Gravitation (Calculus-Based)

Newton's Law of Universal Gravitation

$F = -\frac{GMm}{r^2}\hat{r}$ (Negative sign: force is attractive, directed inward.)

Gravitational Field

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

Inside a uniform spherical shell: $g = 0$.

Inside a solid sphere at radius $r < R$: $g = \frac{GM_r}{r^2}$ where $M_r = M(r/R)^3$, so $g = \frac{GMr}{R^3}$.

Gravitational Potential Energy

$U = -\frac{GMm}{r}$

Derived from: $U = -\int_\infty^r F\,dr' = -\int_\infty^r \frac{GMm}{r'^2}\,dr'$.

Reference: $U = 0$ at $r = \infty$.

Escape Speed

Total energy = 0 at escape: $\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$ $v_e = \sqrt{\frac{2GM}{R}}$

Orbits

Circular orbit: $v = \sqrt{\frac{GM}{r}}, \quad T^2 = \frac{4\pi^2}{GM}r^3$

Total energy in orbit: $E = KE + U = \frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{2r}$

• Bound orbit: $E < 0$.
• Escape: $E \geq 0$.

Kepler's Laws

1. Planets orbit in ellipses with the Sun at one focus.
2. A line from the Sun to a planet sweeps equal areas in equal times (conservation of angular momentum).
3. $T^2 \propto a^3$ where $a$ is the semi-major axis.

Problem: Find the escape speed from Earth's surface. ($M_E = 5.97 \times 10^{24}\ \text{kg}$, $R_E = 6.37 \times 10^6\ \text{m}$)

Solution:

$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2(6.674 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}}$ $= \sqrt{\frac{7.97 \times 10^{14}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 11{,}200\ \text{m/s} \approx 11.2\ \text{km/s}$

• $U = -GMm/r$ with $U = 0$ at infinity.
• Escape speed: $v_e = \sqrt{2GM/R}$.
• Circular orbit: $v = \sqrt{GM/r}$, $E = -GMm/(2r)$.
• Kepler's second law is conservation of angular momentum.
• Inside a uniform sphere, $g \propto r$; inside a shell, $g = 0$.