rich mathematical problems in astronomy

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Sandra Miller and Stephanie Smith Lamar High School Arlington, TX

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Presentation given at SEEC 2014.

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Page 1: Rich Mathematical Problems in Astronomy

Sandra Miller and Stephanie Smith

Lamar High School

Arlington, TX

Page 2: Rich Mathematical Problems in Astronomy
Page 3: Rich Mathematical Problems in Astronomy

This problem is

designed to occur

during a Geometry

unit on circles.

�A line tangent to a

circle forms a right

angle with a radius

drawn at the point

of tangency.

Page 4: Rich Mathematical Problems in Astronomy

�r – radius of the

planet/moon

�h – height of the

observer (eyes)r

r

h

d

observer (eyes)

�d – distance to the

horizon

r

Page 5: Rich Mathematical Problems in Astronomy

�r – radius of the

planet/moon

�h – height of the

observer (eyes)r

r

h

d

observer (eyes)

�d – distance to the

horizon

r

( )= + −2 2d r h r

= + + −2 2 22d r rh h r

( )= +2d h r h

Page 6: Rich Mathematical Problems in Astronomy

Object Radius Horizon

Earth 3959 mi. 3 mi.

Moon 1080 mi.Moon 1080 mi.

Mars 2106 mi.

Jupiter 43,441 mi.

Page 7: Rich Mathematical Problems in Astronomy

Object Radius Horizon

Earth 3959 mi. 3 mi.

Moon 1080 mi. 1.6 mi.Moon 1080 mi. 1.6 mi.

Mars 2106 mi. 2.2 mi.

Jupiter 43,441 mi. 9.9 mi.

Page 8: Rich Mathematical Problems in Astronomy
Page 9: Rich Mathematical Problems in Astronomy

� This problem set is geared toward a Pre-AP

Algebra I class or an Algebra II class.

� By working through this packet, a student

will practicewill practice

� Simplifying literal equations

� Creating formulas

� Unit conversions

� Using formulas to solve problems

Page 10: Rich Mathematical Problems in Astronomy

Sir Isaac Newton developed three equations

that we will use to develop some interesting

information about the solar system.

When a force F acts on a body of mass m, it When a force F acts on a body of mass m, it

produces in it an acceleration a equal to

the force divided by the mass.

The centripetal acceleration a of any body

moving in a circular orbit is equal to the

square of its velocity v divided by the

radius r of the orbit.

The grativational force F between two

objects is proportional to the product of

their two masses, divided by the distance

between them.

=F ma

=2v

ar

= 1 2

2

GmmF

r

Page 11: Rich Mathematical Problems in Astronomy

� If we substitute the formula for centripetal

acceleration into the F = ma equation, we

have an equation for the orbital force:

= =

2 2v mvF m

� The gravitational force that the object being

orbited exerts on its satellite is

= =

v mvF m

r r

=2

GmMF

r

Page 12: Rich Mathematical Problems in Astronomy

�Objects that are in orbit stay in orbit

because the force required to keep them

there is equal to the gravitational force that

the object being orbited exerts on its

satellite.satellite.

� If we set our two equations equal to each

other and solve for v, we end up with a

formula that will give us the orbital speed of

the satellite.

Page 13: Rich Mathematical Problems in Astronomy

� Simplify the equation and solve for v:

=2

2

mv GmM

r r

Page 14: Rich Mathematical Problems in Astronomy

� Simplify the equation and solve for v:

=2

2

mv GmM

r r

=2 GmMmv =2 GmMmv

r

=2 Gmv

r

=GM

vr

Page 15: Rich Mathematical Problems in Astronomy

� Because the mass of the satellite m

cancelled out of the equation, if we know

the orbital velocity and the radius of the

orbit, we can find the mass of the object

being orbited.being orbited.

Page 16: Rich Mathematical Problems in Astronomy

� Rewrite the velocity equation and solve for

M:

=2 GMv

rr

Page 17: Rich Mathematical Problems in Astronomy

� Rewrite the velocity equation and solve for

M:

=2 GMv

rr

=2v r GM

=2v r

MG

Page 18: Rich Mathematical Problems in Astronomy

� Example: Use the Moon to calculate the

mass of the Earth.

�Orbital radius: = × 83.84 10 mr

� Period: T = 27.3 days

�Orbital velocity: =circumference of orbit

period of orbitv

Page 19: Rich Mathematical Problems in Astronomy

� Example: Use the Moon to calculate the

mass of the Earth.

π=

2 rv

TT( )π ×

=

82 3.84 10

24 hours 3600 seconds27.3

1 day 1 hour

= m1023 s

Page 20: Rich Mathematical Problems in Astronomy

� Example: Use the Moon to calculate the

mass of the Earth.

=2v r

MG

− = ×

2112

m6.67 10 N kg

GMG

kg

= × 246.02 10 kg

Page 21: Rich Mathematical Problems in Astronomy

� To calculate escape velocity, we set the

equation for kinetic energy to the equation

for gravitational force and solve for v:

Kinetic energy > Force × distanceKinetic energy > Force × distance

> i2

2

1

2

GmMmv r

r

>2 2GMv

r

>2GM

vr

Page 22: Rich Mathematical Problems in Astronomy

Calculate Earth’s escape velocity in km/s.

� Earth’s mass: 6.02 × 1024 kg

� Earth’s radius: 6.38 × 106 m� Earth’s radius: 6.38 × 10 m

> km11.22 s

v

Page 23: Rich Mathematical Problems in Astronomy

�Now that we’ve worked through the different

equations, we can calculate the mass and

escape velocity of Mars as well as the mass

of the Sun.

Page 24: Rich Mathematical Problems in Astronomy

One of my favorite

sites for possible

astronomy-related astronomy-related

math problems has

been Space Math athttp://spacemath.gsfc.nasa.gov.

Unfortunately, because of cutbacks in

NASA’s education budget, it will not be

updated as frequently.

Page 25: Rich Mathematical Problems in Astronomy

Invert the problem

Ask for prediction

Ask for an explanation:

oral or written

Original (Standard) Problem

Break into multiple parts

Ask for multiple

representation

Ask questions that require qualitative reasoning

Automaticity practice

Ask for generalization

Examples or counter-examples

James Epperson, Ph.D.

Page 26: Rich Mathematical Problems in Astronomy

� The powerpoint and the worksheets will be

posted on my blog at

tothemathlimit.wordpress.com.