Newton's Law of Gravity and Orbital Motion: Study Notes

Newton's Law of Gravity and Orbital Motion

Section 6.5: Newton's Law of Gravity

Newton's Law of Gravity describes the universal force of attraction between all masses in the universe. This law is fundamental to understanding planetary motion, satellite orbits, and the behavior of objects under gravitational influence.

• Gravity as a Universal Force: Gravity acts between all objects with mass, regardless of their location in the universe.

• Inverse-Square Law: The gravitational force between two objects is inversely proportional to the square of the distance between their centers.

• Direct Proportionality to Mass: The force is directly proportional to the product of the masses of the two objects.

Mathematical Formulation:

The gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by:

$F_{12} = F_{21} = G \frac{m_1 m_2}{r^2}$

• $G$ is the gravitational constant: $G = 6.67 \times 10^{-11} \ \mathrm{N \cdot m^2 / kg^2}$

• The forces are equal in magnitude and opposite in direction (Newton's Third Law).

• Doubling the distance between two masses reduces the force by a factor of 4.

Example: Gravitational Force Between Two Spheres

• Given: Force between two spheres is $0.010\ \mathrm{N}$ at $20\ \mathrm{m}$ apart. What is the distance when the force is $0.160\ \mathrm{N}$?

• Solution uses the ratio of forces and the inverse-square law:

$\frac{F_{\text{far}}}{F_{\text{close}}} = \left(\frac{d_{\text{close}}}{d_{\text{far}}}\right)^2$

$\frac{0.010}{0.160} = \left(\frac{d_{\text{close}}}{20}\right)^2 \implies d_{\text{close}} = 5\ \mathrm{m}$

• Key Point: The force increases by a factor of 16 when the distance decreases by a factor of 4.

Example: Gravitational Force Between Two People

• Estimate the force between two people (each $65\ \mathrm{kg}$) sitting $0.60\ \mathrm{m}$ apart:

$F = G \frac{m_1 m_2}{r^2} = (6.67 \times 10^{-11}) \frac{(65)(65)}{(0.60)^2} = 7.8 \times 10^{-7}\ \mathrm{N}$

• Interpretation: This force is extremely small, about the weight of a single hair.

Conceptual Questions (QuickCheck)

• Newton's Third Law: The force of Planet Y on Planet X is equal in magnitude to the force of Planet X on Planet Y.

• Effect of Distance: If the distance between two asteroids is doubled, the gravitational force becomes one-fourth as large.

Gravity on Other Worlds

Weight depends on the gravitational field of the planet or moon, not just on mass. The acceleration due to gravity at the surface of a planet is determined by its mass and radius.

• Weight on Another Planet: $w = mg$

• Using Newton's Law: $F = G \frac{M_{\text{planet}} m}{R_{\text{planet}}^2}$

• Equating the two gives the surface gravity:

$g_{\text{planet}} = G \frac{M_{\text{planet}}}{R_{\text{planet}}^2}$

• Example: On the Moon, $g_{\text{moon}} = 1.62\ \mathrm{m/s^2}$

• A 70-kg astronaut with an 80-kg suit weighs 330 lb on Earth, but only 54 lb on the Moon.

QuickCheck Examples

• If Planet Y has twice the mass and twice the radius of Planet X, its surface gravity is half as much.

• For a given mass, the greatest weight is on the planet with the smallest radius (for the same mass).

Section 6.6: Gravity and Orbits

Newton's law of gravity explains the motion of satellites and planets in orbit. The gravitational force provides the necessary centripetal force for circular orbits.

• Newton's Second Law for Circular Orbits:

$F_{\text{gravity}} = ma = m \frac{v^2}{r}$

$G \frac{Mm}{r^2} = m \frac{v^2}{r}$

• Orbital Speed:

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

• Orbital Period: The time $T$ to complete one orbit is related to the radius and speed:

$v = \frac{2\pi r}{T}$

$T^2 = \frac{4\pi^2}{GM} r^3$

• This is a form of Kepler's Third Law for circular orbits.

Example: Speed Required to Orbit Deimos

• Given: $M_{\text{Deimos}} = 1.8 \times 10^{15}\ \mathrm{kg}$, $R_{\text{Deimos}} = 6.3 \times 10^3\ \mathrm{m}$

• Surface gravity: $g_{\text{Deimos}} = 0.0030\ \mathrm{m/s^2}$

• Orbital speed: $v_{\text{orbit}} = \sqrt{g r} = 4.3\ \mathrm{m/s}$

• Interpretation: This is a very low speed; a strong jump could launch you into orbit around Deimos.

Example: Locating a Geostationary Satellite

• For a satellite to remain stationary above the equator, its period must be 24 hours ($T = 8.64 \times 10^4\ \mathrm{s}$).

• Solving for the orbital radius:

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

$M_{\text{Earth}} = 5.98 \times 10^{24}\ \mathrm{kg}$

$r \approx 4.22 \times 10^7\ \mathrm{m}$ (about 7 times Earth's radius)

• The International Space Station orbits at a radius only about 5% larger than Earth's radius.

Gravity on a Grand Scale

• Gravity acts over astronomical distances, holding galaxies together.

• Stars in a galaxy orbit the center with different periods depending on their distance from the center.

Example: Orbital Period of Phobos

• Given: $r = 9.4 \times 10^6\ \mathrm{m}$, $M_{\text{Mars}} = 6.4 \times 10^{23}\ \mathrm{kg}$

• Using $T^2 = \frac{4\pi^2}{GM} r^3$, the period $T \approx 7.8$ hours.

• This is much shorter than a Martian day (about 25 hours).

Summary of Key Concepts

• Uniform Circular Motion: An object moving in a circle at constant speed has a centripetal acceleration directed toward the center: $a = \frac{v^2}{r}$

• Net Force for Circular Motion: $F_{\text{net}} = m a = m \frac{v^2}{r}$

• Universal Gravitation: $F = G \frac{m_1 m_2}{r^2}$

• Period and Frequency: $T$ is the time for one revolution; $f = \frac{1}{T}$

• Speed in Circular Motion: $v = 2\pi r f = \frac{2\pi r}{T}$

• Planetary Gravity: $g_{\text{planet}} = G \frac{M_{\text{planet}}}{R_{\text{planet}}^2}$

• Apparent Weight: In orbital motion, the net force is gravity alone, so astronauts feel weightless.

• Orbital Motion: For a satellite in a circular orbit:

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

$T^2 = \frac{4\pi^2}{GM} r^3$

Table: Key Equations and Concepts

Concept	Equation	Description
Newton's Law of Gravity	$F = G \frac{m_1 m_2}{r^2}$	Force between two masses
Surface Gravity	$g = G \frac{M}{R^2}$	Acceleration due to gravity at a planet's surface
Orbital Speed	$v = \sqrt{\frac{GM}{r}}$	Speed for a stable circular orbit
Orbital Period	$T^2 = \frac{4\pi^2}{GM} r^3$	Time for one complete orbit
Centripetal Acceleration	$a = \frac{v^2}{r}$	Acceleration toward the center in circular motion
Frequency	$f = \frac{1}{T}$	Number of revolutions per second

Additional info: These notes cover the core concepts of Newtonian gravity, applications to planetary and satellite motion, and the mathematical relationships governing orbits and gravitational forces. They are suitable for exam preparation in a college-level introductory physics course.