This set of Spaceflight Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Lunar Trajectories – Patched Conic Approximations”.
1. Transition from a geocentric motion to selenocentric motion makes use of patched conic approximation to determine an approximate trajectory.
a) True
b) False
View Answer
Explanation: Transition from a geocentric motion to selenocentric motion makes use of patched conic approximation. The gradual process takes plane over a finite arc of the trajectory where Earth and Moon affect path equally. Patched conics give a nice approximation for preliminary analysis.
2. Patched-conic approximation can be used for calculating the return trajectory to Earth from Moon.
a) True
b) False
View Answer
Explanation: There is a high perturbation which is encountered when the spacecraft is in the moon’s sphere of influence. Due to this reason patched–conic approximation cannot be used for return trajectories back to the earth.
3. Which of these assumptions are not employed for using patched-conic approximation for a selenocentric orbit?
a) The Earth is fixed in space
b) The eccentricity of the Moon orbit around the Earth is neglected
c) The flight of the space vehicle takes place in the Moon orbital plane
d) The Moon is fixed in space
View Answer
Explanation: There are various assumptions employed for using patched–conic approximation for a selenocentric orbit. Out of these, the assumption that is invalid is that the moon is not fixed in space.
4. To compute the efficiency of the patched- conic approximation, in how many stages is the trajectory divided into?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: To compute the efficiency of the patched- conic approximation, we divide the Hohmann transfer into three stages. First stage is the hyperbolic departure orbit, second is the elliptic orbit about the sun and finally the reentry hyperbolic orbit.
5. Which of these formulas is used to compute the heliocentric velocity to leave the Earth’s sphere of influence as a result of patched- conic approximation?
a) v1 = \(\sqrt{\frac{2\mu_\odot}{r_e + r_m} \Big(\frac{r_m}{r_e}\Big)}\)
b) v1 = \(\sqrt{\frac{2\mu_\odot}{r_e + r_m}}\)
c) v1 = \(\sqrt{\frac{2\mu_\odot}{r_e – r_m}}\)
d) v1 = \(\sqrt{\frac{2\mu_\odot}{r_e – r_m} \Big(\frac{r_m}{r_e}\Big)}\)
View Answer
Explanation: One of the application of patched- conic approximation is to compute the delta-v which is required to carry out the complete transfer. For this, w first need to find out v1 which is the heliocentric velocity which is required to leave the earth’s sphere of influence. This is calculate using the formula as below:
v1 = \(\sqrt{\frac{2\mu_\odot}{r_e + r_m} \Big(\frac{r_m}{r_e}\Big)}\)
Where, μ⊙ is sun’s gravitational coefficient
re is radius of Earth’s mean orbit
rm is radius of Mars’ mean orbit (It can be any other planet as well).
6. If a spacecraft travels from Earth to Mars, and it leaves the sphere of influence of Earth, then which celestial body’s gravitational field is prevalent?
a) Earth
b) Mars
c) Sun
d) Moon
View Answer
Explanation: Using patched- conic approximation, each planet has its own sphere of influence. When the spacecraft is inside that respective planet’s SOI, the gravitational field of only that particular planet is accounted for. Once it leaves its SOI, then sun’s gravity dominates.
7. What kind of orbit is followed during the elliptic transfer, when the spacecraft is under sun’s gravitational influence?
a) Perturbed Keplerian orbit
b) Unperturbed Keplerian orbit
c) Osculating Keplerian orbit
d) Non-Keplerian orbit
View Answer
Explanation: In the second stage of the trajectory which is approximated by the patched-conic method, the trajectory is elliptical. It is not under any planet’s sphere of influence and only sun’s gravity is dominant. Thus is follows an unperturbed Keplerian orbit around the sun.
8. With the help of patched-conic approximation, the trajectory is resolved into which of these problems?
a) Two-body problem
b) N-body problem
c) Three- body problem
d) Euler’s problem
View Answer
Explanation: Using patched- conic approximation, we simplify the trajectory into three stages so that we can consider each region as a two0body problem and eventually patch them together to get the final result.
9. Which frame of reference is used when the spacecraft leaves Earth’s gravitational field along the hyperbolic escape trajectory?
a) Hipparcos celestial reference frame
b) Barycentric celestial reference frame
c) Earth-centered inertial frame
d) Geocentric frame of reference
View Answer
Explanation: When the spacecraft leaves the Earth’s gravitational field along the hyperbolic escape trajectory, the patched-conic approximation is carried out by using two-body motion. The frame of reference used to carry out this analysis is Earth-centered inertial frame.
10. Which of these transformations is required to perform patched- conic approximation at the patch points?
a) Conformal coordinate transformation
b) Affine coordinate transformation
c) Projective coordinate transformation
d) Coordinate transformation
View Answer
Explanation: In order to use patched- conic method for preliminary analysis, there must be coordinate transformation between the various body-centered frames at the sphere of influence patch point. The reference frame for each stage of the trajectory changes, thus without the coordinate transformation, the results will be inaccurate.
11. What is the definition of the reference velocity used for the heliocentric orbits?
a) Circular orbital speed in heliocentric orbit at 1AU
b) Ratio of velocity at patch points
c) Square of ratio of velocity at patch points
d) Ratio of gravitational parameter and heliocentric velocity
View Answer
Explanation: When heliocentric orbits are involved, we make use of reference distance which is 1AU. The reference velocity is defined at the circular orbital speed in a heliocentric orbit at 1AU. This velocity is mean velocity of Earth in its orbit about the Sun.
The formula is: vref = \(\sqrt{\frac{\mu_s}{r_{ref}}}\) = 29.7847 km/s
Where, μs is the gravitational parameter of the sun = 1.327124 × 1011 km3/s2.
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