100+ Free OBA Practice Questions

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Sample OBA Practice Questions

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1What is the primary astronomical cause of the annual cycle of seasons on Earth?

A.The constant 23.44° tilt of Earth's rotational axis relative to the ecliptic plane as it revolves around the Sun

B.The periodic variation in distance between the Earth and the Sun due to Earth's orbital eccentricity (perihelion vs aphelion)

C.The periodic fluctuation in the Sun's total solar irradiance and magnetic activity over an 11-year cycle

D.The gravitational precession of Earth's rotational axis relative to the background constellations

Explanation: Seasons on Earth are caused by Earth's axial tilt (obliquity) of approximately 23.44° relative to its orbital plane (the ecliptic). As Earth orbits the Sun, this fixed tilt causes each hemisphere to tilt toward the Sun during its summer (receiving more direct sunlight and longer daylight hours) and away from the Sun during its winter.

2A solar day on Earth lasts exactly 24 hours (86,400 seconds), whereas a sidereal day lasts approximately 23 hours, 56 minutes, and 4 seconds. What explains this 3 minute and 56 second difference?

A.Atmospheric refraction delays the perceived setting of the Sun relative to distant background stars

B.Earth advances approximately 1° along its orbit around the Sun each day, requiring an extra ~4 minutes of rotation to realign with the Sun

C.Tidal braking from the Moon's gravitational pull slows down Earth's rotation during daylight hours

D.Earth's orbital velocity increases as it moves from aphelion to perihelion according to Kepler's second law

Explanation: A sidereal day is the time Earth takes to complete one full 360° rotation relative to distant stars (23h 56m 4s). Because Earth simultaneously orbits the Sun by about 360°/365.25 ≈ 0.986° per day in the same prograde direction, Earth must rotate an extra ~0.986° (taking 24h / 360 ≈ 3.93 minutes or 3m 56s) for the Sun to return to the local celestial meridian (a solar day of 24h).

3During an equinox, what approximate solar pattern is observed at locations away from the geographic poles when atmospheric refraction is ignored?

A.The Sun reaches its maximum altitude of exactly 90° (zenith) at solar noon from every latitude on Earth

B.The Sun remains continuously above the horizon for 24 hours at both the North and South geographic poles

C.The Sun rises due east and sets due west, with day and night each lasting approximately 12 hours

D.The shadow cast by a vertical gnomon at local solar noon completely disappears across the entire Southern Hemisphere

Explanation: At an equinox the Sun's declination is approximately 0°, so its daily path intersects an ordinary horizon near the east and west cardinal points. The geometric day and night are each about 12 hours away from the poles; atmospheric refraction, the Sun's finite disk, and polar geometry complicate a claim of exact equality everywhere.

4The city of São Paulo has a geographic latitude of approximately 23°33' S. How many times per year does the Sun reach the exact zenith (altitude 90°) at solar noon in São Paulo?

A.Twice per year, once in late spring and once in late summer

B.Once per year, precisely at the December summer solstice

C.Four times per year, during the equinoxes and cross-quarter days

D.Zero times per year, because São Paulo lies south of the Tropic of Capricorn (23°26' S)

Explanation: The Sun can only reach the zenith (an altitude of 90°) at locations situated between the Tropic of Cancer (23°26' N) and the Tropic of Capricorn (23°26' S), known as the intertropical zone. Because São Paulo is located at latitude 23°33' S, which is south of the Tropic of Capricorn by about 7 arcminutes (~13 km), the Sun never reaches the zenith (maximum altitude at summer solstice is 90° - |23°33' - 23°26'| = 89°53').

5What physical mechanism causes the precession of the equinoxes, which has a cycle period of approximately 25,772 years?

A.Gravitational torque exerted by the Moon and the Sun on Earth's equatorial oblateness (bulge)

B.Magnetic torque generated by the interaction between the solar wind and Earth's liquid outer core dynamo

C.Relativistic frame dragging of spacetime caused by the rapid rotation of the Sun

D.Tidal friction between the oceanic water masses and the ocean floor shifting Earth's center of mass

Explanation: Because Earth is an oblate spheroid with an equatorial bulge (equatorial radius ~21 km larger than polar radius), the gravitational pulls of the Moon and the Sun on this tilted bulge exert a net torque that attempts to pull Earth's equatorial plane into alignment with the ecliptic. Because Earth is spinning like a gyroscope, this torque causes its rotational axis to precess in a cone with an opening half-angle of 23.44° over ~25,772 years.

6An astronomer in Fortaleza, Brazil (latitude φ = 3°43' S) observes a star at upper culmination (meridian transit) with a celestial declination of δ = -15°00'. What is the star's altitude above the horizon at culmination?

A.56°17' above the northern horizon

B.78°43' above the southern horizon

C.86°17' above the southern horizon

D.71°17' above the northern horizon

Explanation: The meridian altitude of a celestial body is calculated as h = 90° - |φ - δ|. Here, latitude φ = -3°43' and declination δ = -15°00'. The angular separation from the zenith is |φ - δ| = |-3°43' - (-15°00')| = |+11°17'| = 11°17'. Therefore, altitude h = 90° - 11°17' = 78°43'. Because the star's declination (-15°) is south of the observer's zenith (-3°43'), the culmination occurs toward the South.

7Astronomically, the Sun passes through 13 constellations along the ecliptic during the course of a year. Which constellation is traversed by the Sun between Scorpius and Sagittarius?

A.Centaurus (Centauro)

B.Orion (Órion)

C.Ophiuchus (Serpentário / Ofiúco)

D.Pegasus (Pégaso)

Explanation: The ecliptic passes through Ophiuchus between the IAU boundaries of Scorpius and Sagittarius. This makes Ophiuchus the thirteenth astronomical ecliptic constellation, even though it is not included among the twelve traditional astrological zodiac signs.

8Observers in the Southern Hemisphere use the constellation Crux (Cruzeiro do Sul) to locate the South Celestial Pole. What is the standard geometric method for finding the pole using Crux?

A.Extend the horizontal arm from Delta Crucis through Beta Crucis by 2.5 times toward the West

B.Find the midpoint along the line connecting Alpha Centauri and Beta Centauri (the Pointers)

C.Bisect the angle formed by the Intrometida (Epsilon Crucis) and Gacrux

D.Extend the long vertical axis from Gacrux (Gamma Crucis) through Acrux (Alpha Crucis) by 4.5 times in that direction

Explanation: To locate the South Celestial Pole (around which all southern sky stars appear to rotate), observers extend an imaginary line along the major axis of the Southern Cross, starting from Gacrux (Gamma Crucis) through Acrux (Alpha Crucis), by approximately 4.5 times the distance between these two stars. Dropping a perpendicular line from that point to the horizon identifies true geographic South.

9At what approximate local solar time does the First Quarter (Quarto Crescente) Moon cross the local meridian (highest point in the sky)?

A.6:00 PM (18:00, sunset)

B.12:00 PM (12:00, solar noon)

C.12:00 AM (00:00, midnight)

D.6:00 AM (06:00, sunrise)

Explanation: In the First Quarter phase, the Moon is at an elongation of 90° East of the Sun. As Earth rotates, the First Quarter Moon rises around solar noon (12:00 PM), culminates on the local meridian around sunset (6:00 PM), and sets around midnight (12:00 AM).

10The Moon's sidereal orbital period is T_sid = 27.32 days, while its synodic period (phase cycle) is T_syn = 29.53 days. What relationship connects these two periods with Earth's orbital period T_earth = 365.25 days?

A.1 / T_syn = 1 / T_sid + 1 / T_earth

B.1 / T_syn = 1 / T_sid - 1 / T_earth

C.T_syn = T_sid · (1 + e_moon)

D.T_syn = sqrt(T_sid · T_earth)

Explanation: The synodic angular frequency is the difference between the Moon's sidereal orbital angular speed and Earth's heliocentric orbital angular speed: omega_syn = omega_sid - omega_earth. Expressed in terms of periods (omega = 2pi/T), this gives: 1/T_syn = 1/T_sid - 1/T_earth. Substituting values: 1/27.32166 - 1/365.2564 = 0.036601 - 0.002738 = 0.033863 = 1/29.5306 days.

About the OBA Exam

OBA is an annual in-person school competition for students from the first year of Ensino Fundamental through Ensino Médio, with four level-specific papers in Portuguese. The 2026 paper has seven Astronomy and three Astronautics questions and may award partial credit in multiple-Certo/Errado items. This English-language four-option MCQ study adaptation is not an official translation, does not reproduce partial-credit scoring, and does not replace the separate international-team selection process.

Assessment

Seven Astronomy and three Astronautics questions, with separate papers for Levels 1–4.

Time Limit

2 hours for Levels 1–3; up to 3 hours for Level 4

Passing Score

No fixed pass mark; ranking, medals, and selection invitations apply

Exam Fee

Free (Sociedade Astronômica Brasileira (SAB) and Agência Espacial Brasileira (AEB), through CO/OBA)

OBA Exam Content Outline

70% of official question count

Astronomy

Level-appropriate Earth-sky relations, the Solar System, stars, galaxies, observation, and astrophysics.

30% of official question count

Astronautics

Rockets, satellites, space missions, Brazilian institutions, exploration, and applications.

Level-dependent

Quantitative and observational reasoning

Angles, scales, orbits, time, light, and interpretation increase in depth by level.

How to Pass the OBA Exam

What You Need to Know

Passing score: No fixed pass mark; ranking, medals, and selection invitations apply
Assessment: Seven Astronomy and three Astronautics questions, with separate papers for Levels 1–4.
Time limit: 2 hours for Levels 1–3; up to 3 hours for Level 4
Exam fee: Free

Keys to Passing

Work through all 100 available questions
Review every answer and explanation
Track weak areas and revisit them
Use our AI tutor for tough concepts

OBA Study Tips from Top Performers

1Use the official content list for your OBA level and solve the corresponding Portuguese past papers.

2Practice explaining physical causes, not just recalling names or numerical facts.

Frequently Asked Questions

Is OBA an online exam in 2026?

No. The 2026 OBA paper is administered in person at the school; later international-selection tests are separate.

Does this bank simulate OBA scoring?

No. It uses four English options and does not reproduce the official Portuguese level papers or partial-credit Certo/Errado items.