# The Physics, Astronomy and Mathematics of the Solar System

## Mathematics at the Frontier of Astronomy

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## SECTION 3

Lesson Plans: Connecting the unit to the curriculum

### Lesson Plan I

Topic: Measurement in Astronomy

Purpose

To investigate the units of measurement used in astronomy

To express/ convert distances in astronomical units

Background Information:

The vastness of space requires us to describe distances and other measurements that are either very large or very minute. These sizes are beyond human comprehension. Therefore for us to grasp this vastness Astronomers developed units of measure that are suitable to describe the situations. Interstellar distance is measured in light years or parsecs, and the mass of the sum, or other stars expressed as
*
M
*
. Scientific notation is used to express very large and very small units. This makes easier to write, read and calculate.

When expressing the measure of human sizes the either SI units of conventional units of ft and inches are used. To express very small units micrometer are used.

*
1 micrometer = 1um = 10
^{
- 6
}
m
*

*
1 nanometer = 1 nm = 10
^{
- 9
}
m
*

To express mass Astronomers use grams, kilogram or solar masses, the following scales are used:

*
1kg = 1000g
*

*
1 M = 1.99 × 10
^{
30
}
kg
*

Astronomers also differentiate between mass and weight. Mass is defined as the measure of material in an object and weight is defined as the pull or force that gravity acts on a body. It is usually expressed in pounds or new tons (
*
1 Newton = 0.225 pounds
*
).

Speed is measured in SI units as meter / seconds. It is also expressed km/s as well as mi/h. The following is the conversion table:

*
1km/s = 10
^{
3
}
m/s
*

*
1km/s = 2237m /h
*

*
1mp/h = 0.447m/s
*

*
1 mp/h = 1.47 ft/s
*

To measure the vast distances across the solar system astronomers use a unit of length called the astronomical unit (AU).

*
1 AU = 1.496 × 10
^{
8
}
km
*

*= 92.96 million miles*

1 AU represents the average distance between the Earth and the Sun.

To measure the distance to the stars astronomers use two different units

- a) Light year (ly): This is the distance that light travels in empty space. The speed of light in empty space has a value of 3.00 ×10
^{ 5 }km/s on 1.86 × 10^{ 5 }mph/ s. Therefore one light year can be expressed as1Ly = 9.46 ×10^{ 1 2 }km= 63,240 AU.- b) Parsec (pc): This unit of length is 3.26 light years.

*1 pc = 3.09 × 10*

^{ 1 3 }km
=
*
3.26 ly
*

*
1 kiloparsec = 1kpc = 1000 pc = 10
^{
3
}
pc
*

*
1 Megaparsec = 1 Mpc = 1,000, 000 pc = 10
^{
6
}
pc
*

Classroom Problems

- 1. The average distance from Earth to the Sun is 1.496 × 10
^{ 8 }km. Express this distance in a) light years b) in parsecs. Express in powers of ten.- 2. The universe is estimated to be about 13 billion years express this in seconds.

### Lesson Plan II

Topic: Using regression to explain Kepler's Law that there is a relationship between the planets' orbits and their distance from the sun.

Materials:

Graph paper, Graphing calculator

Objectives: Students will be able to

- a) Create a scatter plot from the given data
- b) Identify the variables
- c) Write the equation that can be used to model the data

Table 2: Planets distance from the sun and their periods

(table available in print form)

Classroom Problems/ Activities.

1. Draw a scatter plot to show the relationship between the distance from the Sun and the period.

2. Describe the relationship

3. Perform the following transformations of the data. Then explain the graphs

4. Use the result in (ii) to generate a power model.

- i) Distance v/s log (period)
- ii) Log (distance) v/s log (period)

Activity II

The table given shows the relationship between the angle of parallax and the distance of a star in light years.

Table 3: Relationship of Angle to Parallax

Angle of Parallax | Distance in Ly

0.5 sec | 6.5 ly

0.25 sec | 13.0 ly

0.1 sec | 32.6 ly

Use the equation
*
d
*
= 1/p where
*
d
*
= distance and
*
p
*
= parallax angle of a star in arcseconds to add 4 pieces to the table.

- a) Use the table to explain the relationship between the angle of parallax and the distance in light years.
- b) Generate an equation that could be used to model the relationship between angle of Parallax and the distance to a star.
- c) Draw a graph that could be used to model this data.

### Lesson Plan III

Topic: Kepler's contribution to the study of astronomy.

Purpose:

To apply Kepler's laws to a problem situation

To investigate the properties of Ellipses

Make the connection to Newton's Laws

Background Information

Kepler's first law states that the path, or orbit of a planet around the sun is an ellipse, the position of the sun being at a focus of the ellipse.

Definition of an ellipse: A set of all points
*
(x, y)
*
in a plane. The sum of whose distance from the foci is a constant.

A line through the foci intersects the ellipse at two points called the vertices. The chord joining the vertices is the major axis, and its midpoint is the center. The chord perpendicular to the major axis at the center is the minor axis.

Kepler's second law a line joining the planets and the sun sweeps out equal areas during equal intervals of time.

Kepler's third law "the squares of the orbital periods of planets are directly proportional to the cubes of the semi - major axis of the orbits.

*
P
^{
2
}
= a
^{
3
}
*

*
P
*
= the orbital period in years

*
a
*
= semi major axis of orbit

Strategy

Review the meaning of sidereal period for a given planet.

Review the parts of the ellipse

Review the meaning of astronomical units

Discuss Kepler's third law

Worked Sample problem

Significant Task/ Problems

- 1. A piece of solar debris in said to be orbiting the sun. It takes 9 years to complete the orbit. Determine the semimajor axis of the orbit
- 2. The average distance from Mercury to the Sun is 0.39 AU. Determine the sidereal period of Mercury.

Newton's form of Kepler's third law

Kepler's law can only applied to objects in orbit, but Newton's form can be applied to any situation in which two bodies of different masses orbit each other. The equation is given as:

*
P
^{
2
}
*
= [4 Pi

^{ 2 }/ G (m

_{ 1 }+m

_{ 2 })] * a

^{ 3 }, where

*
P
*
= sidereal period of orbits in seconds

*
a
*
= semimajor axis of orbits, in meters

*
m
_{
1
}
*
= mass of first object, in kilograms

*
m
_{
2
}
*
= mass of second object in kilogram

*
G
*
= universal constant of gravitation
*
= 6.67 × 10
^{
- 11
}
N.m
^{
2
}
/kg
^{
2
}
*

Significant Task

Two Hubble telescope on its mission to outer space discovered a new planet. This planet and its moon orbit each other as in the case of the Earth and the Moon. You are an intern in the NASA Space Agency and you are asked to evaluate the data sent back and report the masses of these two planets. From the data you calculated the following information

The orbital distance is 1.888 days and planets are 294,700 Km apart.

### Lesson Plan IV

Topic: Escape Velocity

Purpose: The students will be able to apply the equation of escape velocity to a problem situation.

To have students research the concept of Escape velocity.

Material

Calculator

Significant Task

You are taking part in a science fair and decided to design a spacecraft or a satellite to orbit one of the planets. The rubric for the evaluation specifically states that the escape velocity for the planet you choose to orbit.

Prepare a presentation explaining your design and the escape velocity necessary for the planet of your choice.

### Lesson Plan V

Topic: Comparing the Brightness of stars

Objectives: Students will be able to

- a) Determine the brightness of a star
- b) Differentiate between apparent brightness and absolute brightness
- c) To compare the brightness of two stars

Background Information

The system that modern astronomers use to compare the brightness of a star is based on the classification developed by Hipparcus and formalized by Ptolemy. Stars were placed in six categories according to how bright they appeared when they are seen in the sky without a telescope. The brightest stars were given first magnitude and the dimmest stars were assigned magnitude six. It was recognized that a first magnitude star was 100 times brighter than a sixth magnitude star. To make the computation easier the magnitude difference of 5 corresponds exactly to a factor of 100.The equation
*
x
^{
5
}
= 100
*
establishes the size of a magnitude with

*x = 2.512*. Therefore each magnitude difference in brightness corresponds to a factor of about 2.5. A star of magnitude n is 2.5 times brighter than a star of magnitude

*n+1*.

Sample Problems

- 1. The star that appears brightest in the sky is Sirius. It has an apparent magnitude of - 1.46. The other bright star is Regulus with an apparent magnitude of 1.35. How many times as bright as Regulus is Sirius?
- 2. Two stars are observed in the sky. Star A has an apparent magnitude of 16.9. The other star B has an apparent brightness of 5.0. Which is the brightest and by how much?