# Presentation: 693462

##  Exploring Mars 

***.***

-  Exploring Mars

***Mr. K. NASA/GRC/LTP  ***
***Edited: Ruth Petersen******   ***

**Part 5**
**Pathfinder’s Path II**

**Notes:**
Pathfinder’s Path – II
In “Exploring Mars, Part 4,” the fundaments of orbital kinematics were presented with reference to the Pathfinder mission to Mars. In “Exploring Mars, Part 5,” several ‘real-life’ orbital problems are presented in the preliminary activities and worked through in the pages that follow. These problems are typical of those faced by mission planners whose objective it is to send a spacecraft to a far-off destination in the solar system. 
You are strongly encouraged to put on the mission planner’s hat and try to solve the problems on your own before looking at my solutions.

## 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***_***!***

**Notes:**
Right or wrong, YOUR answers are important. Remember: We learn just as much from our mistakes as from our successes. 
The important thing is to try!

## 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 “Pathfinder’s Path I”. Be sure that you understand the ellipse and the Vis-Viva equation!***

**Notes:**
 The sun is about 1,400,000 km in diameter. If the sun were reduced to the size of a grape (diameter = 1 cm), then earth would have a diameter of 0.01 cm or one tenth of a millimeter (1 millimeter is roughly the thickness of a dime). This nearly microscopic mote (speck) would orbit at a mean distance of 107 cm from the grape representing the sun (107 cm is about 1 m or 3.3 ft.). Mars would be half the size of earth and would orbit at a mean distance of 164 cm or about 5.4 ft. It helps to visualize the solar system by building scale models before attempting such problems as these. Having a scale model in hand greatly helps interpreting results later.

## #1

***Semi-major axis = a***

***Semi-minor axis = b***

***Focal length = f***

***f***

********Remember: f******2****** = a******2****** - b******2****** ***

***#1***

***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).***

**Notes:**
Physicists and engineers routinely match mathematics to physical reality. The most basic match is done in the process of counting. First, a set of counting numbers is defined:  {1, 2, 3, ..., }.
Then the objects to be counted are assembled in some sort of order. For example, let’s say we want to count the letters A through E:  {A, B, C, D, E}.
The set of letters is then matched with the set of counting numbers to form a one-to-one correspondence:
								{1, 2, 3, 4, 5, 6, 7, 8, 9, ..., }
								          
								{A, B, C, D, E}.
The largest counting number in the set of numbers corresponding to the letters A through E is 5. Hence, we say that there are five letters in the set. This number is called the cardinality of the set.
In an analogous manner, mathematical object (such as the conic sections) are matched to the physical reality we call planetary orbits. The match is another type of correspondence:
Ellipse  Closed Orbit = Path Followed by the Planet in Space
One Focal Point  Gravitational Center of the Orbit = Center of the Sun
The Other Focal Point  [Not Used]
Coordinate Axis  System of Reference and Measurement used in All Observations and Calculations (After all, you HAVE to know where you are!!!)
Focal Length, Semi Major Axis, etc.  Convenient Terms with which to develop a Language about Planetary Orbits
(After all, you also HAVE to communicate results!!!)
This process underlies all thinking in the sciences and in philosophy. It is a very good idea to begin mastering it as early as possible! In this presentation, we will make good use of the terms defined above.

## #2

-  Hohmann Transfer Ellipse

-  Earth's Orbit

-  Mars' Orbit

***#2***

***Remember:***
***Every planet travels in an ellipse ***_***with the sun at one focus***_***.***

**Notes:**
When the basic language has been decided upon, the elements of the language are then used to construct models. In this model, an elliptical orbit is constructed as a path that leads from the orbit of earth to the orbit of Mars. The Pathfinder spacecraft followed just such a path on its seven month journey to Mars. This particular path - a Hohmann Transfer Ellipse (Named for Walter Hohmann (1880–1945), a German physicist) – uses only the sun’s gravitation for almost the entire trip. The exceptions are the launch phase (at Earth) and arrival/entry phase (at Mars), with minor trajectory corrections along the way. The directions (arrows) are chosen to make use of the fact that Earth is actually moving at a considerable velocity through space. This velocity becomes part of the spacecraft velocity, with rocket boosters merely adding the difference (V or Delta-Vee) needed to achieve orbit. The Hohmann transfer orbit requires a minimal expenditure of fuel compared to other possible (usually quicker) paths, which require the use of thrusters while en-route. Timing the launch is of the essence when using a Hohmann orbit because it is desirable to actually have Mars at the arrival point when the spacecraft arrives there! The dates on the calendar when it is possible to launch and achieve this objective, given the mission resources available, is called a launch window.

## 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. ***

**Notes:**
OK. Your knowledge of basic terms along with the material from the previous presentation should enable you to accomplish the calculations being asked for here...

## E

***E***

***H***

***M***

******

******

**Notes:**
 Think about what you know and what you need to know:
You should have already calculated the mean orbital velocities of Earth and Mars.
You should be able to calculate the parameters of the Hohmann orbit (i.e., semi-major axis, semi-minor axis, etc.) from simple geometry and a knowledge of the planets’ respective orbits.
The vis-viva equation will allow calculation of the spacecraft velocities at points  and .
Comparison of the spacecraft velocities with the planetary velocities will give the required delta-vees at each of the points  and .

## Solutions to Problems in Preliminary Activities 1 - 5.

**Notes:**
OK, stop here!

Did you work everything through for yourself? The following pages will mean much more to you if you did!!!

If you’re ready......let’s go!!!

## 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***

**Notes:**
The orbits of the planets around the sun are not perfect circles but ellipses. However, they are nearly circular enough that we can approximate them by circles and gain important physical insights. We must realize, however, that our approximation imposes limitations on accuracy. If we were planning a real mission, the same basic ideas would be used. But their mathematical expression would take into account several factors that we are ignoring here. Hence, while we will try to rough out some numbers with a hand calculator, some algebra, and a little intuition, NASA mission planners must actually use high speed computers and elegant numerical programs to attain accuracy. Nonetheless, their first steps are often identical to what follows as they seek to rough out their calculations in the earliest stages of work.
Shown above, the setup for problem 1 shows our approximation along with steps for starting a solution.

## Earth

***d = 150 X 10******6****** km***
***2************d = 9.4 X 10******8 ******km***
***T = 365.25 days = 3.2 X 10******7****** sec***
***Orbital speed = 29 km/sec***
***~ 67,000 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***

**Notes:**
The radius of Earth’s orbit is:  d = 1.50 X 108 km.
Using basic geometry, the circumference of the orbit is then:  2d = 9.4 X 108 km.
Since Earth completes one such circumference in one year (~365.25 days), its speed is simply:  2d/ T = 29 km/sec ~ 67,000 mph.
A similar calculation is done for Mars. Notice that Mars’ orbit is larger than Earth’s, and that its orbital speed is less.

## 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 10******10****** (1/r - 1/2a)******1/2****** ... (***_***MKS Units***_***!******)***

**Notes:**
The total energy of the planet in its orbit is given above without proof:  K = - GMm/2a.
In order to obtain this term from Newton’s equations of motion and his inverse square gravitational field, a second order differential equation must be set up and solved, subject to the condition that potential energy vanishes at r = . (Remember, in Part 4 we talked about the importance of negative energy?) Doing so requires a knowledge of calculus that exceeds the limitations of this session.
Once this term is obtained, however, the velocity of the planet at any point along its orbit may be solved by using basic algebra. In the last expression above, the numerical constant is obtained by using MKS units on the numbers G and M; hence, MKS units must also be used for r and a, as well, to keep the bookkeeping consistent. Given the result above, does the speed in orbit vary as claimed in slide 12 of Part 4? Try running some numbers...

## E

***E***

***M***

***H***

***Activity 4***

***150 Mkm***

***230 Mkm***

***230Mkm + 150Mkm***
***2***

***Setup***

***The Hohmann Transfer Ellipse***

**Notes:**
We now obtain the semi-major axis of the Hohmann transfer orbit. Since we launch when Earth is ~180o from the spacecraft’s arrival point at Mars, the geometry is nice and simple.
Remember: The sun is at one focus of the transfer orbit ellipse! (What is at the other focus?)

## Semi-major axis:

***a = 1⁄2(230Mkm + 150Mkm) = 190Mkm***
***Focal length:***
***f = 190Mkm -  150Mkm = 40Mkm***
***Semi-minor axis:***
***b = *****(*****a******2****** - f******2 *****)*****1/2****** = 186Mkm***

***Calculations***

**Notes:**
From the geometry shown on the previous page, and a little common sense, we are able to obtain all the necessary parameters!

## Activity 5

***Pathfinder Velocities at ************ and ************. ***
***1. At ************:***
***r = 150 Mkm***
***a = 190Mkm***
*** ******1.6 X 10******10****** (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***

**Notes:**
Using the expression for velocity we derived earlier from the vis-viva equation, we obtain our spacecraft’s departure and arrival speeds. Can you determine the total energy K of the Earth’s orbit? Mars’ orbit? The Hohmann transfer orbit? You should be able to at this point! Please give it a try. If you have questions, feel free to e-mail me (my address is on the last page). I will be happy to help you!

## Follow-Up Activities

***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.)***** ***
***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.***

**Notes:**
The remainder of the material is great for discussion. Think about the results you have obtained. Why are they significant? And finally, welcome to the world of orbital kinematics! I hope that you’ve enjoyed the ride!!!

## 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 ? 

***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 ************?****** ***

##   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?

***v******Escape****** = (2GM******Earth******/r******Earth******)******1/2******.***
***  ******With ******M******Earth****** = 6 X 10******24****** 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?***

## joseph.c.kolecki@grc.nasa.gov