This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Single Impulse Orbital Maneuver”.
1. When the change in velocity represents a change in magnitude, what kind of maneuver is it called?
a) Cranking maneuver
b) Pumping maneuver
c) Orbital raising maneuver
d) Orbital lowering maneuver
View Answer
Explanation: Impulsive maneuvers are carried out by brief firings of rocket motor. A velocity shift of the spacecraft results from each impulsive maneuver. If Δv reflects a magnitude change, it is referred to as a pumping maneuver.
2. When the change in velocity represents a change in direction, what kind of maneuver is it called?
a) Cranking maneuver
b) Pumping maneuver
c) Orbital raising maneuver
d) Orbital lowering maneuver
View Answer
Explanation: Each impulsive maneuver results in a velocity change of the spacecraft. There’s change in both the magnitude and direction of the velocity vector. When Δv represents a change in direction it is referred as a crankingmaneuver.
3. What is the approximate mass of propellant consumed if the mass of the spacecraft before burn is 500,000 kg with the specific impulse of 230 s and velocity change of 1.25 km/s?
a) 250 kg
b) 277 kg
c) 300 kg
d) 272 kg
View Answer
Explanation:The relation between change in velocity and mass of propellant is given by-
\(\frac{\Delta m}{m}\) = 1 – e\(^{-\frac{\Delta v}{I_{sp} g_ο}}\)
Given- m = 500,000 kg, Δv = 1.25 km/s, Isp = 230 s
\(\frac{\Delta m}{500,000}\) = 1 – e\(^{-\frac{1.25}{230 × 9.81}}\)
Δm = 500,000 × (1 – 0.99446) = 276.925 kg ≅ 277 kg.
4. What is the change in velocity required to make the final circular orbit’s radius equal to the perigee radius of the initial elliptic orbit of the satellite moving in a geocentric orbit of e = 0.5 and semi major axis of a = 8000 km?
a) 2.44 km/s
b) 12.226 km/s
c) 9.982 km/s
d) 0.25 km/s
View Answer
Explanation: Given- semi major axis (a) = 8,000 km, eccentricity (e) = 0.5
The impulse has to be applied at the perigee of the elliptical orbit to circularize it.
To find the perigee radius: rp = a(1 – e) = 8,000(1 – 0.5) = 4,0000 km
The velocity at the perigee vp = \(\sqrt{\frac{-\mu}{a} + \frac{2\mu}{r_p}} = \sqrt{\frac{-398.600}{8,000} + \frac{2 × 398,600}{4,000}}\) = 12.226 km/s
The velocity at the circular orbit with its radius equals rp is: vc = \(\sqrt{\frac{\mu}{r_p}} = \sqrt{\frac{398,600}{4,000}}\) = 9.982 km/s
The final circular velocity is less than the velocity at the perigee of the elliptical orbit, thus an impulse in the opposite direction has to be applied.
Δv = vp – vc = 12.226 – 9.982 = 2.44 km/s
5. What is the approximate burn time for the single- impulse maneuver with the mass of propellant as 350 kg and mass flow rate of the engine as 1.5 kg/s?
a) 100 s
b) 525 s
c) 0.0043 s
d) 233 s
View Answer
Explanation: The formula to compute the burn time for a single impulse maneuver is given by:
tburn = \(\frac{m_p}{m}\)
̇
Given- mass of propellant (mp) = 350 kg, mass flow rate (m) = 1.5
tburn = \(\frac{350}{1.5}\) = 233.33 s ≅ 233 s
6. For raising or lowering the apogee/perigee radius, single impulse maneuver is carries out.
a) True
b) False
View Answer
Explanation: One of the uses of single impulse maneuver is to raise or lower the apogee and perigee radius. This is done by creating an impulse/burn at the perigee or apogee. This burn should occur at a flight path angle of 0 degrees to have the maximum efficiency.
7. What is the impulse required for plane change from a circular orbit at 20,000 km of inclination 35 degrees to an orbit of 46 degrees?
a) 0.314 km/s
b) 0.549 km/s
c) 0.745 km/s
d) 0.845 km/s
View Answer
Explanation: The formula to compute the velocity impulse is: Δv = 2vsin \(\frac{\Delta i}{2}\)
Given- Altitude of circular orbit (r) = 20,000 km, i1 = 35°, i2 = 35°
The difference in orbital inclination is: Δi = 46° – 35° = 11°
The velocity in the circular orbit is: v = \(\sqrt{\frac{μ}{r}} = \sqrt{\frac{398,600}{6378 + 20,000}}\) = 3.887 km/s
(where altitude = Radius of Earth + r)
Δv = 2vsin \(\frac{\Delta i}{2}\) = 2 × 3.887 × sin \(\frac{11^°}{2}\) = 0.745 km/s
8. What is the minimum magnitude of change in velocity for a single impulsive maneuver?
a) 0.1 km/s
b) 0.25 km/s
c) 0.5 km/s
d) 1 km/s
View Answer
Explanation: The change in velocity for the single impulse maneuver is always higher than 1km/s as there are no re- fueling stations in the space for the spacecraft, so the Δv has to be calculated beforehand to minimize propellant mass carried for the payload.
9. What is the inclination-change impulse required by the satellite with a velocity of 8.5 km/s for the change in orbital inclination of 3 degrees and flight path angle of 30 degrees?
a) 0.385 km/s
b) 0.254 km/s
c) 0.412 km/s
d) 0.614 km/s
View Answer
Explanation: Given- v = 8.5 km/s, Δi = 3°, γ= 30°
The formula to compute the inclination change impulse is:
Δv = 2v cosγ sin \(\frac{\Delta i}{2}\) = 2 × 8.5 cos(30) sin \(\frac{3}{2}\)
Δv = 0.385 km/s
Where vthe velocity of the satellite at the ascending node is, γ is the flight path angle and Δi is the change of inclination between the two orbits.
10. Which of these is not a type of a single-impulse maneuver?
a) Energy change
b) Simple plane change
c) Adjustment of perigee and apogee height
d) Combined change of apsis altitude and plane orientation
View Answer
Explanation: There are four types of single impulse maneuver-Adjustment of perigee and apogee height, simple rotation of the line of apsides, simple plane change and combined change of apsis altitude with plane orientation. There is no change in energy involved by applying single impulse.
Sanfoundry Global Education & Learning Series – Orbital Mechanics.
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