where $G = 6.67 \times 10^{-11}$ N·m²/kg² is gravitational constant.

Vector form: $\vec{F}_{12} = -G\frac{m_1m_2}{r^2}\hat{r}_{12}$

Gravitational Potential Energy

Choosing $U = 0$ at $r = \infty$:

$U(r) = -G\frac{m_1m_2}{r}$

Work to bring masses from infinity to separation $r$:

$W = \int_\infty^r F \, dr' = -G\frac{m_1m_2}{r}$

Gravitational Field

$\vec{g} = -\frac{GM}{r^2}\hat{r}$

Gravitational potential: $\Phi = -\frac{GM}{r}$

$\vec{g} = -\nabla\Phi$

Escape Velocity

Minimum velocity to escape gravitational field:

Set total energy = 0: $\frac{1}{2}mv_{esc}^2 - G\frac{Mm}{R} = 0$

$v_{esc} = \sqrt{\frac{2GM}{R}}$

For Earth: $v_{esc} \approx 11.2$ km/s

Circular Orbits

For circular orbit of radius $r$:

Centripetal force = Gravitational force: $\frac{mv^2}{r} = G\frac{Mm}{r^2}$

Orbital velocity: $v = \sqrt{\frac{GM}{r}}$

Orbital period: $T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}$

Orbital energy: $E = KE + PE = \frac{1}{2}mv^2 - G\frac{Mm}{r}$

$E = -\frac{GMm}{2r}$

(Negative: bound orbit)

Note: $|E| = KE$ (virial theorem)

Kepler's Laws

First Law (Law of Ellipses): Planets move in elliptical orbits with the Sun at one focus.

Second Law (Law of Equal Areas): Line from Sun to planet sweeps equal areas in equal times.

This follows from angular momentum conservation: $\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$

Third Law (Harmonic Law): $T^2 \propto a^3$

where $a$ is semi-major axis.

For circular orbit ($a = r$): $T^2 = \frac{4\pi^2}{GM}r^3$

Elliptical Orbits

Energy: $E = -\frac{GMm}{2a}$

where $a$ is semi-major axis.

Semi-major axis from energy: $a = -\frac{GMm}{2E}$

Angular momentum: $L = m\sqrt{GMa(1-e^2)}$

where $e$ is eccentricity.

Eccentricity: $e = \sqrt{1 + \frac{2EL^2}{m(GM)^2}}$

Perihelion and Aphelion

Perihelion (closest): $r_p = a(1-e)$

Aphelion (farthest): $r_a = a(1+e)$

Velocities: $v_p = \sqrt{\frac{GM}{a}\frac{1+e}{1-e}}, \quad v_a = \sqrt{\frac{GM}{a}\frac{1-e}{1+e}}$

(From energy and angular momentum conservation)

Hohmann Transfer Orbit

Efficient orbit change between circular orbits:

Transfer orbit energy: $E_t = -\frac{GM m}{r_1 + r_2}$

Velocity changes: $\Delta v_1 = \sqrt{\frac{GM}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}} - 1\right)$

$\Delta v_2 = \sqrt{\frac{GM}{r_2}}\left(1 - \sqrt{\frac{2r_1}{r_1+r_2}}\right)$

Reduced Mass Problem

Two-body problem reduces to one-body with reduced mass:

$\mu = \frac{m_1m_2}{m_1+m_2}$

Both orbit common center of mass.

Gravitational Force Inside Sphere

For uniform sphere, only mass at $r' < r$ contributes:

$M_{enclosed} = M\frac{r^3}{R^3}$

$F(r) = G\frac{Mm}{R^3}r$

(Linear with $r$, like spring force!)

At center: $F = 0$

Shell Theorem

1. Uniform spherical shell exerts no force on particle inside
2. Shell acts as point mass for particle outside

These allow us to treat planets as point masses for external objects.

(a) Orbital speed:

For circular orbit: $\frac{mv^2}{r} = \frac{GM_E m}{r^2}$

$v = \sqrt{\frac{GM_E}{r}}$

Using $g = \frac{GM_E}{R_E^2}$ → $GM_E = gR_E^2$:

$v = \sqrt{\frac{gR_E^2}{r}} = \sqrt{g}\frac{R_E}{\sqrt{r}}$

$v = \sqrt{9.8} \cdot \frac{6.37 \times 10^6}{\sqrt{6.77 \times 10^6}}$

$v = 3.13 \times \frac{6.37 \times 10^6}{2.60 \times 10^3}$

$\boxed{v = 7670 \text{ m/s} = 7.67 \text{ km/s}}$

(b) Orbital period:

$T = \frac{2\pi r}{v} = \frac{2\pi(6.77 \times 10^6)}{7670}$

$T = \frac{4.25 \times 10^7}{7670} = 5546 \text{ s}$

$\boxed{T = 92.4 \text{ min} = 1.54 \text{ hours}}$

(c) Acceleration:

Centripetal acceleration: $a = \frac{v^2}{r} = \frac{(7670)^2}{6.77 \times 10^6}$

$a = \frac{5.88 \times 10^7}{6.77 \times 10^6}$

$\boxed{a = 8.69 \text{ m/s}^2}$

This is 89% of g at surface (slightly weaker due to altitude).

At infinity (just escaping): $E_f = 0 + 0 = 0$

Energy conservation: $E_i = E_f$

$\frac{1}{2}mv_{esc}^2 - \frac{GM_E m}{R_E} = 0$

$v_{esc} = \sqrt{\frac{2GM_E}{R_E}}$

Using $g = GM_E/R_E^2$:

$\boxed{v_{esc} = \sqrt{2gR_E}}$

From Earth:

$v_{esc} = \sqrt{2(9.8)(6.37 \times 10^6)}$

$v_{esc} = \sqrt{1.25 \times 10^8} = 1.12 \times 10^4$

$\boxed{v_{esc,Earth} = 11,200 \text{ m/s} = 11.2 \text{ km/s}}$

From Moon:

Surface gravity on Moon: $g_M = \frac{GM_M}{R_M^2} = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}$

$g_M = 1.62 \text{ m/s}^2$

$v_{esc,Moon} = \sqrt{2(1.62)(1.74 \times 10^6)}$

$v_{esc,Moon} = \sqrt{5.64 \times 10^6}$

$\boxed{v_{esc,Moon} = 2370 \text{ m/s} = 2.37 \text{ km/s}}$

Moon's lower mass and smaller radius make escape much easier!

Fun fact: This is why Moon has no atmosphere - gas molecules can reach v_esc and escape.