mr. k. nasa/grc/ltp part 5 pathfinder’s path ii
TRANSCRIPT
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Mr. K. NASA/GRC/LTP
Part 5
Pathfinder’s Path II
Preliminary Activities
In the following preliminary activities, five real-world problems are given
that will require mathematical thought. The lecture portion of the lesson will go over each of these
problems with the students and allow them to compare and discuss their
results with the presenter. Discussion is encouraged during the lecture
portion!
1. The Earth is 150 million km from the sun. It completes one orbit in a
period of approximately 365.25 days. Calculate its orbital speed in km/sec
and mph.
2. Mars is 230 million km from the sun. It completes one orbit in a period of approximately 687 days. Calculate its orbital speed in km/sec and mph.
3. Review the material from “Pathfinder’s Path I”. Be sure that you
understand the ellipse, and the Vis-Viva equation!
Semi-major axis = a
Semi-minor axis = b
Focal length = f
f
*Remember: f2 = a2 - b2
#1
4. Using the following two sketches and Kepler’s law of orbits, identify the semi-major axis, the
semi-minor axis, and the focal length of Pathfinder’s Hohmann transfer ellipse. Given the astronomical information in problems 1 and 2, and
your knowledge of the ellipse*, specify each of these quantities for Pathfinder in millions of km
(Mkm).
#2
Remember:
Every planet travels in an ellipse with the sun at one focus.
5. Use the Vis-Viva equation to predict the orbital velocity (in km/sec and
mph) of the Pathfinder spacecraft at the point of departure (marked with an in the accompanying diagram)
and the point of arrival (marked with a in the accompanying diagram).
Completion of these five exercises will give you a rough idea of some of the basic orbit calculations necessary for
sending the Pathfinder to Mars.
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Solutions to Problems in Preliminary Activities 1 - 5.
Activities 1 & 2Orbital velocities of Earth and Mars
1. Distance from planet to sun = d
2. Circumference = 2d
3. Orbital speed =
2d
Period
Sun’s mass = M
Setup
Planet or spacecraft mass = m
Earth
d = 150 X 106 km2d = 9.4 X 108 km
T = 365.25 days = 3.2 X 107 secOrbital speed = 29 km/sec
= 66,000 mph
Mars
d = 230 X 106 km2d = 1.5 X 109 km
T = 687 days = 5.9 X 107 secOrbital speed = 25 km/sec
= 57,000 mph
Calculations
Activity 3: The Vis-Viva Equation (Only)
The vis viva equation isv = {2(K + GMm/r)/m }1/2
This equation gives the velocity of an object at various ponts on an elliptical orbit. I need to tell you that, with differential equations, we show that
K = -GMm/2a
Thereforev = {2GM (1/r - 1/2a) }1/2
= 1.6 X 1010 (1/r - 1/2a)1/2 … (MKS Units!)
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Activity 4
150 Mkm
230 Mkm
230Mkm + 150Mkm
2
Setup
The Hohmann Transfer Ellipse
Semi-major axis:a = ½(230Mkm + 150Mkm) = 190Mkm
Focal length:f = 190Mkm - 150Mkm = 40Mkm
Semi-minor axis:b = (a2 - f2 )1/2 = 186Mkm
Calculations
Activity 5Pathfinder Velocities at and .
1. At :r = 150 Mkma = 190Mkm
1.6 X 1010 (1/r - 1/2a)1/2 … (MKS Units!) v = 32 km/sec
= 72,000 mph
2. At :r = 230Mkma = 190Mkm
v = 21 km/sec= 47,000 mph
Follow-Up Activities
1. Compare the Pathfinder velocity at with earth’s orbital velocity at . What is the difference and why?
(Express your answer in terms of total orbital energy.)
2. Compare the Pathfinder velocity at with mars’ orbital velocity at .
Again, what is the difference and why? (Express your answer in terms of total
orbital energy.)
3. The earth has a radius of 6400 km and spins once on its axis in 24 hours. Calculate the velocity of a point at the
equator in km/sec and mph.
4. When viewed from celestial north, the Earth both rotates and revolves
counter-clockwise. Do the orbital and rotational velocities add or subtract at
local midnight? How about at local noon? What considerations might
affect the time of day for a launch? Why did NASA launch the Pathfinder
spacecraft eastward?
5. In spacecraft design, energy is sometimes expressed in terms of
change in velocity required to achieve orbit ( “delta-vee” or v). Given what
we’ve just done, what v does the Pathfinder require at ?
6. Actually, additional energy (velocity) is required for a spacecraft
just to escape the Earth’s gravitational field. This velocity is
given by the expression
vEscape = (2GMEarth/rEarth)1/2.
With MEarth = 6 X 1024 kg, calculate this velocity in km/sec and mph. This velocity must be added to the v
calculated in Problem 5. How much, as a percent, does the result change compared to the value obtained in
Problem 5? Does leaving the Earth’s gravity well cost a lot in fuel?
