Chapter 13 Gravitation

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# Chapter 13 Gravitation - PowerPoint PPT Presentation

Chapter 13 Gravitation. Newton’s law of gravitation Any two (or more) massive bodies attract each other Gravitational force (Newton\'s law of gravitation) Gravitational constant G = 6.67*10 –11 N*m 2 /kg 2 = 6.67*10 –11 m 3 /(kg*s 2 ) – universal constant.

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Chapter 13

Gravitation

Newton’s law of gravitation

• Any two (or more) massive bodies attract each other
• Gravitational force (Newton\'s law of gravitation)
• Gravitational constantG= 6.67*10 –11 N*m2/kg2 = 6.67*10 –11 m3/(kg*s2) – universal constant

Gravitation and the superposition principle

• For a group of interacting particles, the net gravitational force on one of the particles is
• For a particle interacting with a continuous arrangement of masses (a massive finite object) the sum is replaced with an integral

Chapter 13

Problem 9

Shell theorem

• For a particle interacting with a uniform spherical shell of matter
• Result of integration: a uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell\'s mass were concentrated at its center

Gravity force near the surface of Earth

• Earth can be though of as a nest of shells, one within another and each attracting a particle outside the Earth’s surface
• Thus Earth behaves like a particle located at the center of Earth with a mass equal to that of Earth
• g = 9.8 m/s2
• This formula is derived for stationary Earth of ideal spherical shape and uniform density

Gravity force near the surface of Earth

In reality gis not a constant because:

Earth is rotating,

Earth is approximately an ellipsoid

with a non-uniform density

Gravity force near the surface of Earth

Weight of a crate measured at the equator:

Gravitation inside Earth

• For a particle inside a uniform spherical shell of matter
• Result of integration: a uniform spherical shell of matter exerts no net gravitational force on a particle located inside it

Gravitation inside Earth

• Earth can be though of as a nest of shells, one within another and each attracting a particle only outside its surface
• The density of Earth is non-uniform and increasing towards the center
• Result of integration: the force reaches a maximum at a certain depth and then decreases to zero as the particle reaches the center

Chapter 13

Problem 20

Gravitational potential energy

• Gravitation is a conservative force (work done by it is path-independent)
• For conservative forces (Ch. 8):

Gravitational potential energy

• To remove a particle from initial position to infinity
• Assuming U∞ = 0

Escape speed

• Accounting for the shape of Earth, projectile motion (Ch. 4) has to be modified:

Escape speed

• Escape speed: speed required for a particle to escape from the planet into infinity (and stop there)

Escape speed

• If for some astronomical object
• Nothing (even light) can escape from the surface of this object – a black hole

Chapter 13

Problem 33

Johannes Kepler

(1571-1630)

Tycho Brahe/

Tyge Ottesen

Brahe de Knudstrup

(1546-1601)

• Kepler’s laws
• Three Kepler’s laws
• 1. The law of orbits: All planets move in elliptical orbits, with the Sun at one focus
• 2. The law of areas: A line that connects the planet to the Sun sweeps out equal areas in the plane of the planet’s orbit in equal time intervals
• 3. The law of periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit

First Kepler’s law

• Elliptical orbits of planets are described by a semimajor axisa and an eccentricitye
• For most planets, the eccentricities are very small (Earth\'s e is 0.00167)

Second Kepler’s law

• For a star-planet system, the total angular momentum is constant (no external torques)
• For the elementary area swept by vector

Third Kepler’s law

• For a circular orbit and the Newton’s Second law
• From the definition of a period
• For elliptic orbits

Satellites

• For a circular orbit and the Newton’s Second law
• Kinetic energy of a satellite
• Total mechanical energy of a satellite

Satellites

• For an elliptic orbit it can be shown
• Orbits with different ebut the same a have the same total mechanical energy

Chapter 13

Problem 50

Chapter 13:

Problem 2

2.16

• Chapter 13:
• Problem 4
• 2.13 × 10−8 N;
• (b) 60.6º

• Chapter 13:
• Problem 20
• G(M1 +M2)m/a2;
• (b) GM1m/b2;
• (c) 0

• Chapter 13:
• Problem 32
• 2.2 × 107 J;
• (b) 6.9 × 107 J