. 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 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: f 2 = a 2 - b 2 #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.

E H M  

Solutions to Problems in Preliminary Activities

Activities 1 & 2 Orbital velocities of Earth and Mars 1. Distance from planet to sun = d 2. Circumference = 2  d 3. Orbital speed = 2  d Period Sun’s mass = M Setup Planet or spacecraft mass = m

Earth d = 150 X 10 6 km 2  d = 9.4 X 10 8 km T = days = 3.2 X 10 7 sec Orbital speed = 29 km/sec = 66,000 mph Mars d = 230 X 10 6 km 2  d = 1.5 X 10 9 km T = 687 days = 5.9 X 10 7 sec Orbital speed = 25 km/sec = 57,000 mph Calculations

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 ponts 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 (1/r - 1/2a) 1/2 … (MKS Units!)

E M H Activity 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 = (a 2 - f 2 ) 1/2 = 186Mkm Calculations

Activity 5 Pathfinder Velocities at  and . 1. At  : r = 150 Mkm a = 190Mkm 1.6 X (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

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 v Escape = (2GM Earth /r Earth ) 1/2. With M Earth = 6 X 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?