With Altitude ($h$):
- $g_h = \frac{GM_e}{(R_e+h)^2} = g \left(1 + \frac{h}{R_e}\right)^{-2}$
- For $h \ll R_e$: $g_h \approx g \left(1 - \frac{2h}{R_e}\right)$
With Depth ($d$):
- $g_d = \frac{GM_e(R_e-d)}{R_e^3} = g \left(1 - \frac{d}{R_e}\right)$
- At center ($d=R_e$): $g_c = 0$.
With Latitude ($\lambda$) due to Earth's Rotation:
- $g' = g - R_e \omega^2 \cos^2\lambda$
- At poles ($\lambda = 90^\circ$): $g' = g$.
- At equator ($\lambda = 0^\circ$): $g' = g - R_e \omega^2$.

Minimum velocity required for a body to escape Earth's gravitational field. $$v_e = \sqrt{\frac{2GM_e}{R_e}} = \sqrt{2gR_e} \approx 11.2 \text{ km/s}$$
Does not depend on the mass of the escaping body.

Velocity required for a satellite to orbit in a stable circular path at radius $r$ from center of Earth. $$v_o = \sqrt{\frac{GM_e}{r}}$$ where $r = R_e + h$.
For orbit close to Earth's surface ($r \approx R_e$): $v_o = \sqrt{gR_e} \approx 7.9 \text{ km/s}$.
Relation: $v_e = \sqrt{2} v_o$ (for same radius).

For a circular orbit: $T = \frac{2\pi r}{v_o} = 2\pi r \sqrt{\frac{r}{GM_e}} = 2\pi \sqrt{\frac{r^3}{GM_e}}$.
Geostationary satellite: $T = 24 \text{ hours}$, orbits at $h \approx 36000 \text{ km}$ above equator.

Kinetic Energy: $K = \frac{1}{2}mv_o^2 = \frac{1}{2}m \frac{GM_e}{r} = \frac{GM_e m}{2r}$.
Potential Energy: $U = -\frac{GM_e m}{r}$.
Total Energy: $E = K + U = \frac{GM_e m}{2r} - \frac{GM_e m}{r} = -\frac{GM_e m}{2r}$.
Total energy is negative, indicating a bound system.
Binding Energy: $-E = \frac{GM_e m}{2r}$.
To change orbit, energy must be added/removed.

The line joining any planet to the Sun sweeps out equal areas in equal intervals of time.
Angular momentum of the planet about the Sun is conserved ($L = constant$). $$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$$
A planet moves faster when closer to the Sun (perihelion) and slower when farther (aphelion).

The square of the orbital period ($T$) of any planet is proportional to the cube of the semi-major axis ($a$) of its elliptical orbit. $$T^2 \propto a^3 \quad \text{or} \quad \frac{T^2}{a^3} = \text{constant} = \frac{4\pi^2}{GM_{central}}$$
For circular orbit, $a = r$.

Work done to assemble the sphere from infinitesimal layers: $$U_{self} = -\frac{3}{5} \frac{GM^2}{R}$$