Chapter 13: Gravitation

Chronological History of Gravitation and Orbital Mechanics

The development of gravitational theory is characterized by a progression of observations and mathematical refinements from ancient Greece to the early 20th century:

• Aristotle (c. -350, Greece): Proposed that objects have "Natural tendencies" to move toward their natural place in the universe.
• Nicolaus Copernicus (1530, Poland): Introduced the model of Heliocentrism, placing the Sun at the center of the solar system.
• Tycho Brahe (1580, Denmark): Emphasized that "Observations rule!" and provided the detailed astronomical data necessary for later laws.
• Johannes Kepler (1610, Germany): Formulated the 3 Planetary Laws based on Brahe's data.
• Galileo Galilei (1610, Italy): Utilized the telescope for astronomical observation and developed early concepts of Relativity.
• Isaac Newton (1680, England): Developed the Universal Law of Gravitation, classical Mechanics, and Calculus.
• Albert Einstein (1910, Germany): Revolutionized the field with the General Theory of Relativity, Quantum Theory, and a new understanding of Gravitation.

Johannes Kepler and the Laws of Orbital Motion

Based on exhaustive studies of the apparent motions of planets over many years, Kepler formulated three empirical laws:

1. First Law: Planets follow elliptical orbits, with the Sun situated at one focus of the ellipse.
2. Second Law: As a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time.
3. Third Law: The ratio of the square of the period, $T$, of any planet's orbit is proportional to the cube of its average distance from the Sun, $r$:
   • $\frac{T^2}{r^3} = \text{constant}$

Newton's Law of Universal Gravitation

Newton bridged celestial and terrestrial mechanics by realizing that the force accelerating an apple downward on Earth is the same force that keeps the Moon in its orbit. This led to his universal law:

• Action-Reaction Pairs: The two gravitational forces between objects are an action-reaction pair; they are equal in magnitude and opposite in direction.
• Force Magnitude: The magnitude of the gravitational force between two particles with masses $m_1$ and $m_2$ separated by a distance $r$ is:
  • $F_g = \frac{G m_1 m_2}{r^2}$
• Direction of Force: The gravitational force is always attractive and acts along the line connecting the centers of the two masses.

Case Study: Earth and Moon as Seen from OSIRIS-Rex

Data from the NASA OSIRIS-Rex mission provides a perspective on the Earth-Moon system:

• Earth to Spacecraft Distance: $5,420,000\,km$
• Earth to Moon Distance: $390,000\,km$
• Moon to Spacecraft Distance: $5,120,000\,km$

The Principle of Superposition

Objects are often subject to gravitational forces from multiple masses simultaneously. The net gravitational force on an object is the vector sum of all individual gravitational forces acting upon it:

• $\mathbf{F}_{net} = \mathbf{F}_1 + \mathbf{F}_2 + \dots + \mathbf{F}_n$

Gravitation Near Spherical Bodies and Surfaces

• Point Mass and Sphere: The gravitational force attracting a point mass to a sphere (of mass $M$) is the same as if the entire mass of the sphere were concentrated at its geometric center.
• Surface Gravity Calculation: For an object of mass $m$ at the surface of a planet of mass $M$ and radius $R$, the force is calculated as if the mass were a point one radius away:
  • $F_g = \frac{G M m}{R^2}$
  • Since $F_g = mg$, the acceleration due to gravity is: $g = \frac{G M}{R^2}$

Newtonian Explanation of Kepler’s Laws

All three of Kepler’s laws are a direct consequence of the inverse square nature of the gravitational force. For the special case of a circular orbit:

• The centripetal force is provided by gravity: $\frac{m v^2}{r} = \frac{G M m}{r^2}$
• Simplifying reveals the orbital speed: $v = \sqrt{\frac{G M}{r}}$
• Because $v = r \omega$ and $\omega = \frac{2\pi}{T}$, we derive Kepler's 3rd Law:
  • $T^2 = \frac{4\pi^2 r^3}{G M}$
• Note: This relationship is independent of the mass of the orbiting body ($m$).

Gravitational Potential Energy ($U$)

• Definition: Gravitational potential energy between two masses is: $U = -\frac{G M m}{r}$
• Zero Reference: The choice of zero potential energy is at infinity ($r = \infty$), resulting in negative values for all finite distances. This is arbitrary; physical significance lies in the difference specifically: $\Delta U$.
• Distance Relationship: The potential energy of a system (e.g., Earth-astronaut) increases (becomes less negative) as the distance between the two bodies increases.
• Uniform Spherical Shell: The gravitational potential energy of a mass outside a uniform spherical shell is the same as if all the shell's mass were concentrated at the center.

Mathematical Derivation: $U$ for a Uniform Ring

For a point mass $m$ at distance $x$ on the axis of a uniform ring of mass $M$ and radius $r$:

1. Distance from element $dM$ to mass $m$ is $s = \sqrt{x^2 + r^2}$.
2. Potential contribution: $dU = -\frac{G m (dM)}{s}$.
3. Total potential: $U = -\int \frac{G m \, dM}{\sqrt{x^2 + r^2}}$.
4. Since $\sqrt{x^2 + r^2}$ is constant for the whole ring:
   • $U = -\frac{G M m}{\sqrt{x^2 + r^2}}$

The Motion of Satellites and Projectiles

• Trajectory: The path of a projectile fired from a great height depends entirely on its initial speed (ignoring air resistance).
• Circular Satellite Orbits:
  • The speed ($v$) is precisely calibrated so the distance from the Earth's center remains constant.
  • Gravity ($F_g$) provides the necessary centripetal acceleration ($a_{rad}$).
  • A satellite is considered to be constantly "falling around" the Earth.
  • Apparent Weightlessness: Astronauts experience this because they and the satellite are falling together with the same acceleration.
• Orbit Types (Conic Sections): Under an inverse square force, orbits following classical paths are conic sections: circles, ellipses, parabolas, or hyperbolas.

Apparent Weight and Earth's Rotation

Rotation affects the measurement of weight on Earth:

• At the North or South Pole: Apparent weight is equal to true weight because there is no radial acceleration from rotation.
• Away from the Poles: Apparent weight ($w$) is not equal to true weight ($w_0$) due to centripetal acceleration ($a_{rad}$).
• Formulas:
  • True weight: $w_0 = m g_0$
  • Apparent weight: $w = m g$
  • Vector relationship: $\mathbf{g} = \mathbf{g}_0 - \mathbf{a}_{rad}$

Variations in Acceleration Due to Gravity ($g$)

Questions & Discussion (Checkpoints)

Checkpoint 1: Gravitational Force Comparison The mass of the Moon is $1/81$ of the mass of the Earth. Compared to the gravitational force the Earth exerts on the Moon, the force the Moon exerts on the Earth is:

• A. $81^2 = 6561$ times greater.
• B. $81$ times greater.
• C. equally strong.
• D. $1/81$ as great.
• E. $(1/81)^2 = 1/6561$ as great.
• Answer: E (as per transcript notation).

Checkpoint 2: Planet X Surface Gravity Compared to Earth, Planet X has twice the mass and twice the radius. This means the surface gravity on Planet X is:

• A. four times as much.
• B. twice as much.
• C. the same.
• D. half as much.
• E. one-quarter as much.
• Answer: E (as per transcript notation).

Checkpoint 3: Shrinking Sun If the Sun were to shrink to half its radius while maintaining the same mass, what happens to the radius $r$ and period $T$ of Earth's orbit?

• A. $r$ decrease, $T$ decrease.
• B. $r$ increase, $T$ increase.
• C. $r$ decrease, $T$ increase.
• D. $r$ increase, $T$ decrease.
• E. $r$ and $T$ are unchanged.
• Answer: E.

Checkpoint 4: Orbit Torque As a planet moves in an elliptical orbit, the Sun exerts a force pointing directly toward the Sun. What is the torque that the Sun exerts on the planet?

• A. Constant and nonzero.
• B. Greatest when the planet is closest.
• C. Least but not zero when the planet is closest.
• D. Zero when the planet is closest.
• E. Zero at all times.
• Answer: E.

Problem: Geosynchronous Satellites

Definition: A satellite with an orbital period of exactly one day ($T = 1\,day$). These stay over the same point on Earth if they orbit in the equatorial plane in the direction of rotation.

Calculation parameters provided:

• Moon orbit radius: $384,399\,km$
• Moon orbital period: $27.321\,days$
• Task: Calculate the radius of a geosynchronous orbit using the provided Moon data (applying Kepler's 3rd Law).