ASE 366K Spacecraft Dynamics

Transcription

1 ASE 366K Spacecraft Dynamics Homework 2 Solutions 50 Points Total: 10 points each for 1.16, 1.19, 2.6, 2.7, and 10 points for completing the rest Show that the position vector is a min or max at the apses of the orbit. (a) We can use the basic conic equation to show that this is true. dr dν r p 1 + e cos ν pe sin ν (1 + e cos ν) 2 Since an extreme of a function occurs at the points where the derivative is zero, we can say that r is min/max when dr/dν is zero. This happens when sin ν is zero, which is at ν 0 and ν 180, which are the apses of the orbit (a) Prove that: m k,j1 j k Gm j m k r 2 jk r j r k r jk 0 The easiest way to prove this is to expand the double sum out for a few terms and show that stuff cancels out. Lets look at the terms where k 1, j 2 and k 2, j 1: Gm 2 m 1 r 2 21 r 2 r 1 + Gm 1m 2 r 1 r 2 r 21 r12 2 r 12 We know that r 2 r 1 (r 1 r 2 ) and also that, in magnitude, r 12 r 21. Therefore we can see that those two terms cancel each other out. Likewise, for the general double sum, any given term will have an equal and opposite term somewhere else which will cancel it out, and therefore the entire double sum will be zero. A more intuitive way of looking at this problem would be to realize that the given double sum simply adds up all of the forces internal to the system. For any two bodies a and b, the force that body b exerts on body a is equal in magnitude but opposite in direction to the force that body a exerts on body b. When these equal but opposite force vectors are added, they cancel each other out. Therefore for all pair of bodies, the mutual force vectors cancel out and the total internal force in the system is zero. (b) To show that r cm at + b, we will assume its truth and work our way down to something that we know is true. r cm k1 m kr k k1 m k at + b 1

2 dr cm dt k1 m kṙk k1 m k a d 2 r cm dt 2 k1 m k r k k1 m k 0 Since we already proved in part (a) that k1 m k r k 0, we can say that the last equation is true and therefore all 3 equations are true. (c) The equation that we proved in part (b) is linear with respect to the time t. From this we can say that the center of mass moves at a constant rate and therefore does not accelerate. In other words, no matter how all the other bodies in the system move, the center of mass will either stay still or move at a constant speed in a straight line We know that at any point in an ellipse, the distance from the attracting focus to that point r plus the distance from the vacant focus to that point r is equal to twice the semi-major axis a. At the end of the semi-minor axis, r r and therefore r a. For the satellite s original elliptical orbit, we can calculate the energy to be E µ /2a 1/8. From this we can find the speed V right before the impulsive velocity change. E 1 (V ) 2 µ 8 2 a V 1/2 [DU/T U] Right after the impulse, we know that the new speed V + is just equal to escape speed. V + 2µ /a 1/2 [DU/T U] The cheapest burn would be if the velocity impulse was applied in the same direction as the original velocity. Therefore, V V + V [DU/T U] In problem 1.16 we showed that r a at the end of the semi-minor axis. Using the energy equations, we get: E µ 2a V 2 V 2 µ a µ a This is also the speed for a satellite in circular orbit at distance r a, so we have shown that the speed at the ends of the semi-minor axis is equal to the local circular speed Some dudes saw a thing in space with parameters r 1.5 [DU], V 2/3 [DU/T U], and φ 30. What s goin on with it? If the thing were escaping the Earth, its speed would have to be at least escape speed. V esc 2µ r [DU/T U] > V 3 2

3 Since its speed is less than escape speed, it must be on an elliptical orbit. If the thing was a ballistic missile, then at some point in its trajectory it would hit the surface of the earth. Another way of saying this is that the periapsis of its orbit is below the surface of the earth. E V 2 2 µ r 1 3 [DU 2 /T U 2 ] a µ 2E 3 2 [DU] h rv cos φ 3 2 [DU 2 /T U] e 1 + 2Eh2 µ r p a(1 e) 3 [DU] < r Since the periapsis distance is less than the radius of the earth (1 [DU]), the thing is gonna nail the earth. Run for your lives A vehicle destined for Mars is first launched into a [km] circular orbit around earth. (a) The speed of the circular orbit is V c µ /a. V c [km/s]. (b) After some time the speed is increased to V [km/s]. E V µ r E.939 [km2 /s 2 ] h rv 2 cos φ (φ 0 for a circular orbit) h [km 2 /s] e 1 + 2Eh2 µ 2 e Since e > 1, the orbit is hyperbolic which is expected since the vehicle should escape the earth to get to mars. (c) Since the energy is constant, the speed at r [km] is: V 2E + 2µ [km/s] r The hyperbolic excess speed for the vehicle is: V 2E 3.13 [km/s] These two speeds are very similar because the distance is so large that the vehicle is almost at infinity with respect to the earth. 2.1 Given r 1 ˆK [DU] and v 1Î [DU/T U], find the orbital elements. h r v 1Ĵ [DU 2 /T U] p h 2 /µ 1 [DU] e ( v h ) /µ r/r. So e 0. Since the orbit is circular, a p. a 1 [DU]. cos i h K /h 0, so i ±90. Since we are assuming that inclination is always between 0 and 180, we have i 90. 3

4 n ˆK h 1Î cos Ω n I /n 1. So Ω 180. For a circular orbit, ω and ν 0 are undefined. 2.5 (a) r 2Ĵ [DU], Direct equatorial circular orbit. Circular, so e 0 and ω and ν 0 are undefined and p r 2 [DU]. Equatorial, so i 0 and Ω is undefined. (b) r 1 ˆK [DU], v v esc Î. Since object has escape speed, e 1. Also, since the object starts out on the ˆK axis with an initial velocity in the Î direction, its motion will be in the Î- ˆK plane and it will eventually cross the Î axis, ascending. Therefore, the ascending node is on the Î axis where Ω 180 and Î- ˆK motion implies that i 90. We can see from inspection that the current position is at periapsis, so ν 0 0 and the angle from the ascending node (on the Î axis) to periapsis (on the ˆK axis) is ω 90. We know that at periapsis of a parabola r p/2. Therefore p 2 [DU]. (c) r 1 ˆK [DU], v 2Î + 2Ĵ [DU/T U]. Since the current position is along the ˆK axis, we know that the orbit must be polar and therefore i 90. Also, the given initial velocity is in the Î-Ĵ plane with equal magnitudes in both Î and Ĵ directions. This means that it will cross the Î-Ĵ plane while descending at an angle of 5 (measured counterclockwise from the Î axis). After this intersection, it will again cross the Î-Ĵ plane, but this time ascending and at an angle of So, Ω 225. This is more or less all that can be calculated purely by inspection. 2.6 r 1.2 ˆK [DU], v 0.Î 0.3 ˆK [DU/T U]. h r v 0.8Ĵ [DU 2 /T U] p h 2 /µ. p [DU]. e v h/µ r/r e 0.1Î ˆK. So e Inclination is the angle between ˆK and h, and since h coincides with Ĵ, i 90. n ˆK h 0.8Î cos u 0 n r/nr 0. Therefore u 0 ±90. Since the ˆK component of r is greater than zero, we know that u 0 will be less than 180. So u cos Ω n I /n 1, so Ω 180. cos ω n e/ne ω ± Since e K < 0, ω > 180 so ω cos ν 0 e r/er ν 0 ± Since r v 0.36 < 0, ν 0 > 180 so ν l 0 Ω + u 0 l Impact occurs at r 1 [DU]. From the conic equation, we have p r 1 + e cos ν cos ν impact p/r e ν impact ±159.7 Since the object is currently at ν , the impact will occur at the ν value right after this. So, ν impact Measuring latitude (degrees up from Î-Ĵ plane) is made easy by the fact

5 that the object is in a polar orbit. We know that ω is measured from Î (ascending node location) to e. Furthermore, we know that ν impact is measured from e to the position vector at impact. If we add these two angles, we will get the angular distance from the Î axis to the impact point. Doing so, we get θ impact Therefore, impact occurs at a latitude of θ impact 79.8 North. 2.7 r 0 Î Ĵ ˆK [DU], v 0 1/3(Î Ĵ + ˆK) [DU/T U]. h 0 r 0 v 0 2/3( Î + ˆK) [DU 2 /T U] p h 2 /µ. So p 8/9 [DU]. e v h/µ r/r Î Ĵ ˆK. So e cos i h k /h 2/2 i ±5. Since inclination is always positive by convention, i 5. n ˆK h 2/3Ĵ cos Ω n I /n 0 Ω ±90. Since n J < 0, Ω > 180 so Ω 90. cos ω n e/ne ω ±10.8. Since e K > 0, ω < 180 so ω cos ν 0 e r/er ν 0 ± Since r v 1/3 < 0, ν 0 > 180 so ν