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Canary Islands PAU Physics Examination — ULPGC & ULL (Física 2º Bachillerato 2026) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Canary PAU Physics Exam
90 Minutes
Exam duration
COPAU / ULPGC / ULL
EUR 76.12
Base registration fee for Access Phase
Gobierno de Canarias
0–10 Scale
Grading scale for Bachillerato and PAU exams
Spanish Ministry of Education
5 Core Blocks
Gravitation, Electromagnetism, Waves, Optics, Modern Physics
LOMLOE Physics Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Canary Islands PAU Physics (ULPGC & ULL 2026) is a 90-minute university entrance exam assessing 2º Bachillerato Physics across 5 core subject blocks.
Sample Canary PAU Physics Practice Questions
Try these sample questions to test your Canary PAU Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1An artificial satellite moves in a circular path around Earth at an orbital distance r1. If the radius of its orbit is quadrupled (r2 = 4 r1), what is the new orbital period T2 in terms of the initial period T1?
A.T2 = 2 T1
B.T2 = 4 T1
C.T2 = 8 T1
D.T2 = 16 T1
Explanation: By Kepler's third law, T^2 / r^3 is constant for any body orbiting the same central mass. Therefore, (T2 / T1)^2 = (r2 / r1)^3 = (4)^3 = 64. Taking the square root gives T2 / T1 = sqrt(64) = 8, so T2 = 8 T1.
2An object of mass m is at an altitude h equal to Earth's radius (h = R_E) above Earth's surface. If the gravitational field strength at the surface is g0 = 9.80 m/s², what is the gravitational field strength g at that altitude?
A.4.90 m/s²
B.1.225 m/s²
C.9.80 m/s²
D.2.45 m/s²
Explanation: Gravitational field strength varies inversely with the square of the distance to Earth's center: g = G M_E / r². At the surface r1 = R_E, while at an altitude h = R_E the distance from the center is r2 = R_E + h = 2 R_E. Thus, g = G M_E / (2 R_E)² = g0 / 4 = 9.80 / 4 = 2.45 m/s².
3A mass m = 100 kg is moved between two points A and B located on the same equipotential surface of Earth's gravitational field. What work does the gravitational force do during this displacement?
A.9800 J
B.0 J
C.Depends on the path taken between A and B
D.980 J
Explanation: The work done by a conservative force such as gravity is W = -m (V_B - V_A). Since both points lie on the same equipotential surface, V_A = V_B, so Delta V = 0 and the work is exactly 0 J.
4A satellite follows a circular orbit around Earth of radius r = 4 R_E, both radii measured from Earth's centre. If v0 is the orbital speed for a circular orbit of radius r0 = R_E, what is the satellite's orbital speed v?
A.v = 2 v0
B.v = v0
C.v = v0 / 4
D.v = v0 / 2
Explanation: The speed in a circular orbit of radius r is found by equating the gravitational force to the centripetal force: G M m / r² = m v² / r => v = sqrt(G M / r). If the radius goes from R_E to 4 R_E, the speed decreases by a factor of sqrt(4) = 2, so v = v0 / 2.
5Which of the following statements correctly characterizes the conservative nature of the gravitational field?
A.The work done along any closed path is zero.
B.The gravitational force always does positive work on any moving mass.
C.The gravitational potential increases when approaching the central mass.
D.A satellite's kinetic energy stays constant in any elliptical orbit.
Explanation: By definition, a force field is conservative if the circulation of the force around any closed loop is zero (oint F · dr = 0), or equivalently, if the work done between two points depends only on the initial and final positions.
6Taking the convention that gravitational potential energy is zero at infinity (U(infinity) = 0), what is the sign and expression for the potential energy of a mass m at a distance r from Earth?
A.Positive, U(r) = +G M_E m / r²
B.Positive, U(r) = +G M_E m / r
C.Negative, U(r) = -G M_E m / r²
D.Negative, U(r) = -G M_E m / r
Explanation: Because the gravitational force is attractive, positive work must be done against the field to move a mass from r out to infinity. With U(infinity) = 0, the potential at any finite point r is negative: U(r) = -G M_E m / r.
7Calculate the escape velocity from Earth's surface (M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m, G = 6.674 × 10⁻¹¹ N·m²/kg²).
A.7.91 km/s
B.11.18 km/s
C.15.81 km/s
D.22.36 km/s
Explanation: Escape velocity is found from conservation of mechanical energy, setting the total energy to zero at infinity: 1/2 m ve² - G M_E m / R_E = 0 => ve = sqrt(2 G M_E / R_E) = sqrt(2 × 6.674×10⁻¹¹ × 5.97×10²⁴ / 6.37×10⁶) = sqrt(1.250×10⁸) = 11180 m/s = 11.18 km/s.
8A satellite of mass m = 500 kg is moved from a circular orbit of radius r1 = 2 R_E to another circular orbit of radius r2 = 3 R_E. What is the minimum energy that must be supplied to the satellite? (Data: g0 = 9.80 m/s², R_E = 6.37 × 10⁶ m).
A.1.56 × 10¹⁰ J
B.5.20 × 10⁹ J
C.1.30 × 10⁹ J
D.2.60 × 10⁹ J
Explanation: The total energy of a satellite in a circular orbit of radius r is E = -G M m / (2 r) = -m g0 R_E² / (2 r). At r1 = 2 R_E, E1 = -m g0 R_E / 4. At r2 = 3 R_E, E2 = -m g0 R_E / 6. The energy that must be supplied is Delta E = E2 - E1 = m g0 R_E (1/4 - 1/6) = m g0 R_E / 12 = 500 × 9.80 × 6.37×10⁶ / 12 = 2.60 × 10⁹ J.
9What is the relationship between the escape velocity ve and the orbital speed vorb for a circular orbit of the same radius r around a celestial body?
A.ve = sqrt(2) * vorb
B.ve = vorb / sqrt(2)
C.ve = 2 * vorb
D.ve = vorb
Explanation: The formulas are vorb = sqrt(G M / r) and ve = sqrt(2 G M / r). Dividing the two expressions gives ve / vorb = sqrt(2), so ve = sqrt(2) * vorb ≈ 1.414 vorb.
10A geostationary satellite orbits in the equatorial plane with a period T = 24 hours (86400 s). Calculate the orbital radius r measured from Earth's center (M_E = 5.97 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N·m²/kg²).
A.4.22 × 10⁷ m
B.5.45 × 10⁷ m
C.3.58 × 10⁷ m
D.6.37 × 10⁶ m
Explanation: Applying Kepler's third law T² = (4 pi² / (G M_E)) r³, we solve for r = (G M_E T² / (4 pi²))^(1/3). Substituting with G M_E = 3.984×10¹⁴ and T² = 7.465×10⁹: r = (3.984×10¹⁴ × 7.465×10⁹ / 39.478)^(1/3) = (7.534×10²²)^(1/3) = 4.22 × 10⁷ m (about 42,200 km from Earth's center, or roughly 35,800 km above the surface).
About the Canary PAU Physics Practice Questions
Verified exam format metadata for Canary Islands PAU Physics Examination — ULPGC & ULL (Física 2º Bachillerato 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.