This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Impulsive & Non-Impulsive Maneuvers”.
1. What does impulsive maneuver mean?
a) Rocket fire short bursts for required delta-v
b) Rocket fires long bursts for long time for the required delta-v
c) Rockets with instantaneous bursts
d) Maneuvers which are analyzed by Lambert’s equation
View Answer
Explanation: Maneuvers are required by the spacecrafts to move from one orbit to another. Impulsive maneuvers are the ones in which the rockets fire in short bursts for small duration to produce the required velocity change. This changes the magnitude and direction of the velocity vector.
2. 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 maneuver is carried out by firing brief on-board rocket motor. This results in a velocity shift of the spacecraft. If Δv reflects a magnitude change, it is referred to as a pumping maneuver.
3. 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. This velocity change yield both magnitude and direction change of the velocity vector. When Δv represents a change in direction it is referred as a cranking maneuver.
4. What does non- impulsive maneuver mean?
a) Rocket fire short bursts for required delta-v
b) Rocket fires long bursts for long time for the required delta-v
c) Rockets with instantaneous bursts
d) Maneuvers which are analyzed by Lambert’s equation
View Answer
Explanation: When the thrust applied by the rocket is done over a longer period to obtain the required delta-v, it is known as the non-impulsive maneuver. Unlike impulsive maneuver it does not involve applying impulse over a short period.
5. For each delta-v applied during the impulsive maneuver, what is the mass of propellant consumed?
a) \(\frac{\Delta m}{m}\) = e\(^{-\frac{\Delta v}{I_{sp} g_ο}}\)
b) Δm = \(\frac{1 – e^{-\frac{\Delta v}{I_{sp} g_ο}}}{m}\)
c) Δm = m – e\(^{-\frac{\Delta v}{I_{sp} g_ο}}\)
d) \(\frac{\Delta m}{m}\) = 1 – e\(^{-\frac{\Delta v}{I_{sp} g_ο}}\)
View Answer
Explanation: When an impulsive maneuver is carried out, there’s a change in delta-v applied. This results in change in the mass of propellant because of the consumption. The relation is given by:
\(\frac{\Delta m}{m}\) = 1 – e\(^{-\frac{\Delta v}{I_{sp} g_ο}}\)
Where, m is mass of spacecraft before the impulsive burn
Δm is the mass of propellant consumed
Δv is the impulsive burn
Isp is the specific impulse of the propellant.
6. During a non-impulsive maneuver, the spacecraft’s mass reduces. What is the formula to compute the rate of mass decrease due to combustion?
a) \(\frac{dm}{dt}\) = \(\frac{T}{I_{sp} g_0}\)
b) \(\frac{dm}{dt}\) = \(\frac{I_{sp} g_0}{T}\)
c) \(\frac{dm}{dt}\) = Isp g0
d) \(\frac{dm}{dt}\) = TIsp g0
View Answer
Explanation: When the rocket motor is fired during a non-impulsive maneuver to obtain a delta-v, there’s combustion of propellants. This results in decrease of mass of the spacecraft which his given by:
\(\frac{dm}{dt}\) = \(\frac{T}{I_{sp} g_0}\)
Where, T is thrust
Isp is the specific impulse of the propellant.
7. When impulsive maneuver is done at a point other than an apse line, we need to include the change in direction and magnitude of velocity vector.
a) True
b) False
View Answer
Explanation: When a delta-v calculation is done for an impulsive maneuver at a point that is not on the apse line, care must be taken to include the change in direction as well as the magnitude of the velocity vector. The formula thus becomes:
Δv = \(\sqrt{v_1^2 + v_2^2 – 2v_1 v_2 cos\Delta \gamma}\)
Where, Δγ is the difference in flight path angles.
8. What is the approximate mass of propellant consumed if the mass of the spacecraft before burn is 100,000 kg with the specific impulse of 150 s and velocity change of 1 km/s?
a) 50 kg
b) 68 kg
c) 100 kg
d) 150 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 = 100,000 kg, Δv = 1 km/s, Isp = 150 s
\(\frac{\Delta m}{100,000}\) = 1 – e\(^{-\frac{1}{150 × 9.81}}\)
Δm = 100,000 × (1 – 0.99932) = 67.935 kg ≅ 68 kg.
9. Which of these is not a type of an impulsive 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.
10. What is the formula used to compute the time elapsed for a non-impulsive maneuver?
a) t = \(\frac{m_0 g_0 I_{sp}}{T} \Bigg[e^{-\frac{\Delta v}{I_{sp} g_ο} \Big(\sqrt{\frac{\mu}{r}} – \sqrt{\frac{\mu}{r_0}}\Big)} \Bigg]\)
b) t = \(\frac{T}{m_0 g_0 I_{sp}} \Bigg[1 – e^{-\frac{\Delta v}{I_{sp} g_ο} \Big(\sqrt{\frac{\mu}{r}} – \sqrt{\frac{\mu}{r_0}}\Big)} \Bigg]\)
c) t = \(\frac{m_0 g_0 I_{sp}}{T} \Bigg[1 – e^{-\frac{\Delta v}{I_{sp} g_ο} \Big(\sqrt{\frac{\mu}{r}} – \sqrt{\frac{\mu}{r_0}}\Big)} \Bigg]\)
d) t = \(\frac{m_0 g_0 I_{sp}}{T} \Bigg[1 – e^{-\frac{\Delta v}{I_{sp} g_ο}} \Bigg]\)
View Answer
Explanation: The formula to compute the time elapsed for a non-impulsive maneuver is given by:
t = \(\frac{m_0 g_0 I_{sp}}{T} \Bigg[1 – e^{-\frac{\Delta v}{I_{sp} g_ο} \Big(\sqrt{\frac{\mu}{r}} – \sqrt{\frac{\mu}{r_0}}\Big)} \Bigg]\)
Where, m0 is the initial mass of spacecraft before the burn
Isp is the specific impulse of the propellant
T is the tangential thrust
Δv is the impulse
r is initial radius of the orbit
r0 is the radius of the final orbit.
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