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ORBITAL MOTION

ORBITAL MOTION. Formulas. Aphelion, perihelion refer to farthest distance, and closest to the Sun. Apogee, perigee refer to farthest distance and closest to the Earth. (peri is the closest). Finding the semi major axis. This is the average distance from the orbiting body.

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ORBITAL MOTION

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  1. ORBITAL MOTION

  2. Formulas Aphelion, perihelion refer to farthest distance, and closest to the Sun. Apogee, perigee refer to farthest distance and closest to the Earth. (peri is the closest) Finding the semi major axis. This is the average distance from the orbiting body How much it departs from a circle between o and 1 Velocity at peri (closest point), use correct Velocity at ap (farthest point), use correct Period using a in solar system only use if orbiting the Sun.

  3. Period of orbiting object in seconds, use correct The escape velocity from a body , use correct Velocity for object orbiting in a circular orbit . Use correct Closet distance the orbiting object comes to the object being orbited in an elliptical orbit Farthest distance the orbiting object comes to the object being orbited in an elliptical orbit

  4. Conversions To change from km/sec to miles/hr Km/sec (3600 sec/hr)(0.62137 miles/km= miles/hr If the period is in seconds / 3600 for hours To change from au to km multiply by 150,000,000 or km To change from miles to km multiply by 1.61 km/mile e has no units. There is a review of math with scientific notation on slide 15 ** Be sure you use the of the body that is being orbited *** I did not always follow exactly significant figures, but I did not use all the digits. Answers may vary slightly depending upon how you round off the decimals, and that’s ok.

  5. Body μ (km3s-2) Sun 132,712,440,018 Mercury 22,032 Venus 324,859 Earth 398,600 Mars 42,828 Jupiter 126,686,534 Saturn 37,931,187 Uranus 5,793,947 Neptune 6,836,529 Pluto 1,001 The gravitational parameter • In Astrodynamics,the standard gravitational parameter   of a celestial body is the product of the gravitational constant    and the mass   : • The units of the standard gravitational parameter are km3 s-2 [ **Be sure to use the right for the object you are orbiting ! ****

  6. Eccentricity of ellipse eccentricity (e) 0=circle, 1 = line aphelion - perihelion e = ___________________________ aphelion + perihelion e=0 e=0.98

  7. Properties of Ellipses • semi-major axis = a • 1/2 length of major axis b a a = Semi-major axis b = Semi-minor axis

  8. a = aphelion + perihelion _______________________ 2 aphelion perihelion P2= a3

  9. Using the Kepler’sThird Law P2µ a3 P2= a3 if : P measured in earth years, and a in AU. A planet’s avg distance from the sun is 4 au, what is the period of the planet ?

  10. Astronomers use the metric system, whereas we are all more familiar with the English system. I will use conversion so that you can be more familiar with the answers. I. The space shuttle is in a circular orbit 200 miles above the earth. Find the period and velocity of the shuttle. 200miles(1.61 km/mile) =322 km above the Earth. The height of the satellite in the problem must be the radius of the earth + height of object. Radius of earth 6378km r+h = 6378+322= 6700 km

  11. You can’t use because it circles the earth, not the Sun. a is the distance from center of Earth to the shuttle II. This problem covers a lot of formulas. An asteroid’s closest approach to the sun is 2 au, and its farthest distance from the Sun is 4.5 au. Find a, the eccentricity, distance at perihelion, distance at aphelion, period, velocity at perihelion, and aphelion.

  12. Find the perihelion, and aphelion distances. = 3.25au (1 - 0.385) = (3.25)(.615) = 1.99 au = 3.25au (1+ 0.385) = 4.43 au Find the period. Perihelion, and aphelion must be changer to km, since contains km . To change multiply au by 150,000,000 km/au , or 24.8 km/sec(3600)(.6213) = 55,469.7 miles/hour

  13. = 3.504(3600)(0.62137)= 7,838.8 mils/hour III. An Earth satellite is in an elliptical orbit around the Earth; its perigee is 160 km, and the apogee is 800 km. Find e, period, and velocity at perigee, and apogee. Radius of the Earth = 6380 km =800 +6380 = 7180km = 160 + 6380 = 6540 km

  14. We need a /3600 = 1.57 hrs

  15. I had to include my Halley’s comet problem. Halleys has a period of ~76 years and e=0.967 Find the velocity at perihelion (close), and aphelion (far). = 17.94(0.033) = .592 au ( ) = = 17.94(1.967) = 35.45 au ( ) =

  16. A Review of Scientific Notation Math • Multiplication: • The digit terms are multiplied in the normal way and the exponents are added. The end result is changed so that there is only one nonzero digit to the left of the decimal. • Example: (3.4 x 106)(4.2 x 103) = (3.4)(4.2) x 10(6+3) = 14.28 x 109 = 1.4 x 1010(to 2 significant figures) • Example: (6.73 x 10-5)(2.91 x 102) = (6.73)(2.91) x 10(-5+2) = 19.58 x 10-3 = 1.96 x 10-2(to 3 significant figures)

  17. Division: • The digit terms are divided in the normal way and the exponents are subtracted. The quotient is changed (if necessary) so that there is only one nonzero digit to the left of the decimal. • Example: (6.4 x 106)/(8.9 x 102) = (6.4)/(8.9) x 10(6-2) = 0.719 x 104 = 7.2 x 103(to 2 significant figures) • Example: (3.2 x 103)/(5.7 x 10-2) = (3.2)/(5.7) x 103-(-2) = 0.561 x 105 = 5.6 x 104(to 2 significant figures)

  18. Powers of Exponentials: • The digit term is raised to the indicated power and the exponent is multiplied by the number that indicates the power. • Example: (2.4 x 104)3 = (2.4)3 x 10(4x3) = 13.824 x 1012 = 1.4 x 1013(to 2 significant figures) • Example: (6.53 x 10-3)2 = (6.53)2 x 10(-3)x2 = 42.64 x 10-6 = 4.26 x 10-5(to 3 significant figures) • Roots of Exponentials: • Change the exponent if necessary so that the number is divisible by the root. Remember that taking the square root is the same as raising the number to the one-half power. • Example:

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