3. 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.
4. d
r
h
r
r – radius of the
planet/moon
h – height of the
observer (eyes)
d – distance to the
horizon
5. d
r
h
r
r – radius of the
planet/moon
h – height of the
observer (eyes)
d – distance to the
horizon
d=
r + h) − r 2
(
2
d = r 2 + 2rh + h2 − r 2
d = h ( 2r + h )
9. 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 practice
Simplifying literal equations
Creating formulas
Unit conversions
Using formulas to solve problems
10. Sir Isaac Newton developed three equations
that we will use to develop some interesting
information about the solar system.
F = ma
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.
v
a=
r
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.
Gm1m2
F=
r2
The grativational force F between two
objects is proportional to the product of
their two masses, divided by the distance
between them.
2
11. If we substitute the formula for centripetal
acceleration into the F = ma equation, we
have an equation for the orbital force:
v 2 mv 2
F = m =
r
r
The gravitational force that the object being
orbited exerts on its satellite is
GmM
F= 2
r
12. 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.
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.
14. Simplify the equation and solve for v:
mv 2 GmM
= 2
r
r
GmM
2
mv =
r
Gm
v =
r
GM
v=
r
2
15. 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.
18. Example: Use the Moon to calculate the
mass of the Earth.
Orbital radius: r = 3.84 × 10 8 m
Period: T = 27.3 days
circumference of orbit
Orbital velocity: v =
period of orbit
19. Example: Use the Moon to calculate the
mass of the Earth.
2πr
v=
T
=
2π ( 3.84 × 108 )
24 hours 3600 seconds
27.3
1 hour
1 day
= 1023 m s
20. Example: Use the Moon to calculate the
mass of the Earth.
G = 6.67 × 10 −11 N m2
kg2
v 2r
M=
G
= 6.02 × 10
24
kg
21. To calculate escape velocity, we set the
equation for kinetic energy to the equation
for gravitational force and solve for v:
Kinetic energy > Force × distance
1
GmM
2
mv > 2 i r
2
r
2GM
2
v >
r
v>
2GM
r
22. Calculate Earth’s escape velocity in km/s.
Earth’s mass:
Earth’s radius:
6.02 × 1024 kg
6.38 × 106 m
v > 11.22 km s
23. 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.
24. One of my favorite
sites for possible
astronomy-related
math problems has
been Space Math at
http://spacemath.gsfc.nasa.gov.
Unfortunately, because of cutbacks in
NASA’s education budget, it will not be
updated as frequently.
25. Invert the
problem
Ask for an
explanation:
oral or written
Examples or
counterexamples
Ask for
prediction
Break into
multiple parts
Original
(Standard)
Problem
Ask for
multiple
representation
Ask for
generalization
Automaticity
practice
Ask questions
that require
qualitative
reasoning
James Epperson, Ph.D.
26. The powerpoint and the worksheets will be
posted on my blog at
tothemathlimit.wordpress.com.