dataset update

astro-mcqa.csv

@@ -50,3 +50,30 @@ An artificial Earth satellite is in an elliptical orbit with a perigee altitude
 An artificial Earth satellite is in an elliptical orbit with a perigee altitude of hp = 250 km and an apogee altitude of ha = 800 km. What is its orbital period expressed in minutes?,"['18.0', '89.5', '95.0', '100.9']","[0, 0, 1, 0]",,True,6ef8b7da-7aea-4809-95ce-c17f64866bee
 A geostationary orbit is defined as an orbit were the satellite is always pointing towards the same point on the Earth's surface. What is its eccentricity?,"['0', '1', '1.3', '0.5']","[1, 0, 0, 0]",,True,5064e3a3-16e9-431c-816e-3d8cd10c9331
 A geostationary orbit is defined as an orbit were the satellite is always pointing towards the same point on the Earth's surface. What is its inclination?,"['7.01°', '0°', '3.2°', '45°']","[0, 1, 0, 0]",,True,6192c62d-489d-45e4-acf6-23083a809273
+"In orbital mechanics, what is the meaning of the mean motion n = sqrt( mu / a^3 )?","['It is the average angular rate over one full orbit, in rad/s', 'It can be defined only for parabolic orbits.', 'It corresponds to the instantaneous angular rate for a non-circular orbit.', 'In the specific case where the eccentricity e tends to zero, the mean motion and the true anomaly tend to be equal.', 'The mean motion relates to the true position of a body moving along a Keplerian orbit.']","[1, 0, 0, 0, 0]",,True,bc2f3517-6e5a-4b39-89c6-fb11c2209392
+What reference frame is the best suited for an interplanetary probe?,"['Geographic coordinate system', 'Geo-centric coordinate system', 'Heliocentric-inertial coordinate system']","[0, 0, 1]","The center of the heliocentric-inertial coordinate system is the Sun. The two other coordinate systems use the Earth as center, and are not appropriate to for an interplanetary probe.",True,8d7cc580-c7b9-4875-9ba6-bf468df0a61d
+The complete precessional cycle of the Earth lasts about 26000 years. What is the precession rate? (in degrees per year),"['0.00384', '0.01384', '0.02384', '0.03384', '0.04384', '0.05384']","[0, 1, 0, 0, 0, 0]",The precession rate per year is 360°/26000 years = 0.0138°/years.,True,e4fda0e7-b147-48fd-bcea-fbc007d2ee9b
+What is the RAAN?,"['The inclination of the orbital plane.', 'The time of the periapsis transit.', 'The angle from a reference direction to the point where the satellite crosses the plane of reference, towards the south.', 'The angle from a reference direction to the point where the satellite crosses the plane of reference, towards the north.']","[0, 0, 0, 1]","The longitude of the ascending node (Ω) is one of the orbital elements used to specify the orbit of an object in space. It is the angle from a reference direction, called the origin of longitude, to the direction of the ascending node, measured in a reference plane.",True,7c7013ed-cb14-4298-a435-259072390c40
+How long is a sideral day?,"['24h00min', '24h04min', '11h56min', '23h56min']","[0, 0, 0, 1]","The sidereal day, is the time it takes for the Earth to make one full rotation with respect to the stars, 23h56'04''.",True,e5939e55-2802-4753-b849-2efdd4aafa9a
+Where is an orbital maneuver Delta-V adding energy to a spacecraft on orbit best performed?,"['Along the orbital velocity', 'Radially', 'Along the orbit angular momentum', 'Any direction of the Delta-V vector will add energy to the spacectaft on orbit.']","[1, 0, 0, 0]","In the impulsive case, only maneuvers along the velocity will impact the energy. Radial and cross-track maneuvers will only change the direction of the speed without changing its norm.",True,3e8adab1-90a8-4d92-a636-f514f12d90cf
+What is the Hohmann transfer for small Delta-V useful for?,"['Interplanetary trajectories', 'Earth-Moon transfer', 'Manoeuvers around small bodies (e.g. comets)', 'Orbital rendez-vous']","[0, 0, 0, 1]",,True,b706586d-50b8-4d17-b9f2-890c679ff6e8
+Where is the Delta-V for an inclination change smaller?,"['at low altitudes', 'at high altitudes', 'at the equator', 'at the poles']","[0, 1, 0, 0]","The Delta-V for inclination change is proportional to the orbital velocity. Thus, the smaller the orbital velocity is, the less is required to change the inclination and RAAN. The orbital velocity decreases with the square of the distance from the center of the body, thus higher altitudes are better.",True,a972601f-0015-48b8-8747-5e5dd8e92e12
+What is the difference between geostationary and geosynchronous orbits?,"['Geostationary has an orbital period of 24h whereas geosynchronous orbits is 23h56', 'The orbital parameters for geostationary are inclination = 0° and eccentricity = 0', ""A geosynchronous satellite orbits the same location over Earth's surface while a geostationary satellite remains on average at the equator"", 'All of the above']","[0, 1, 0, 0]","A geostationary satellite orbits the same location over Earth's surface while a geosynchronous satellite remains on average at the equator. The definition of geostationary orbits is orbital period is Earth's sideral day, e = 0 and i = 0°",True,e723f537-9123-4794-99c4-6dc0f39f9e8f
+The ground tracks for a satellite in LEO (typical orbital period is 90 min) is shifting to the west from one equator crossing to the next. What is the typical value of the shift in degrees?,"['12.5', '15', '17.5', '20', '22.5']","[0, 0, 0, 0, 1]","At LEO altitudes, the typical orbital period is 90 min. In an hour the Earth rotates by 15° (= 360° / 24h), thus in 90 min, we have a shift of 22.5°",True,60f3877b-24ba-45fe-af9e-d67d14170174
+"For a circular orbit of a given inclination, at what altitude will a satellite be impacted the most by nodal regression?","['200 km', '2000 km', '500 km', '12000 km']","[0, 0, 0, 0]","The lower the satellite, the stronger the effects of the nodal regression.",True,f9c603e5-f517-4636-849f-350a2759fcdc
+"Among the following assertion about Lagrange points, which one is true?","['The Sun-Earth system does not have any Lagrange points because of the Moon', 'The distance between the Lagrange points and the Earth varies drastically with the seasons', ""The Lagrange points orbit the Sun with the same orbital period as the Earth's"", 'None of the above']","[0, 0, 1, 0]",The Lagrange points orbit the Sun with the same orbital period as the Earth's. The mass of the satellite on a Lagrange point is adjusted (lessened in the case of L1 and highthened for L2) such that the orbital period remains one year.,True,41c719d0-9f62-40da-abeb-8bd99a86ef8b
+"A spacecraft is below and behind the ISS, both are on circular orbits. What will happen to their relative position?","['The spacecraft will overtake the ISS', 'The ISS has a larger velocity than the spacecraft, so it will leave the spacecraft behind', 'Nothing, they will stay at their initial distance', 'The spacecraft will get to an higher altitude than the ISS']","[1, 0, 0, 0]",The orbital velocity decreases with altitude (with the square root of the inverse to the distance to Earth's center). Thus the spacecraft moves faster and will catch-up with the ISS and eventually overtake it.,True,a2db30ce-3528-4548-a228-c27b103a545c
+What is the effect of a posigrade burn on a circular orbit?,"['There is no effect', 'The altitude 90° from the burn point is decreased', 'The altitude 180° from the burn point is decreased', 'The altitude 180° from the burn point is increased']","[0, 0, 0, 1]","At the burn point, the velocity is increased. It means that the energy of the spacecraft changes and the geometry of the orbit changes. The altitude of the burn point will not change, but becomes the minimal altitude if the initial orbit is circular. Its maximum will be reached at the apogee, 180° from the perigee. Thus the altitude 180° from the burn point is increased.",True,11665c59-aacf-437d-b83e-16aaa6921ced
+What is the reference point of the rendezvous profile diagram?,"['The chaser', 'The target', 'The center of the Earth', 'The control center']","[0, 1, 0, 0]",The rendezvous profile diagram is centered and relative to the target.,True,a28c168f-3f6a-4edb-b0c0-2ae974963858
+"What is the shape of a circular orbit, for the chaser, in a rendezvous profile diagram? (assuming the target is higher than the chaser).","['A point', 'A periodic wavy and pointy curve', 'A line', 'A sinusoidal', 'A cycloid']","[0, 0, 1, 0, 0]","On a circular orbit, the altitude does not change, so the value of the R axis does not change. The V axis does not stay the same as the velocity of the two spacecraft are not the same. The chaser will move along an horizontal line in the diagram.",True,e8b2cae4-92f6-48e6-9436-c5f33a5ac832
+A spacecraft on a circular orbit at 400 km does a retrograde burn of 1 m/s. What will be the change in the altitude (in km) of the perigee?,"['1.5', '2.5', '3.5', '4.5']","[0, 0, 1, 0]",This can be computed in a straightforward way using the equation Delta-r is close to 3.5 * Delta-V where Delta-r is in km and the burn value in m/s.,True,61e3c07e-fd5e-4429-aeb1-66afe0919eda
+"A chaser is on a circular orbit at the same altitude as the target and few kilometers behind.
+Where will the chaser be one orbit after a posigrade burn ?","['The chaser will be much further ahead', 'The chaser will be behind the target, further away than it was initally', 'The chaser will be higher than the target', 'The chaser will be lower than the target']","[0, 1, 0, 0]","Since there is a posigrade burn, the altitude of the semi-major axis is increased, the orbital velocity is decreased. If the change of altitude is sufficient, the chaser will drift behind the target further away than it was initially.",True,9fc96902-8398-4630-b53b-6d3049c8254e
+What is the definition of the Astronomical Unit (AU)?,"['It is the average distance between the Earth and the Sun', 'It is the average radius of the solar system', 'It is the average radius of the Sun', 'It is the average distance betwteen the Moon and the Earth']","[1, 0, 0, 0]","An Astronomical Unit is the mean distance between the Earth and the Sun. In 2012, the International Astronomical Union defined the distance to be 149,597,870,700 meters.",True,3fedeb91-5ab5-45de-be70-a2ce8ad59e10
+What is the sphere of influence?,"['A region in space that can be controlled by a spacecraft', 'The region around the Sun in which only planets have a gravitational influence', 'A sphere around each planet inside which the motion of a spacecraft must be considered a three-body Keplerian problem', 'A sphere around each planet inside which the motion of a spacecraft is considered to be two-body Keplerian']","[0, 0, 0, 1]","For the Earth, the radius of the sphere of influence is about 924 000 km.",True,6a0ff5bb-46cb-4b10-a12b-4573589bb59f
+What is the hyperbolic excess velocity?,"['The velocity required to get into a LEO orbit from the surface of the Earth', 'The velocity at which we cross the sphere of influence', 'The velocity needed to reach the arriving planet', 'The velocity needed to reach the sphere of influence']","[0, 1, 0, 0]",It is the speed in excess of the minimum velocity required to reach the sphere of influence.,True,f274dd2f-9811-4424-abca-f3c6e4c61bd8
+"A interplanetary probe is in the sphere of influence of the Earth, in transit towards an inner planet. What is its energy with respect to the Earth?","['E <= 0', 'E < 0', 'E = 0', 'E > 0']","[0, 0, 0, 1]","The energy for the hyperbola is epsilon = + mu / 2a (energy per unit mass), thus positive. Another way to see this, is that the probe is no longer gravitationally bounded to the Earth, ergo its energy is positive.",True,42d75d3d-7aed-4cc3-a784-814bfc3e0fd6
+"What is the ideal shape of the transfer trajectory between two planets, and why?","['A Hohmann transfer orbit, because it is the less expensive orbit in terms of energy', 'A straight line, because is it the shortest path', 'A Hohmann transfer orbit, because it is the shortest path', 'A straight line, because it is the less expensive orbit in terms of energy']","[1, 0, 0, 0]","The Hohmann transfer is the optimal and ideal trajectory in terms of time of flight and fuel. However, in reality, different trajectories are preferred. For Mars transfer, more energetic trajectories are usually preferred to shorten the time of flight.",True,b2345b77-cb06-476f-82f4-092f001b8356
+"An interplanetary probe is in transit towards an inner planet, very close to the arrival planet. How is the velocity of the planet with respect to the velocity of the probe?","['faster', 'identifical', 'slower']","[0, 0, 1]","The transfer ellipse ressemble a Hohmann transfer to go from a high to a lower orbit, thus the velocity of the probe is higher than the arrival planet. When leaving the Earth, the probe should also reduce its heliocentric speed. This can be done by crossing the sphere of influence with the velocity vector of the spacecraft in the opposite direction of Earth's velocity vector.",True,e344b8fb-a2cc-42b2-979c-544644a2a5b0
+There is no burn during the slingshot maneuvers and yet the speed in the heliocentric reference frame increases. How is that possible?,"['At the periapsis of the slingshot, some massive object is jettisoned', 'An ion thruster is systematically switched on and this type of propulsion is only efficient in planetary neighborhood', 'The velocity vector of the probe is not changed in norm, but in direction in the planetocentric reference frame']","[0, 0, 1]","The norm of the velocity vector stays constant, there is no acceleration in the direction of the velocity vector. However, the direction of the trajectory of the spacecraft in the sphere of influence of the planet is a hyperbola. The direction of the velocity vector changes when the spacecraft is inside the sphere of influence.",True,6012cf0f-5cb3-440a-a35b-91c61539f393
+What can Gravity-assist be used for? (more than one answer possible),"['Increase the heliocentric velocity', 'Decrease the heliocentric velocity', 'Make communications between the probe and Earth easier', 'Explorations of ""worlds"" during the transit flight to the destination']","[1, 1, 0, 1]","The most common application of gravity assist is to increase the heliocentric velocity. However, it can be reduced if the direction of the spacecraft velocity vector is well chosen when entering the sphere of influence. Voyager and Pioneer probes took advantage of their multiple gravity assist maneuvers to explore and take close-up images of all the planets in our solar system. This was made possible due to a particularly favorable alignement of the planets.",True,bb3def0d-b962-4777-84f9-bdd55055516a