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Announcements

- Exam 1 is two weeks away. Will be similar to homework problems. Bring a calculator.
- Homework Set 3: Chapter 3 #27, 29 & 32 + Supplemental Problems

Kepler’s Laws of Planetary Motion

Empirical laws developed by Johannes Kepler based on the observational data of Tycho Brahe

Kepler’s 1st Law

The Law of Ellipses

The planets move in elliptical orbits with the Sun located at one focus

Kepler’s 2nd Law

Law of Areas

A line drawn from a planet to the Sun will sweep out equal areas in equal time periods

Kepler’s 3rd Law

The Law of Harmonies

The ratio of the square of the orbital period to the cube of the orbital semimajor axis (the radius) is the same for all the planets

For objects orbiting the Sun, if the orbital period (P) is measured in years and the semimajor axis (A) is measured in AU then

If units other than AU and years are used or the object is orbiting something other than the Sun, then a conversion constant must be added

Using Kepler’s 3rd Law

The constant k can be determined by looking at the orbital properties of other bodies orbiting the same parent body. Be sure to use consistent units.

For the Sun: Earth orbits the Sun in 1 year with a semimajor axis of 1 AU so

Or, using 1 AU =1.496x1011m and 1 year = 3.156x107s

Using Kepler’s 3rd Law

A new asteroid is discovered orbiting the Sun in a circular orbit at a distance of 8 AU. What is it’s orbital period?

Jupiter’s moon Io orbits Jupiter at an average distance of 421,600 km with an orbital period of 1.769 days. If a new moon is discovered orbiting Jupiter at an average distance of 2,350,000 km, what will it’s orbital period be?

Solution 1

Given: orbital radius (A) = 8AU Also know orbital radius of Earth (1.0AU) and orbital period of Earth (1year)

Equation to use:

In this case, using orbital radius in AU and period in years, the constant is 1AU/year so

Solution 2

Given: Io orbital radius = 421,600 km Io orbital period = 1.769 days

New moon orbital radius = 2,350,000 km

Rather than calculating what the constant is, set up the problem as equal ratios

Newton’s 1st Law of Motion

The Law of InertiaAn object in straight line uniform motion will continue that motion unchanged unless some external force acts on it

This was law based entirely on the work of Galileo

Newton’s Second Law

The Force Law: F = maThe acceleration a body experiences is directly proportional to the net force acting on it and inversely proportional to its mass

Acceleration and force are actually vectors

Vectors have both magnitude and direction

A vector can be broken into horizontal and vertical components using a few simple trigonometric relationships. These relationships come from the Pythagorean Theorem relating the sides and angles of a right triangle.

Newton’s Third Law

The Action-Reaction LawFor every force there is an equal and opposite reaction force

Using Newton’s Laws

A 215 kg sailboat experiences a force from the wind of 75.0 N at 30° north of east while the current pushes it with a force of 50.0 N due south. What is the magnitude and direction of the acceleration of the boat?

Solution

North-South Components

Wind: (75.0N)sin30° = 37.5N

Current: (50.0N)sin270° = -50.0N

Net = 37.5N – 50.0N = -12.5N

East-West Components

Wind: (75.0N)cos30° = 64.45N

Current: (50.0N)cos270° = 0.0N

Net = 64.45N + 0.0N = 64.45N

The acceleration is 0.31 m/s2 at 10° south of east

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