Mid Term 2-2019 Solution(1)

AE 423 Fall 2019 Mid Term 1 1 Name: AERO 423 Mid Term 1 Thursday 14 November, 2019 1. (40 pts) A satellite is in a circular orbit at an altitude of 900 km and an inclination of 63 degrees. It has a sensor with a beam width of 110 deg. (the beam width is twice the nadir pointing angle). a) What is the grazing angle? b) What is the Earth central angle of the coverage with this sensor? c) A requirement is that the satellite provide coverage up to 78 degrees latutude. What is the minimum allowable inclination that will satisfy this requirement. d) What is the swath width? e) If we want continuous coverage of every point along the flight path (ground track), that is every point in the orbit plane is within view of a satellite, what is the minimum number of satellites required to satisfy this requirement? Solution Compute the various angles a) b) c) d e) The minimum number of satellites is R = R e + 900 = 7278.137 η = BW / 2 = 55deg cos ε = 1 + h / R e ( ) sin η ε = 20.814 deg λ max = 90 − η + ε ( ) = 14.186 deg λ 0 = cos − 1 R e R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 28.796 deg grazing angle = ε = 20.81deg Earth Central angle = λ max = 90 − η − ε = 14.186 deg i min == 78 − λ max = 63.814 deg SW = 2 R e λ max = 3158.4 km 2 π / 2 λ max ( ) = 12.7, N = 13

AE 423 Fall 2019 Mid Term 1 2 2. (35 pts) Elon Musk is going to send another car to space, but this time to Jupiter. The car will initially be in a 500 km altitude circular orbit abut the Earth. The alignment of the planets is such that there is no wait time to leave for Jupiter. Determine a) The trip time to Jupiter. b) The wait time for return to Earth. c) for the escape trajectory leaving Earth. d) The D v required to put the car into the transfer orbit to Jupiter 1AU =149,597,870km, R Jup =5.202604 AU Note: Be careful with units if you use radii in AU in your computations. Solution a) a This means that Earth made more than two rotations about the Sun during the transfer. It will show up in the calculation of the wait time and angle b) Subtract 2 years from the wait ime to determine the position of the Earth relative to Jupiter. c) First compute the velocity needed for the Hohmann transfer, and then the velocity needed to escape. v ∞ R 1 = R earth = 1 AU = 149,597,870 km R 2 = R Jup = 5.202604 AU a t = 0.5 R 1 + R 2 ( ) = 3.013 AU = 4.64 × 10 8 km T trip = 0.5 T = π a t 3 μ s = 997.42 days = 2.7307 years (There are 365.26 days/year) φ f . n E = 2 π r / year n J = 0.5295 r / year n syn = n E − n J = 5.7538 r / y T syn = 2 π / n syn = 398.87 days θ 1 = n 1 t 12 = 17.158 radians φ f = π − θ 1 = − 14.016 radians T wait = 2 N π − 2 φ f ( ) n 1 − n 2 ( ) You need N = 5 to make T wait positive T wait = 0.588 years

AE 423 Fall 2019 Mid Term 1 4 3. (30 pts) There is a 3 satellite formation in a leader-follower formation. The chief is in the middle and the 1 st deputy is leading the chief by 3 km and the 2 nd deputy is 7 km behind the chief. The chief is in a circular orbit of radius 7000 km. 3.1 The 1st deputy is to maneuver into a 2x1 ellipse centered on the chief. The size of the ellipse will be 6km x 3 km. That is What is the required D v , that is ,to maneuver into this 2x1 ellipse? Which answer is correct? Show your work, no credit if work is not shown. a) 1.617 m/s b) -1.617 m/s c) 3.23 m/s d) -3.23 m/s Solution a) There are several ways to solve this problem. First we need the mean motion, The initial condition on y is We need to make a maneuver to go into the 2x1 ellipse. For the 2x1 ellipse For the 2x1 ellipse has to be negative when x is a maximum, so has to be positive at t =0, so b = 0. If is initially negative the motion will be a 2x1 ellipse, but not centered on zero. Answer is a) b) For the 2x1 ellipse, starting at y =3, 0.25 orbits later x = 1.5 and y =0. In the CW solution the secular term has to be zero so . This means So Answer is a) 3.2 What is the eccentricity of orbit of the 1 st deputy. Not the eccentricity of the 2x1 ellipse, but the orbit after the maneuver. Which answer is correct? Show your work, no credit if work is not shown. a) 0.000215 b) 0.000430 c) 0.000322 Solution The deputy has a maximum altitude of (7000+1.5). So x 2 + y / 2 ( ) 2 = 1.5 2 = 2.25 ! x 0 n = μ R 3 = μ 7000 3 = 0.001078 y 0 = 3. y 0 = 3, x 0 = 0. Using y = 2 B cos ψ + β ( ) , x = B sin ψ + β ( ) β = 0, π . Therefore, B = 1.5, ! x 0 = Bn cos β = 0.001617cos β ! y ! x ! x A = 2 x 0 + ! y 0 / n = 0 ! y 0 = 0. x ψ ( ) = ! x 0 n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ sin ψ x π / 2 ( ) = 1.5 = ! x 0 n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ sin π / 2 ( ) = ! x 0 n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δ v x = n ! x 0 = 0.001617

AE 423 Fall 2019 Mid Term 1 5 Answer is (a) 3.3 The 2 nd deputy is to change its relative motion to include out-of-plane motion. This is the dog wags tail formation I discussed that is . The out of plane motion is to have an amplitude of 5 km. What is the to achieve this motion. Which answer is correct? Show your work, no credit if work is not shown. a) 2.16 m/s b) 5.39 m/s c) -3.23 m/s Solution Answer is (b) r a = R + x max = R + B a = R e = r a a − 1 = 0.000215 Another way is from Gauss'splanetary equations Δ e = p h sin f ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δ v r = R μ ! x 0 = ! x 0 v c e = 0.000215 A = B = 0 = ! x 0 = ! y 0 = x 0 = 0, C = y 0 ≠ 0. Δ v , that is, ! z 0 z 0 = 0, z max = 5 z = z 0 cos ψ + ! z 0 / n ( ) sin ψ = ! z 0 / n ( ) sin ψ 5 = ! z 0 / n ! z 0 = Δ v z = 5 n = 0.00539 km / s = 5.39 m / s