System engineer Space Ops. Education Coursework Assignment

System Engineer Space OPS
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Question 1
Why does a spacecraft transferring from an initial circular orbit to a larger circular orbit perform two maneuvers, both of which increase it's speed, but end up with a slower orbital velocity in the final orbit than it started with in the original orbit? 
The Hohmann Transfer, which is an orbital maneuver, focuses on the way a spacecraft or satellite transfers from one circular orbit to another circular orbit where the orbits are of different radii. When a spacecraft moves from went a low orbit to a higher orbit, the craft needs to accelerate. When the velocity of the lower circular orbit is higher than velocity of the higher circular orbit, where there is acceleration then the craft moves up in the slower orbit (Sellers& Astore, 2007). In other words, at the lower orbit the forward velocity is higher than at a higher orbit. The instantaneous velocity changes (ΔVs) are tangent to the initial and final orbits (Sellers, & Astore, 2007). In the first maneuver, the velocity increases and then the spacecraft moves into the elliptical orbit where more energy is added. In the second maneuver required to move the spacecraft to the lower orbit, there is a decrease in the semimajor axis as a result the energy and this is achieved as the velocity decreases.
When the satellite or space craft altitude changes this also influences a change in the orbital speed where and moving to a higher satellite altitude slows the orbital velocity. In the first change in velocity, the velocity increases and the spacecraft gets out of the first orbit and moves to the transfer orbit where the velocity decreases at a time when the spacecraft gains more of the potential energy in place of the kinetic energy. A low-altitude orbit requires less energy (per unit mass) than other a high altitude. At the same time, gravitational field weakens at longer distances, while where the orbital speeds decrease as the velocity increases.
Question 2
Why do we take the absolute value of the difference between the two orbital velocities when we compute total ΔV? 
The absolute value of the difference between orbital velocities is used to determine the total ΔV since the velocity change helps to calculate the energy and fuel needed. The spacecraft accelerates to the higher orbits and decelerates t when moving into a lower orbit. The ΔV then from one orbit to another depends on the velocity transfer at orbit 1 and the velocity of orbit 1. In computing the ΔV total, the energy equations used are from orbital mechanics and orbital maneuvering reflects the relationship between the velocity vector, gravitation parameter and the spacecraft position vector, which influence the mechanical energy (Sellers& Astore, 2007).
Question 3
What velocity change (ΔV) needed to from the initial circular orbit into the transfer orbit?
ε orbit 1= 3.986 × 105(km3/s2 / 2(500 km) = 398.6
ε orbit 2= 3.986 × 105(km3/s2 / 2(150 km) = 1328.67
ε transfer= - 3.986 × 105(km3/s2 / 2(650 km)= 613.23
Vorbit 1= 48.90
V orbit at transfer 1= 53.11
r=500 km and a=500 km
vpark= 398600 *((2/500-1/500))^0.5
Vpark= 28.23 km/ s
a = (aTarget + aParking)/2 = (500+ 150)/2= 325km
V transfer= 398600 *((2/500-1/325)) ^0.5
V transfer where r=500 km and a =325, V= 19.18
Earth's gravitational parameter ≈ 3.986 × 105(km3/s2) = 3.986∙1014 m3/s or 3.986∙* 10^14
ΔV= 19.18km/s- 28.23 km/ s= 9.05
ΔV1= -9.05
What velocity change (ΔV) needed to from the transfer orbit into the final orbit?
r=150 km and a=150 km
Vpark= 51.55 km/ s
a = (aTarget + aParking)/2 = (500+ 150)/2= 325km
V transfer where r=500 km and a =325, V= 63.94
ΔV= 51.55-63.94= -12.39 km/s
ΔV2= -12.39 km/s
TOF required for the transfer
Transfer orbit's time of flight (TOF)
TOF= π * √325km 3/ 3.986 × 105(km3/s2 =29.15 seconds
Question 4
A spacecraft deployed into circular orbit, inclined 57 ° at 200km altitude, needs to change to a polar orbit at the same altitude. What ΔV does this maneuver require?
R = 6378,000 m
r = R + 200,000 m
z=r = 6,578,140 m ≈ 6,578,000 m
GM = 3.986∙1014 m3/
V = √ (GM/r)
V =7,784.34 m/s
i = 57°
cos (57) = 0.5446
vx = - V cos i= 7,784.34*0.5446 = -4,239.65 m/s
Δ vx= 4,239.65m/s
sin(57) = 0.8387
vz= Vsin i = 0.8387* 7,784.34= 6,528.50 m/s
ΔVz =7,784.34 m/s - 6,528.50 m/s= 1,255.84 m/s
ΔV= √ [(ΔVx)²+(ΔVy)²+(ΔVz)²] = 4,421.7 m/s
Question 5
Now that the spacecraft from the precious problem is in a polar orbit, what Δν will change the right ascension of the ascending node by 35 °
z=r = 6,578,140 m ≈ 6,578,000 m
V y= -7,784.3 m/s
i= 35°
V x= -7,784.3 m/s (– sin 35°), where sin 35is 0.5736 =4,464.9 m/s
Vy=7,784.3 m/s (cos 35°), where cos 3 is 0.8192 = -6,376.56 m/s
ΔVx= 4,464.9m/s
ΔVy= 7,784.34 - 6,376.56 = 1407.78 m/ s
ΔV = √ [(ΔVx) ²+ (ΔVy) ²+ (ΔVz ) ²] = 4,681.6...