Developed by Jim Beasley – Mathematics & Science Center – mathscience.k12.va

1. A Spinning Earth and Rocket Launches A Lesson From the Mathematics and Science Center By Mr. Beasley Developed by Jim Beasley – Mathematics & Science Center – http://mathscience.k12.va.us

2. Space Flight • Basis for modern space flight had it’s origin in ancient times • Until about 50 years ago, space flight was just the stuff of fiction • Jules Verne’s From the Earth to the Moon • Buck Rogers and Flash Gordon movies and cartoons • Russians launched the first artificial Earth satellite in 1957, starting the race into space

3. Once in motion satellites • remain in motion. Newton’s Laws of Motion First Law An object at rest will remain at rest and an object in motion will remain in motion unless acted upon by an outside force.

4. Example of Inertia

5. Newton’s Laws of MotionFirst Law (Inertia)

6. Newton’s Laws of Motion First law explains why: • Once in motion satellites remain in motion.

7. Newton’s Laws of Motion Second Law If a force is applied to an object, it will cause the object to accelerate in the direction the force is applied. F = ma Let’s Put Some Numbers to the Equation

8. Newton’s Laws of Motion(Second Law) F= M x A

9. υ m Explains why satellites move in circular orbits. The acceleration is towards the center of the circle - called centripetal acceleration. r Centripetal Force M Newton’s Laws of Motion Second Law F = ma

10. Rocket - Exhaust gases in one direction; Rocket is propelled in opposite direction. Newton’s Laws of Motion Third Law For every action there is an equal and opposite reaction.

11. Newton’s Laws of Motion(Third Law) Atlas 5 Launch

12. r m M M Universal Law of Gravity Force = G x (M x m / r2) G is the universal constant of gravitation.

13. Cannonball Newton’s Concept of Space Flight The Cannonball Analogy

14. Newton’s Concept of Space Flight The Cannonball Analogy

15. Newton’s Concept of Space Flight The Cannonball Analogy

16. Cannonball Newton’s Concept of Space Flight The Cannonball Analogy

17. If you walked around a merry-go-round whose diameter was 10 meters, how far would you walk? Things That Are Round C • What do we know about round things? • We know Pi (π); The ratio of the circumference of a circle to its diameter. d C = π x d = 3.14159…x 10 m = Approx. 31 m

18. Spinning Objects • What do we know about spinning things? • If you stand on the edge of the 10 meter diameter merry-go-round, every time it rotates you move π x 10 meters or about 31 meters.

19. Spinning Objects • How far do you move each second if it rotates once a second? • π x 10 m x 1 = ~31 m • Exercise: How far do you move each second if it rotates twice a second? Three times a second? Five?

20. Spinning Objects • The speed of an object is the distance it travels divided by the amount of time it takes. • How fast would you be traveling on a 10 meter diameter merry-go-round if it rotated twice a second? • Speed = distance ÷ time = 2 x (π x d) ÷ 1 Sec. = 2 x (π x 10)/1 = 63 m/sec

21. Spinning Objects • The speed of an object is the distance it travels divided by the amount of time it takes. • Double the size of the merry-go-round. If it rotates once per second, how fast will you be moving? Speed = 1 x (π x 20)/1 = 63 m/sec

22. Spinning Objects The period of rotation is how long it takes to complete one rotation of the merry-go-round. • What is the period of rotation if the 20 meter diameter merry-go-round is traveling at a speed of 10 m/sec? Period = 63 m 10 m/sec = 6.3 seconds

23. Spinning Objects The frequency of rotation is the number of times the merry-go-round rotates per sec. • The inverse of period or, 1  Period • What is the frequency of rotation if the merry-go-round has a period of 3.1 seconds? Frequency = 1  3.1 = 0.32 rotations per second

24. Review of Terms • Distancea point on the outside edge of a merry-go-round travels in one revolution is: π x diameter • Speedof rotation is the distance traveled divided by the amount of time it takes. • Period of rotation is how long it takes to complete one rotation of the merry-go-round. • Frequency of rotation is the number of times the merry-go-round rotates per second. • Frequency = 1 / Period (an inverse relationship)

25. Review Questions • Period = Distance / Speed • If you double the size of the merry-go-round, what happens to the period? • Answer: It doubles • If you double the speed, what happens to the period? • Answer: It halves • Frequency = 1 / Period • If you double the frequency of rotation, what happens to the period? • Answer: It halves

26. A Spinning Earth • Is the Earth perfectly round? • No - Because it spins (rotates on its axis). • For our lab, we’ll assume that it is perfectly round.

27. A Spinning Earth • A large merry-go round with a diameter of approximately 12526 kilometers, at the equator. Knowing its period of rotation is 24 hours, what is the speed of a point on the equator? • Speed = Circumference/Period = π x 12526/24 = 1639.7 kph

28. A Spinning Earth • Suppose you are at Cape Farewell, Greenland, how fast would you be moving? • First, we need to review some geography. • Latitude of a point on the earth’s surface is the angle, in degrees, between a line drawn from the earth’s center and a plane of the equator.

29. c b 37° a A Spinning Earth • Next, we have to consider the geometry. • What do we know about a right triangle? • If you know two sides, you can calculate the third using the Pythagorean Theorem (c2=a2+b2). If a = 4, b=3, then c = 5 Divide a by c: a/c = 4/5 = 0.8 Now, strike the “cos” key on your calculator and type “37”. Enter. What is your answer? You have discovered a very useful function.

30. A Spinning Earth • Finally, we can use the “cos” function to calculate the diameter of a circle around the earth at the 60º latitude. • The diameter of a circle at any latitude is found by the following equation: • Diameter (at any latitude) = Diameter of Earth x Cos (Latitude) • Diameter (at 60º latitude) = 12526 x cos(60º) = 12526 x 0.5 = 6262km

31. A Spinning Earth • Then we solve for the speed of that point at a latitude of 60º. • Remember, Speed = Distance / Time • Speed = (π x 6263)/24 = 19675/24 = 819.8 kph • However, there is an easier way to do this. • Speed (at any latitude) = speed at equator multiplied by cos(latitude) • Speed (at 60º) = 1639.7 x cos (60) = 819.8 kph

32. A Spinning Earth Practice Exercise: Calculate the speed of a point on the earth at latitude of 47.5ºN. • Speed = 1639.7 x cos(47.5) = 1108 kph

33. A Spinning Earth • Practice Exercise: What speed would a point in the center of Guatemala be spinning? Use the globe to find the latitude. • Center of Guatemala is approx. latitude of 15°N • Answer: Speed (Guatemala) = 1639.7 x cos(15) = 1584 kph

34. Launching a Rocket Into Earth Orbit • A speed of about 30,000 kilometers per hour is required to escape the earth’s gravity. • The average cost of a Space Shuttle launch is $400M. • Approximately $13,333 per kph

35. Launching a Rocket Into Earth Orbit NASA uses the spin of the earth to help save rocket fuel and cost. How much is saved in launching from Cape Kennedy as opposed to a launch site at Halifax, Nova Scotia?

36. Activity • State your hypothesis - • Which of the following locations would be the least costly for launching a rocket into earth orbit? • London, England • Cape Town, South Africa • Los Angeles, California • Miami, Florida • Quito, Ecuador • Which is the best direction to launch (east or west)?

37. Activity • Test your hypothesis. • Find these cities on the globe. • Estimate the latitude to nearest whole number of degrees (for example, 17°N). • Perform the calculations for each location.