G \int_V \rho dV $$
  3. Equate the two resulting volume integrals: $$ \int_V (\nabla \cdot \mathbf{g}) dV = - 4 \pi G \int_V \rho dV $$
  4. Since this equality must hold for any arbitrary volume $V$, the integrands must be equal, yielding the differential form of Gauss's Law: $$ \nabla \cdot \mathbf{g} = - 4 \pi G \rho \tag{3.2} $$
  5. Substitute the relation $\mathbf{g} = -\nabla V$ (Equation 2.3) into the result: $$ \nabla \cdot (-\nabla V) = - 4 \pi G \rho $$ $$ -\nabla^2 V = - 4 \pi G \rho $$

This final result is Poisson's Equation for the gravitational potential $V$:

$$ \nabla^2 V = 4 \pi G \rho \tag{3.3} $$

Where $\nabla^2$ is the Laplacian operator ($\nabla^2 = \nabla \cdot \nabla$).

4. Orbital Dynamics: Derivation of Kepler's Laws

4.1. Proof of Kepler's Second Law (Conservation of Angular Momentum)

Kepler's Second Law is a direct consequence of the conservation of angular momentum in a central force field.

Proof:

  1. The gravitational force is a central force, acting only along the radial direction: $\mathbf{F} = f(r) \hat{\mathbf{r}}$.
  2. The torque $\boldsymbol{\tau}$ exerted on the orbiting body about the center of force is $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$: $$ \boldsymbol{\tau} = \mathbf{r} \times \left( - G \frac{M m}{r^2} \hat{\mathbf{r}} \right) $$
  3. Since $\mathbf{r}$ and $\hat{\mathbf{r}}$ are parallel vectors (differing only in magnitude), their cross product is zero: $$ \boldsymbol{\tau} = \mathbf{0} $$
  4. By Newton's Second Law for rotation, the torque equals the rate of change of angular momentum $\mathbf{L}$: $\boldsymbol{\tau} = \frac{d\mathbf{L}}{dt}$. $$ \frac{d\mathbf{L}}{dt} = \mathbf{0} \implies \mathbf{L} = \text{constant} $$
  5. Angular momentum $L$ relates to the rate of area sweeping $\frac{dA}{dt}$. In polar coordinates, the differential area $dA$ swept in time $dt$ is $dA = \frac{1}{2} r^2 d\theta$. $$ \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} $$
  6. Since the magnitude of angular momentum is $L = m r^2 \frac{d\theta}{dt}$, we can substitute: $$ \frac{dA}{dt} = \frac{L}{2m} $$

Since $\mathbf{L}$ and $m$ are constants, the area rate is constant, proving Kepler's Second Law:

$$ \frac{dA}{dt} = \text{constant} \tag{4.1} $$

4.2. Derivation of Kepler's Third Law

Kepler's Third Law relates the orbital period $T$ to the semi-major axis $a$. We derive it for the simpler case of a circular orbit of radius $r$, where $r=a$.

Proof (Circular Orbit):

  1. For a stable circular orbit, the gravitational force $F_g$ provides the centripetal force $F_c$: $$ F_g = F_c $$ $$ G \frac{M m}{r^2} = m a_c = m \omega^2 r $$
  2. Cancel the orbiting mass $m$: $$ G \frac{M}{r^2} = \omega^2 r $$
  3. The angular velocity $\omega$ is related to the period $T$ by $\omega = \frac{2\pi}{T}$. Substitute this into the equation: $$ G \frac{M}{r^2} = \left( \frac{2\pi}{T} \right)^2 r = \frac{4\pi^2}{T^2} r $$
  4. Rearrange the equation to solve for $T^2$: $$ T^2 = \left( \frac{4\pi^2}{G M} \right) r^3 $$

Replacing $r$ with the general semi-major axis $a$ for elliptical orbits, we get Kepler's Third Law:

$$ T^2 = \left( \frac{4\pi^2}{G M} \right) a^3 \tag{4.2} $$

5. Escape Velocity (Energy Principle)

The escape velocity $v_e$ is the minimum speed required for an object to completely overcome the gravitational field of a massive body $M$. This condition requires the object's final velocity to be zero at infinite distance ($r \to \infty$).

Proof:

The principle of conservation of mechanical energy is used, where $E_{\text{total}} = K + U = \text{constant}$.

  1. Initial state ($i$): Object of mass $m$ starts at distance $R$ with escape velocity $v_e$. $$ E_i = K_i + U_i = \frac{1}{2} m v_e^2 - G \frac{M m}{R} $$
  2. Final state ($f$): Object reaches infinity ($\infty$) with minimum required speed, $v_f=0$. $$ E_f = K_f + U_f = 0 - G \frac{M m}{\infty} = 0 $$
  3. By energy conservation, $E_i = E_f$: $$ \frac{1}{2} m v_e^2 - G \frac{M m}{R} = 0 $$
  4. Solve for $v_e$: $$ v_e^2 = \frac{2 G M}{R} $$

The escape velocity is thus:

$$ v_e = \sqrt{\frac{2 G M}{R}} \tag{5.1} $$

6. Problems for Practice

Intermediate Problems (6.1 - 6.5)

P6.1. Gravitational Potential of a Rod: A thin rod of length $L$ and uniform mass $M$ is placed along the x-axis, with one end at the origin. Find the gravitational potential $V$ at a point $P$ located on the x-axis at a distance $a > L$ from the origin.

P6.2. Work Done by Gravity: A mass $m$ is moved from the Earth's surface (radius $R_E$) to a height $2R_E$ above the surface. Calculate the work done by the gravitational force. Express the answer in terms of $G$, $M_E$, $m$, and $R_E$.

P6.3. Orbital Speed Ratio: A satellite orbits a planet in a circular path at a distance $r_1$. If the satellite is moved to a new circular orbit at a distance $r_2 = 4r_1$, what is the ratio of the new orbital speed $v_2$ to the old orbital speed $v_1$?

P6.4. Field Inside a Spherical Shell: Use Gauss's Law to prove that the gravitational field $\mathbf{g}$ is zero everywhere inside a uniform, hollow spherical shell of mass $M$ and radius $R$.

P6.5. Geosynchronous Orbit: The period of a geosynchronous satellite is $T = 24 \text{ hours}$. Calculate the radius $r$ of the orbit above the Earth's center. Assume $M_E \approx 5.97 \times 10^{24} \text{ kg}$ and $G \approx 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$.

Advanced Problems (6.6 - 6.8)

P6.6. Field of a Uniform Sphere: Find the gravitational field $\mathbf{g}(r)$ inside a solid sphere of uniform mass density $\rho$, radius $R$, and total mass $M$, as a function of the distance $r$ from the center ($r \le R$). Use Gauss's Law and express the answer in terms of $G$, $M$, $R$, and $r$.

P6.7. Small Perturbation of a Circular Orbit: A particle is in a circular orbit of radius $r_0$ around a mass $M$. If the radius is slightly perturbed to $r = r_0 + \delta r$, derive the differential equation for $\delta r(t)$. Show that the motion is oscillatory, and find the period of small radial oscillations.

P6.8. Energy of a System of Masses: Three masses $m_1$, $m_2$, and $m_3$ are placed at the vertices of an equilateral triangle of side $a$. Calculate the total gravitational potential energy $U_{\text{total}}$ of the system.

Irodov-like Problems (6.9 - 6.10)

P6.9. Tunnelling Through the Earth: A straight, smooth tunnel is drilled through the Earth passing through its center. A small object of mass $m$ is dropped into the tunnel from one opening. Assuming the Earth is a perfect sphere of uniform density $\rho$ and radius $R$, show that the motion is simple harmonic motion (SHM). Calculate the time $T$ it takes for the mass to reach the other side of the tunnel. (Express $T$ in terms of $G$ and $R$).

P6.10. Gravitational Slingshot Maneuver: A probe of mass $m$ approaches a planet of mass $M$ (where $M \gg m$) with an initial velocity $\mathbf{v}_i$ far from the planet. The probe executes a close flyby such that the direction of the probe's velocity vector is perfectly reversed (i.e., the scattering angle is $\Theta = 180^\circ$). Derive the expression for the minimum distance $r_{\text{min}}$ of the closest approach to the planet's center during the maneuver, in terms of $G$, $M$, and the magnitude of the initial velocity $v_i$. (Note: For $\Theta=180^\circ$, the impact parameter $b$ is zero, implying a head-on collision. Express the theoretical $r_{\text{min}}$).

7. Detailed Solutions

S7.1. Gravitational Potential of a Rod (P6.1)

Formula First Approach:

The gravitational potential $V$ due to a continuous body is found by integrating the contribution of infinitesimal mass elements $dm$ (Equation 2.4). For a uniform rod of length $L$ and total mass $M$, the linear mass density is $\lambda = M/L$.

$$ V = - \frac{G M}{L} \ln \left( \frac{a}{a-L} \right) \tag{S7.1} $$

Detailed Proof:

  1. Define the infinitesimal mass element $dm$ at position $x$ along the rod: $$ dm = \lambda \, dx = \frac{M}{L} dx $$
  2. The point $P$ is at $x=a$. The distance $r$ from $dm$ (at $x$) to $P$ is $r = a - x$.
  3. Substitute $dm$ and $r$ into the potential integral. The integration limits run from $x=0$ to $x=L$: $$ V = - \int_{0}^{L} \frac{G (M/L) dx}{a - x} = - \frac{G M}{L} \int_{0}^{L} \frac{dx}{a - x} $$
  4. Perform the substitution $u = a - x$, which means $du = -dx$. The limits transform: $x=0 \implies u=a$; $x=L \implies u=a-L$. $$ V = - \frac{G M}{L} \int_{a}^{a-L} \frac{-du}{u} = \frac{G M}{L} \int_{a}^{a-L} \frac{du}{u} $$
  5. Integrate the natural logarithm: $$ V = \frac{G M}{L} \left[ \ln|u| \right]_{a}^{a-L} = \frac{G M}{L} (\ln(a-L) - \ln a) $$
  6. Use logarithm properties ($\ln A - \ln B = \ln (A/B)$ and $-\ln(A/B) = \ln(B/A)$) to reach the final form: $$ V = \frac{G M}{L} \ln \left( \frac{a-L}{a} \right) = - \frac{G M}{L} \ln \left( \frac{a}{a-L} \right) $$

S7.2. Work Done by Gravity (P6.2)

Formula First Approach:

The work done by a conservative force, like gravity ($W_g$), is the negative change in potential energy ($\Delta U$):

$$ W_g = - \Delta U = U_i - U_f $$

Initial position $r_i = R_E$. Final position $r_f = R_E + 2R_E = 3R_E$. Potential energy is $U(r) = - G \frac{M_E m}{r}$ (Equation 2.5).

$$ W_g = - \frac{2}{3} \frac{G M_E m}{R_E} \tag{S7.2} $$

Detailed Proof:

  1. Calculate the initial potential energy $U_i$ at $r_i = R_E$: $$ U_i = - G \frac{M_E m}{R_E} $$
  2. Calculate the final potential energy $U_f$ at $r_f = 3R_E$: $$ U_f = - G \frac{M_E m}{3R_E} $$
  3. Calculate the work done $W_g = U_i - U_f$: $$ W_g = \left( - G \frac{M_E m}{R_E} \right) - \left( - G \frac{M_E m}{3R_E} \right) $$ $$ W_g = G M_E m \left( \frac{1}{3R_E} - \frac{1}{R_E} \right) $$
  4. Combine the fractions: $$ W_g = G M_E m \left( \frac{1 - 3}{3R_E} \right) = G M_E m \left( - \frac{2}{3R_E} \right) $$
  5. Final result: $$ W_g = - \frac{2}{3} \frac{G M_E m}{R_E} $$

S7.3. Orbital Speed Ratio (P6.3)

Formula First Approach:

The orbital speed $v$ for a circular orbit is derived by equating gravitational and centripetal forces ($G \frac{M m}{r^2} = m \frac{v^2}{r}$), yielding:

$$ v = \sqrt{\frac{G M}{r}} $$

Given $r_2 = 4r_1$, the ratio $v_2/v_1$ is:

$$ \frac{v_2}{v_1} = \frac{1}{2} \tag{S7.3} $$

Detailed Proof:

  1. Express the speeds $v_1$ and $v_2$ in terms of their respective radii: $$ v_1 = \sqrt{\frac{G M}{r_1}} \quad \text{and} \quad v_2 = \sqrt{\frac{G M}{r_2}} $$
  2. Form the ratio of the speeds: $$ \frac{v_2}{v_1} = \frac{\sqrt{G M/r_2}}{\sqrt{G M/r_1}} = \sqrt{\frac{r_1}{r_2}} $$
  3. Substitute the given relation $r_2 = 4r_1$: $$ \frac{v_2}{v_1} = \sqrt{\frac{r_1}{4r_1}} = \sqrt{\frac{1}{4}} $$
  4. Final result: $$ \frac{v_2}{v_1} = \frac{1}{2} $$

S7.4. Field Inside a Spherical Shell (P6.4)

Formula First Approach:

Gauss's Law (Equation 3.1) states $\oint_S \mathbf{g} \cdot d\mathbf{A} = - 4 \pi G M_{\text{encl}}$. For any closed Gaussian surface $S$ constructed entirely within the hollow shell, the enclosed mass $M_{\text{encl}}$ is zero.

$$ \oint_S \mathbf{g} \cdot d\mathbf{A} = 0 \implies \mathbf{g} = \mathbf{0} \quad \text{for } r < R \tag{S7.4} $$

Detailed Proof:

  1. Construct a spherical Gaussian surface $S$ concentric with the shell, with radius $r < R$.
  2. Determine the enclosed mass $M_{\text{encl}}$. Since the shell is hollow and uniform, and the surface $S$ is inside the shell, no mass is enclosed: $$ M_{\text{encl}} = 0 $$
  3. Apply Gauss's Law (Equation 3.1): $$ \oint_S \mathbf{g} \cdot d\mathbf{A} = - 4 \pi G (0) = 0 $$
  4. By spherical symmetry, the gravitational field $\mathbf{g}$ must be radial and have a constant magnitude $g(r)$ over the Gaussian surface $S$. Thus, the flux integral simplifies: $$ \oint_S \mathbf{g} \cdot d\mathbf{A} = g(r) \oint_S dA = g(r) (4\pi r^2) $$
  5. Equate the flux expression to the result from Gauss's Law: $$ g(r) (4\pi r^2) = 0 $$
  6. Since $r > 0$, the surface area $4\pi r^2$ is non-zero, which requires the field magnitude to be zero: $$ g(r) = 0 $$

S7.5. Geosynchronous Orbit (P6.5)

Formula First Approach:

We use Kepler's Third Law (Equation 4.2) for the circular orbit, $T^2 = \left( \frac{4\pi^2}{G M_E} \right) r^3$, solving for the radius $r$:

$$ r = \left( \frac{G M_E T^2}{4\pi^2} \right)^{1/3} $$

Using the constants ($T = 86400 \text{ s}$, $M_E$, $G$):

$$ r \approx 4.225 \times 10^7 \text{ m} \quad (\approx 42,250 \text{ km}) \tag{S7.5} $$

Detailed Calculation:

  1. Convert the period to seconds: $T = 24 \text{ hr} \times 3600 \text{ s/hr} = 86400 \text{ s}$.
  2. Cube the radius term from Kepler's Third Law: $$ r^3 = \frac{G M_E T^2}{4\pi^2} $$
  3. Substitute the numerical values: $$ r^3 = \frac{(6.67 \times 10^{-11}) (5.97 \times 10^{24}) (86400)^2}{4\pi^2} \text{ m}^3 $$
  4. Calculate the result: $$ r^3 \approx 7.53 \times 10^{22} \text{ m}^3 $$
  5. Take the cube root to find the orbital radius: $$ r \approx 4.225 \times 10^7 \text{ m} $$

S7.6. Field of a Uniform Sphere (Inside) (P6.6)

Formula First Approach:

The field $\mathbf{g}(r)$ for $r \le R$ is determined by the mass enclosed $M_{\text{encl}} = M(r^3/R^3)$.

$$ g(r) (4\pi r^2) = - 4 \pi G M_{\text{encl}} $$

The field increases linearly with distance from the center:

$$ \mathbf{g}(r) = - \frac{G M}{R^3} \mathbf{r} \tag{S7.6} $$

Detailed Proof:

  1. Define the Gaussian surface $S$ as a concentric sphere of radius $r \le R$.
  2. Calculate the mass $M_{\text{encl}}$ enclosed by $S$. Since the density $\rho$ is uniform, $M_{\text{encl}}$ is proportional to the volume $V_r = 4/3 \pi r^3$: $$ M_{\text{encl}} = \rho V_r = \left( \frac{M}{4/3 \pi R^3} \right) \left( \frac{4}{3} \pi r^3 \right) = M \frac{r^3}{R^3} $$
  3. Apply Gauss's Law (Equation 3.1). Due to symmetry, $\oint_S \mathbf{g} \cdot d\mathbf{A} = g(r) (4\pi r^2)$: $$ g(r) (4\pi r^2) = - 4 \pi G M_{\text{encl}} $$
  4. Substitute $M_{\text{encl}}$ into the equation: $$ g(r) (4\pi r^2) = - 4 \pi G \left( M \frac{r^3}{R^3} \right) $$
  5. Solve for $g(r)$ by cancelling $4\pi$ and $r^2$: $$ g(r) = - \frac{G M r^3}{R^3 r^2} = - \frac{G M r}{R^3} $$
  6. In vector form, recognizing that $r \hat{\mathbf{r}} = \mathbf{r}$, the field is directed radially inward: $$ \mathbf{g}(r) = - \frac{G M}{R^3} \mathbf{r} $$

S7.7. Small Perturbation of a Circular Orbit (P6.7)

Formula First Approach:

The radial motion $\delta r$ is governed by the effective potential $V_{\text{eff}}(r)$. Linearizing the equation of motion leads to the SHM form $\ddot{\delta r} + \omega^2 \delta r = 0$, where $\omega^2 = G M / r_0^3$. The period of small radial oscillations $T_r$ is equal to the orbital period $T_0$:

$$ T_r = T_0 = 2\pi \sqrt{\frac{r_0^3}{G M}} \tag{S7.7} $$

Detailed Proof (Using Effective Force Method):

  1. The radial equation of motion for a particle under a central force $F(r) = - G M m / r^2$ is: $$ m \left( \ddot{r} - r \dot{\theta}^2 \right) = F(r) = - G \frac{M m}{r^2} $$
  2. Use the constant angular momentum $L = m r^2 \dot{\theta}$ to replace $\dot{\theta}^2$ with $\dot{\theta}^2 = \left( \frac{L}{m r^2} \right)^2$: $$ m \ddot{r} - m r \left( \frac{L^2}{m^2 r^4} \right) = - G \frac{M m}{r^2} \implies \ddot{r} = \frac{L^2}{m^2 r^3} - \frac{G M}{r^2} $$
  3. For a circular orbit at radius $r_0$, the equilibrium condition is $\ddot{r}=0$. This gives the condition relating $L$ and $r_0$: $$ 0 = \frac{L^2}{m^2 r_0^3} - \frac{G M}{r_0^2} \implies L^2 = G M m^2 r_0 $$
  4. Introduce a small perturbation $r = r_0 + \delta r$ ($\delta r \ll r_0$). The radial force can be written as $F_{\text{eff}}(r) = \frac{L^2}{m r^3} - G \frac{M m}{r^2}$.
  5. Expand $F_{\text{eff}}(r)$ around $r_0$ using the linear approximation for the equation of motion: $m \ddot{\delta r} = \left( \frac{d F_{\text{eff}}}{d r} \right)_{r_0} \delta r$.
  6. Calculate the derivative of the effective force: $$ \frac{d F_{\text{eff}}}{d r} = - \frac{3 L^2}{m r^4} + \frac{2 G M m}{r^3} $$
  7. Evaluate the derivative at $r_0$ and substitute $L^2 = G M m^2 r_0$: $$ \left( \frac{d F_{\text{eff}}}{d r} \right)_{r_0} = - \frac{3 (G M m^2 r_0)}{m r_0^4} + \frac{2 G M m}{r_0^3} = - \frac{3 G M m}{r_0^3} + \frac{2 G M m}{r_0^3} $$ $$ \left( \frac{d F_{\text{eff}}}{d r} \right)_{r_0} = - \frac{G M m}{r_0^3} $$
  8. The differential equation for the perturbation $\delta r$ becomes: $$ m \ddot{\delta r} = - \left( \frac{G M m}{r_0^3} \right) \delta r \implies \ddot{\delta r} + \left( \frac{G M}{r_0^3} \right) \delta r = 0 $$
  9. This is the SHM equation $\ddot{\delta r} + \omega^2 \delta r = 0$ with $\omega^2 = G M / r_0^3$. The period of radial oscillation $T_r$ is $T_r = 2\pi/\omega$: $$ T_r = 2\pi \sqrt{\frac{r_0^3}{G M}} $$

S7.8. Energy of a System of Masses (P6.8)

Formula First Approach:

The total potential energy $U_{\text{total}}$ of a system is the sum of the potential energies of all unique pairs $U_{ij}$ (Equation 2.5), $U_{\text{total}} = \sum_{i < j} U_{ij}$. For three masses at distance $a$, there are three pairs: $(m_1, m_2)$, $(m_2, m_3)$, and $(m_1, m_3)$.

$$ U_{\text{total}} = - \frac{G}{a} (m_1 m_2 + m_2 m_3 + m_1 m_3) \tag{S7.8a} $$

If $m_1=m_2=m_3=m$:

$$ U_{\text{total}} = - 3 G \frac{m^2}{a} \tag{S7.8b} $$

Detailed Proof:

  1. Identify the potential energy for each unique pair. Since the triangle is equilateral, the distance $r_{ij}$ for all pairs is $a$. $$ U_{12} = - G \frac{m_1 m_2}{a} $$ $$ U_{23} = - G \frac{m_2 m_3}{a} $$ $$ U_{13} = - G \frac{m_1 m_3}{a} $$
  2. Sum the potential energies of the three pairs: $$ U_{\text{total}} = U_{12} + U_{23} + U_{13} = - G \frac{m_1 m_2}{a} - G \frac{m_2 m_3}{a} - G \frac{m_1 m_3}{a} $$
  3. Factor out the common term $G/a$: $$ U_{\text{total}} = - \frac{G}{a} (m_1 m_2 + m_2 m_3 + m_1 m_3) $$
  4. If the masses are equal ($m_1=m_2=m_3=m$): $$ U_{\text{total}} = - \frac{G}{a} (m^2 + m^2 + m^2) = - 3 G \frac{m^2}{a} $$

S7.9. Tunnelling Through the Earth (SHM) (P6.9)

Formula First Approach:

The force $F$ on the mass $m$ at distance $r$ inside the uniform Earth is $F = m g(r)$, where $g(r) = - \frac{G M}{R^3} r$ (from S7.6). This leads to the SHM equation $\ddot{r} + \omega^2 r = 0$ with $\omega^2 = G M / R^3$. The time $T$ to reach the other side is half a period ($\pi/\omega$):

$$ T = \pi \sqrt{\frac{R^3}{G M}} \tag{S7.9} $$

Detailed Proof:

  1. Consider the mass $m$ at a distance $r$ from the center of the Earth. The gravitational field $g(r)$ inside a uniform sphere is: $$ g(r) = - \frac{G M r}{R^3} $$
  2. The force $F$ on the mass $m$ is $F = m g(r)$. This force acts toward the center (restoring force): $$ F = - \left( \frac{G M m}{R^3} \right) r $$
  3. Apply Newton's Second Law: $F = m \ddot{r}$: $$ m \ddot{r} = - \left( \frac{G M m}{R^3} \right) r $$
  4. Divide by $m$ and rearrange to the standard SHM differential equation form $\ddot{r} + \omega^2 r = 0$: $$ \ddot{r} + \left( \frac{G M}{R^3} \right) r = 0 $$
  5. This confirms the motion is Simple Harmonic Motion (SHM), with angular frequency $\omega$: $$ \omega = \sqrt{\frac{G M}{R^3}} $$
  6. The time $T$ to reach the other side is half of the full period $T_{\text{SHM}} = 2\pi/\omega$: $$ T = \frac{1}{2} T_{\text{SHM}} = \frac{\pi}{\omega} = \pi \sqrt{\frac{R^3}{G M}} $$

S7.10. Gravitational Slingshot Maneuver ($\Theta=180^\circ$) (P6.10)

Formula First Approach:

The distance of closest approach $r_{\text{min}}$ for $180^\circ$ deflection (a head-on collision) is theoretically zero, as the impact parameter $b$ must be zero for the $\Theta=180^\circ$ scattering condition $\cot\left(\frac{\Theta}{2}\right) = 0$ to hold.

$$ r_{\text{min}} = 0 \tag{S7.10} $$

Detailed Proof (Using General Hyperbolic Orbit Equation):

  1. The distance of closest approach $r_{\text{min}}$ is given by the turning point of the orbit. For a hyperbolic trajectory, the general solution is: $$ r_{\text{min}} = \frac{1}{2} \left[ - \frac{G M}{v_i^2} + \sqrt{\left( \frac{G M}{v_i^2} \right)^2 + 4 b^2} \right] $$
  2. The scattering angle $\Theta$ is related to the impact parameter $b$ by: $$ \cot\left(\frac{\Theta}{2}\right) = \frac{b v_i^2}{G M} $$
  3. For a perfect velocity reversal, $\Theta = 180^\circ$. Substituting this into the scattering relation: $$ \cot\left(\frac{180^\circ}{2}\right) = \cot(90^\circ) = 0 $$ $$ 0 = \frac{b v_i^2}{G M} \implies b = 0 $$
  4. Substitute the condition $b=0$ into the $r_{\text{min}}$ equation: $$ r_{\text{min}} = \frac{1}{2} \left[ - \frac{G M}{v_i^2} + \sqrt{\left( \frac{G M}{v_i^2} \right)^2 + 4 (0)^2} \right] $$
  5. Simplify the expression: $$ r_{\text{min}} = \frac{1}{2} \left[ - \frac{G M}{v_i^2} + \frac{G M}{v_i^2} \right] $$
  6. The result shows the theoretical point of closest approach for a $180^\circ$ deflection (a direct hit on the center of mass) is: $$ r_{\text{min}} = 0 $$