Exploring Mars

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Exploring Mars
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Part 5 Pathfinders Path II
Mr. K. NASA/GRC/LTP Edited Ruth Petersen
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Preliminary Activities In the following
preliminary activities, five real-world problems
are given that require mathematical thought.
During the session, we will go over each of these
problems and compare and discuss your results.
Discussion is encouraged!
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• 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.
• 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.
• Review the material from Pathfinders Path I.
  Be sure that you understand the ellipse and the
  Vis-Viva equation!

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• Using the following two sketches and Keplers law
  of orbits, identify the semi-major axis, the
  semi-minor axis, and the focal length of
  Pathfinders 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).

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Semi-minor axis b
f
Focal length f
Semi-major axis a
Remember f2 a2 - b2
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Hohmann Transfer Ellipse
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Earth's Orbit
Remember Every planet travels in an ellipse with
the sun at one focus.
Mars' Orbit
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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 orbital
calculations necessary for sending the Pathfinder
to Mars.
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?
E
H
M
?
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Solutions to Problems in Preliminary Activities 1
- 5.
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Activities 1 2 Orbital velocities of Earth and
Mars
1. Distance from planet to sun d
Planet or spacecraft mass m
Suns mass M
2. Circumference 2?d
2?d Period
3. Orbital speed
Setup
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Earth d 150 X 106 km 2?d 9.4 X 108 km T
365.25 days 3.2 X 107 sec Orbital speed 29
km/sec 67,000 mph Mars d 230 X 106 km 2?d
1.5 X 109 km T 687 days 5.9 X 107 sec Orbital
speed 25 km/sec 57,000 mph
Calculations
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Activity 3 The Vis-Viva Equation (Only) The vis
viva equation is v 2(K GMm/r)/m 1/2 This
equation gives the velocity of an object at
various points on an elliptical orbit. I need to
tell you that, with differential equations, we
show that K -GMm/2a Therefore v 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
The Hohmann Transfer Ellipse
230 Mkm
E
H
230Mkm 150Mkm 2
150 Mkm
M
Setup
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Calculations
Semi-major axis a ½(230Mkm 150Mkm)
190Mkm Focal length f 190Mkm - 150Mkm
40Mkm Semi-minor axis b (a2 - f2 )1/2 186Mkm
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Activity 5 Pathfinder Velocities at ? and ?. 1.
At ? r 150 Mkm a 190Mkm 1.6 X 1010 (1/r -
1/2a)1/2 (MKS Units!) ? v? 32 km/sec 72,000
mph 2. At ? r 230Mkm a 190Mkm ?v? 21
km/sec 47,000 mph
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• Follow-Up Activities
• Compare the Pathfinder velocity at ? with earths
  orbital velocity at ?. What is the difference and
  why? (Express your answer in terms of total
  orbital energy.)
• 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.)
• 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.

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• 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?
• In spacecraft design, energy is sometimes
  expressed in terms of change in velocity required
  to achieve orbit ( delta-vee or ?v). Given what
  weve just done, what ?v does the Pathfinder
  require at ??

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• Actually, additional energy (velocity) is
  required for a spacecraft just to escape the
  Earths 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 Earths gravity well cost a lot in
  fuel?

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joseph.c.kolecki_at_grc.nasa.gov