Dr. Neal, WKU MATH 117 Orbital Motion Around Earth According to Newton’s Law of Gravitation, the acceleration due to gravity g of any celestial body is inversely proportional to the square of the distance to the center of mass of the body. For a planet, g =Cr2, where C is a constant that depends on the planet, and r is the distance to the center of the planet. The gravity decreases as r increases. For an object in a circular orbital motion around a planet, such as a satellite or Space Shuttle, the outward centripetal acceleration due to its motion is given by a =v2r. In order for the object to stay in orbit, the centripetal acceleration a must be in equilibrium with the gravity of the planet g. Thus, we must have a = g, or v2r=Cr2. To obtain the speed necessary to stay in orbit, we solve for v: v = Cr The Orbital Constant C Because g =Cr2, if we know the gravity on the surface and the radius of the planet, then we can solve for C = g r2. For our purposes, we will want all units to be in miles and hours. The constant C will be in (miles)3/(hrs)2. To obtain these units, we use C = g r2 = gftsec2 ! r2 mile2 ! 15280mileft ! 36002sechr22. For Earth, g ≈ 32.174 ft / sec2 and r ≈ 3963.2 miles; thus, C ≈ 32.174 !(3963. 2)2 ! 15280 !36002 ≈ 1.24 ! 1012 mileshr23. C = 1.24 ! 1012 mileshr23Dr. Neal, WKU Circular Orbit In order to stay in circular orbit above Earth, the velocity must satisfy v = Cr = 1.24 ! 1012r, where r is the distance to the center of Earth. Using r = 3963.2 + altitude, the orbital velocity then must satisfy v =1. 24 ! 1012(3963.2 + a ) mph. Here is a summary of the equations we need: The radius of the Earth is about 3963.2 miles; g ≈ 32.174 ft / sec2 Earth rotates once in 23h 56m 4.1 s ≈ 23.9345 hrs. (a) Orbital Velocity v =1.24 ! 1012d mph where d is the distance to center of Earth or v =1. 24 ! 1012(3963.2 + a ) mph where a is the altitude above Earth (b) Angular Velocity of Earth (How fast the angle of rotation moves as Earth rotates.) !E=2"23 +5660+4.13600# $ % & ' ( radianshr) 0.262516 radianshr Can multiply by 180º! to get !E" 15. 041º per hour (c) Velocity from Angular Velocity (For ! in rad per hr) v =!r and !=vr For Earth’s spin, !E" 0. 262516 rad/hr and r ! 3963. 2 miles at the equator So the velocity of the equator spin is v =!r = 0. 262516 " 3963.2 # 1040. 4 mph (d) Orbital Angular Velocity of Satellite !S=vd=v(3963.2 + a ) radianshr where v is found from Part (a) Multiply by 180º! to get deg per hr Centripetal Acceleration acc =v2r=Cr2=1.24 ! 1012(3963.2 + a )2mileshr2 !528036002to put in fts2" # $ % & 'Dr. Neal, WKU Example. Suppose the Space Shuttle is 220 miles above Earth orbiting in the plane of the equator. (a) What speed is needed to maintain orbit at 220 miles altitude? (b) What is the centripetal acceleration (in ft/s2 ) at this altitude? What pct. of g? (c) What is the Shuttle’s angular velocity? (d) How long will it take the Shuttle to complete one 360º orbit? (e) How long does it take for the shuttle returns to its starting point above the equator? Solution. (a) To stay in orbit, v =1. 24 ! 1012(3963.2 + a ) = 1. 24 ! 1012(3963.2 + 220) ≈ 17,216.96 mph. (b) acc = v2r = Cr2 = 1. 24 ! 1012(3963. 2 + 220)2mileshr2 (! 5280 /36002 ) ≈ 28.87 ftsec2 (divide by 3.28 to get ≈ 8.8 metsec2). These values give the gravity pull of Earth at the 220 mile altitude, which is 28.87/32.174 = 0.8973, or about 89.73% of g. (c) !=vr=v(3963.2 + a) = 17216.96(3963. 2 + 220)≈ 4.11574 (rad)/hr (which is about 235.81º/hr) (d) The time t for one 360º = 2π radian revolution of the Shuttle, is given by 4.11574 t = 2π ! t = 2!4.11574 ≈ 1.5266 hrs ≈ 1 hr, 31 min, 36 sec. (e) The angular velocity function of the Shuttle at this altitude is 4.11574 t. The angular velocity function of Earth is 2!23 +5660+4.13600" # $ % & ' t ≈ 0.262516 t For shuttle to lap Earth, we must find the time t such that 4.11574 t = 0.262516 t+ 2π. That is, the shuttle’s angle must equal the Earth’s angle plus one revolution. Solving for t we have 3.853224 t = 2π and t ≈ 1.63 hrs ≈ 1 hr, 37 min, 50 sec.Dr. Neal, WKU Exercises 1. A communications satellite is 18,200 miles above the Earth orbiting in the plane of the equator. (a) What speed must it have in order to maintain orbit? (b) What is its centripetal acceleration in ft/s2? (c) What is its angular velocity? (d) How long will it take to complete one 360º orbit? (e) How long until it returns to its starting point above the equator? 2. A probe is 75 miles above Mars orbiting in the plane of the equator. (The radius of Mars is 2108.56 miles, and its orbital constant is C ≈ 1.33 ! 1011 (miles)3/(hrs)2 . Mars rotates in 24h 37m 22.662s. (a) What speed must the probe have in order to maintain orbit? (b) What is its angular velocity? (c) How long until the probe returns to its starting point above Mars’ equator?Dr. Neal, WKU Solutions 1. (a) v = 1.24 ! 10123963.2 + 18, 200 ≈ 7,479.88 mph. (b) acc = 1.24 ! 1012(3963.2 + 18, 200 )2mileshr2 (! 5280 /36002 ) ≈ 1.03 ftsec2. (c) != v / r ≈ 7,479.88 / (3963.2 + 18,200) ≈ 0.33749 (rad)/hr. (≈ 19.337 deg per hr) (d) Set 0. 33749 t = 2π ! t ≈ 18.6174 hrs ≈ 18 hr, 37 min, 3 sec (e) The Earth’s angular velocity function is 0.262516 t. For the satellite to lap Earth, then 0. 33749 t = 0.262516 t + 2π. So 0.074974 t = 2π and t ≈ 83.8 hrs 2. (a) v = ! 1.33 "10112108.56 + 75 ≈ 7804.47 mph. (b) != v / r ≈ 7804.47 / (2108.56 + 75) ≈ 3.5742 (rad)/hr. So the angular velocity function is 3.5742 t. (c) For Mars’ equator, the angular velocity function is 2!24 +3760+22.6623600" # $ % & ' t ≈ 0.255176 (rad)/hr. Thus, set ! 3.5742 t = 0.255176 t + 2π ! t ≈ 1.893 hrs ≈ 1 hr, 53 min, 35