Study Notes on Gravitation

Chapter 13: Gravitation

The Goals of this Chapter

• Understand the following definitions:
    - Geosynchronous Orbit: An orbit where the satellite has a period equal to the rotation of the Earth (24 hours).
    - Aphelion/Apogee: The furthest distance from the Sun/Earth in an orbit.
    - Perihelion/Perigee: The closest distance to the Sun/Earth in an orbit.

• Recognize when to apply:
    - The new formulae concerning gravitational force and gravitational potential energy.

• Apply Kepler’s Laws:
    - To analyze both elliptical and spherical orbits.

• Understand Escape Velocity:
    - The minimum velocity required for an object to break free from a gravitational field without further propulsion.

Newton’s Law of Universal Gravitation

• Conceptual Inquiry:
    - Explore the relationship between gravitational force acting upon an apple and the moon in orbit.
    - When throwing an apple horizontally from a mountain, it falls, illustrating Earth's curvature.
    - The challenge is to throw the apple fast enough to continuously fall without moving closer to the ground—a demonstration of orbital motion.

• Illustration:
    - Newton provided sketches in his work "Principia" of 1687 showing these concepts visually.

The Inverse Square Law

• Fundamentals:
    - The same gravitational force acts on both the apple and the moon, affirming its universal nature.

• Characteristics of Gravitational Force:
    - The gravitational force is:
      - Always attractive.
      - Proportional to the product of the two masses involved.
      - Inversely proportional to the square of the distance between them, expressed mathematically as:
        $F = G \frac{m_1 m_2}{r^2}$

• Value of Gravitational Constant (G):
    - $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$

• Force Vector Notation:
    - $\mathbf{F} = -\frac{G m_1 m_2}{r^2} \hat{r}$

• Key Variables:
    - $m_1$, $m_2$: masses of the two bodies.
    - $r$: distance between the centers of the two masses.

Gravitational Attraction Between Spherical Bodies

• Complex Calculations:
    - For arbitrary objects, gravitational attraction requires integration over the object's mass distribution:
      $\mathbf{F} = \sum_i G \frac{m M_i}{r^2} \hat{r} = \int G m \rho(\mathbf{r}) dV \frac{1}{r^2} dV$

• Simplified Approach for Spherical Bodies:
    - For spherical objects (or nearly spherical), treat all mass as concentrated at the center provided the other mass is located outside the sphere.
    - This simplification is proven through calculus in section 13.6 of the textbook.

Connecting G and g

• Weight Calculation:
    - Relationship between gravitational force and weight can be expressed as:
      $F = mg = G \frac{m M_E}{r^2}$
    - Rearranging gives:
      $g = G \frac{M_E}{r^2}$

• Sea Level Gravitational Acceleration:
    - $g_{Sea Level} = G \frac{M_E}{R_E^2}$
    - Calculation yields:
      $g = (6.67 \times 10^{-11)(5.97 \times 10^{24})}{(6.37 \times 10^{6})^2} = 9.81344 \, m/s^2$

• Gravitational Acceleration at Different Elevations:
    - Example for Mt. St. Helens elevation (2550m):
      $g_{Mt. St. Helens} = G \frac{M_E}{(R_E + h)^2} = 9.80559 \, m/s^2$
      - Sea Level value: $1.0008 imes g$.

Earth’s Gravitational Field at the Surface

• Graphical Representation:
    - Gravitational field representation at various distances from the mass at the origin.
    - The field shows how the mass influences surrounding space; actual force exists only when another mass is present in the field.

Example Problem #1: Weight on Another Planet

• Problem Setup:
    - Weight of person on Earth = 200 lbs.
    - Given data:
      - Mass of Mars: $M_{Mars} = 6.39 \times 10^{23} \, kg$
      - Radius of Mars: $R_{Mars} = 3390 \, km$
      - Mass of Earth: $M_{Earth} = 5.97 \times 10^{24} \, kg$
      - Radius of Earth: $R_{Earth} = 6370 \, km$

• Calculation of Weight:
    - Using:
      $w_{Mars} = w_{Earth} \frac{M_{Mars}}{M_{Earth}} \left( \frac{R_{Earth}}{R_{Mars}} \right)^2$
      - Substituting values yields:
        $w_{Mars} = (200 \, lbs) \times \frac{6.39 \times 10^{23} \, kg}{5.97 \times 10^{24} \, kg} \times \left( \frac{6370 \, km}{3390 \, km} \right)^2 = 75.6 \, lbs$

The Combo Platter: Inside a Solid Sphere

• Gravitational Force at Center of Earth:
    - Condition assumed: human survivability under heat and pressure.
    - Derive mass: $M = \rho \frac{4}{3} \pi r^3$
      - For a portion of Earth: $m_E \left( \frac{r^3}{R_E^3} \right)$.

• Expression for Gravitational Force:
    - At the center:
      $F = G \frac{m m_E}{R_E^3} r$

Gravitational Potential Energy

• Connection to Conservative Forces:
    - Result from previous chapters stating that conservative forces and potential energy are related:
      $\mathbf{F} = -\nabla U$
    - For gravitational potential energy:
      - $U = -G \frac{m_1 m_2}{r}$

Connecting Potential Energy Forms

• Relationship:
    - From different expressions of potential energy:
      $U(R_E + h) - U(R_E) = - G \frac{m_E m}{R_E + h} + G \frac{m_E m}{R_E}$
    - Rearranging leads to the approximation:
      $U(R_E + h) - U(R_E) \approx mg h$ where $g = G \frac{m_E}{R_E^2}$.

Concepts of Gravitational Energy

• Visualizing Energy Differences:
    - Potential energy expressions in different forms:
      - $F: \frac{2}{R_E}, U: \frac{3}{R_E}$ for different altitudes using gravitational relations.

Escape Velocity

• Understanding Escape Velocity:
    - Defined as the speed needed to overcome gravitational influence:
      $0 = U + K <br>ightarrow 0 = -G \frac{m M}{r} + \frac{1}{2} mv_e^2$
    - Rearranging gives:
      $v_e = \sqrt{\frac{2GM}{r}}$
    - Example escape velocity from Earth:
      $v_{e, Earth} = 11,180 \, m/s$ (approximately 25,000 mph).

Example Problem #2: Astronaut in Space

• Scenario:
    - Astronaut of mass 95 kg at rest, stranded 12 m away from a spaceship of mass 880,000 kg.
    - Spaceship radius: 15.4 m.

• Using Conservation of Energy:
    - Initial energy conservation: $U_0 + K_0 = U_f + K_f - G \frac{ma m_s}{r_0}$
    - Calculation involving gravitational attraction between the astronaut and spaceship yields:
      $v = \sqrt{2G \frac{m_s}{(r_f - r_0)}}$

• Time to Return:
    - Estimating time required to return to the ship:
      $t_{est} = 2 \frac{\Delta x}{a_{est}} = 11900 \, s \approx 3.305 \, hr$
    - Comparative analysis of throwing away flashlight with a velocity of 5.20 m/s.

• Final Calculations for Velocity and Acceleration:
    - Using gravitational force formula yield spacecraft moving at a deceleration due to gravitational pull.

Kepler’s Laws of Motion

• 1st Law:
    - Planets move along elliptical paths with the Sun at one focus.
    - Circular orbits are a specific case of this. (e.g., the moon's orbit around Earth).

• Definitions:
    - Semi-major Axis ($a$): Half the ellipse's longest axis.
    - Semi-minor Axis ($b$): Half the ellipse's shortest axis.
    - Eccentricity ($e$): Measure of how much the orbit deviates from circularity. Defined as:
      $e = \sqrt{1 - \left(\frac{b}{a}\right)^2}$
    - Perihelion/Perigee: Closest distance to the Sun/Earth.
    - Aphelion/Apogee: Farthest distance from the Sun/Earth.

Kepler’s Laws of Motion – 2nd Law

• Description:
    - Equal area is swept over equal time, implying that planets move faster when closer to the Sun and slower when further away.
    - Associated changes in kinetic energy.

Kepler’s Laws of Motion – 3rd Law

• Period and Semi-major Axis Relationship:
    - $T \propto a^{3/2}$, where $T$ is the orbital period and $a$ is the semi-major axis.
    - Calculation for the period:
      $T = 2 \pi a^{3/2} \frac{1}{\sqrt{GM}}$

• Astronomical Units:
    - 1 AU is the average distance from the Sun to Earth.
    - Equivalent notation for circular orbits utilizes $r$.

Kepler’s Laws of Motion – 3rd Law Extended

• Graphs of Planetary Periods:

• Various planets, dwarf planets, and asteroids graphed against their semi-major axes, indicating clear relationships reinforcing Kepler's 3rd Law.

Example Problem #4: Work Required for Satellite Change

• Context:
    - Satellite being raised from Low Earth Orbit (~800 km) to geosynchronous orbit.

• Gravitational Calculations:
    - Key values:
      - Mass of Earth: $M_E = 5.97 \times 10^{24} \, kg$.
      - Final calculation indicating the work required in Joules and comparing household energy consumption equivalents.

Halley’s Comet Perihelion and Aphelion Analysis

• Calculating Distances:
    - Based on gravitational potential energy and angular momentum preservation:
      $U_0 + K_0 = U_f + K_f$ for aphelion distance calculations, reinforcing conservation principles relating to eccentric orbits.