Pre-Lesson Instructions:
• Duplicate the Student Sheets ( one per student).
• Use the graph ing-calculator-qua.html">calculator to determine the speed, distance, and time of travel
for the rover and planets.
• Use the protractor to help in find how far the planet has rotated around
the Sun by measuring the angle it sweeps out in a certain amount of
time.
Background Information:

Launching a spacecraft to Mars is not
an easy thing. The Earth is rotating
about its axis while orbiting the Sun.
Mars is also rotating about its own axis
while orbiting the Sun. To
compensate for all this, you need to
know the rotational speed and orbital
speed of both planets, as well as the
speed of your spacecraft. Solving this
problem is what NASA engineers have
been working years in prepa ration of
the Mars rover missions. They are
searching for the launch window—the
time to launch a spacecraft at a known
velocity in a known direction to
intercept the target. To fully understand this, you need to know the physics
of circular motion.
The laws and equations that govern circular motion are
just like the ones that
govern linear motion . In fact, many linear quantities are still useful when
describing circular trajectories. However, the equations become more
complicated when the circles become ellipses . For this lesson, we will assume
that all objects in orbits travel in nice, neat circles. Use Table 1 as a guide
to
see the similarities between linear and circular motion. The equations and
units you need are described in the fol lowing paragraph .

We can describe the linear velocity of an object as
the distance traveled over the time it took. This still
works for circular motion, only we use the angular
distance traveled over the time. Angular distance is
similar to taking the linear distance and bending
around the radius of a circle. An object in orbit will
have a linear velocity and an angular velocity. They
are related by the equation v = rω, where v is the
linear velocity in meters per second (m/s), r is the
radius in meters (m), and ω is the angular velocity
in radians per second (rad/s). We can use the same formula for s = r θ,
where s is the linear distance (arc) around the circle in m, r is the radius,
and
θ is the angle in radians (rad) swept out by the object as it goes around the
circle. If we recall geometry, we know the circumference (C) is 2πr. We can
find the orbital speed using v = 2πr/T, where T is the period or time for the
planet to make one complete revolution around the Sun.
| Table 1 |
| Term |
Linear |
Circular |
| variable |
units |
variable |
variable |
| distance |
d |
m |
θ |
rad |
| velocity |
v |
m/s |
ω |
rad/s |
| acceleration |
a |
m/s2 |
α |
rad/s2 |
In today’s lesson, you will de termine when the proper time
to launch a
spacecraft to Mars is. This will take careful planning and precise calculations.
You will also need some information about the Earth and Mars. Table 2 gives
you all the information you need. However, you will need to solve for other
properties to make solving the scenarios easier.
| Table 2 |
Earth |
Mars |
| length of day |
23.93 hours |
24.62 hours |
| length of year |
365.42 days |
365.42 days |
| distance from Sun |
92,955,820
miles |
149,597,890
kilometers |
141,633,330
miles |
227,936,640
kilometers |
travel time from Earth to
Mars |
6 months |
Guidelines:
1. Read orally the 9-12 NASAexplores article, “A Martian Hole-In-One.”
2. Distribute the Student Sheets.
3. Have each student use the information given to find the launch window
for a Mars-bound spacecraft.
4. Have each student answer the questions at the end.
Discussion/Wrap-up:
• Discuss the complexity of launching a spacecraft to another planet.
• From the article, discuss the analogy of the rotating golfer on a rotating
golf course to the actual launching of a spacecraft.
• For your information, the orbital speed of Earth is 29,786 m/s with an
angular speed of 0.0172 radians per day. The orbital speed of Mars is
24,163 m/s with an angular speed of 0.00916 radians per day. The
radians per day units were used to make the answers come out in days.
• Possible answer to scenario 1:

• Possible answer to scenario 2:

• Answers to the questions:
a. 1.6488 rad, 94.46 degrees, or 375,821,932 kilometers
b. 78,338,750 kilometers or 48,677,510 miles
c. 595.8 days
d. No, because the starting position has changed , so the next best
launch time is when Mars is “6 months” ahead of the Earth would be
around 410 days
e. Easier; as suming a direct flight, Jupiter and Saturn do not move
very much, in relation to the Earth’s orbit, over an Earth year.
Extensions:
• Use the Internet to find the actual path the rovers took to get to Mars.
• Research other satellites and the path they took to get where they were
going. Hint: start with the Cassini probe. See the Related Links section.
• Review the video clip from NASA’s Teaching In Space Program. See the
Related Links section for details.