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2026 Statistics
Key Facts: PAU Physics (Andalusia) Exam
90 min
Time Limit
Distrito Único Andaluz
0–10
Grading Scale
Junta de Andalucía
4.0
Min. Access Phase Mark
PAU Regulations
EUR 58.70
Base Exam Fee
Junta de Andalucía 2026
100
Practice Questions
English Study Adaptation
The Andalusia PAU Physics (Física) exam is a 90-minute examination set by the Distrito Único Andaluz / Comisión Interuniversitaria de Andalucía with a registration fee of EUR 58.70 (Access Phase base fee). Grades are awarded on a 0–10 scale, with a minimum 4.0 required to average with high school GPA. This study portal provides an English-language MCQ study adaptation featuring 100 practice questions with real calculations covering all 5 official curriculum modules.
Sample PAU Physics (Andalusia) Practice Questions
Try these sample questions to test your PAU Physics (Andalusia) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Two point masses of m1 = 500 kg and m2 = 800 kg are separated by a distance of r = 2.0 m in a vacuum. Using Newton's Law of Universal Gravitation with G = 6.67 x 10^-11 N m^2/kg^2, what is the magnitude of the gravitational force between them?
A.6.67 x 10^-6 N
B.1.33 x 10^-5 N
C.3.34 x 10^-6 N
D.2.67 x 10^-5 N
Explanation: Applying Newton's law of universal gravitation, F = G * m1 * m2 / r^2. Substituting the given values gives F = (6.67 x 10^-11) * (500) * (800) / (2.0)^2 = (6.67 x 10^-11) * 400,000 / 4 = 6.67 x 10^-6 N.
2If the acceleration due to gravity at Earth's surface is g0 = 9.80 m/s^2, what is the magnitude of the gravitational field strength g at an altitude h equal to Earth's radius R_E above the surface?
A.4.90 m/s^2
B.2.45 m/s^2
C.1.23 m/s^2
D.9.80 m/s^2
Explanation: Gravitational field intensity follows an inverse-square law with distance from Earth's center: g(r) = G * M_E / r^2. At altitude h = R_E, the radial distance from Earth's center is r = R_E + h = 2 R_E. Thus g = G * M_E / (2 R_E)^2 = g0 / 4 = 9.80 / 4 = 2.45 m/s^2.
3What is the escape velocity from Earth's surface, assuming Earth's mass M_E = 5.97 x 10^24 kg, radius R_E = 6.37 x 10^6 m, and G = 6.67 x 10^-11 N m^2/kg^2?
A.7.91 km/s
B.11.2 km/s
C.15.8 km/s
D.22.4 km/s
Explanation: Escape velocity is obtained by setting total mechanical energy to zero: 0.5 m v_esc^2 - G M_E m / R_E = 0, giving v_esc = sqrt(2 G M_E / R_E). Substituting values: v_esc = sqrt(2 * 6.67 x 10^-11 * 5.97 x 10^24 / 6.37 x 10^6) = sqrt(1.249 x 10^8) = 11,178 m/s = 11.2 km/s.
4A satellite orbits Earth in a circular orbit at an altitude h = 3,600 km above the surface. Given Earth's radius R_E = 6,400 km and mass M_E = 6.0 x 10^24 kg, what is the satellite's orbital speed?
A.6.33 km/s
B.7.91 km/s
C.5.21 km/s
D.9.68 km/s
Explanation: The orbital radius from Earth's center is r = R_E + h = 6,400 km + 3,600 km = 10,000 km = 1.0 x 10^7 m. Equating gravitational force to centripetal force yields v = sqrt(G M_E / r) = sqrt((6.67 x 10^-11 * 6.0 x 10^24) / 1.0 x 10^7) = sqrt(4.002 x 10^7) = 6,326 m/s = 6.33 km/s.
5What is the orbital period T of a satellite in a circular orbit of radius r = 1.0 x 10^7 m around Earth (M_E = 6.0 x 10^24 kg)?
A.1.41 hours
B.2.76 hours
C.5.52 hours
D.24.0 hours
Explanation: The period is T = 2 pi r / v. With v = sqrt(G M_E / r) = 6326 m/s, T = 2 * pi * 1.0 x 10^7 / 6326 = 9,933 seconds. Converting to hours: T = 9,933 / 3600 = 2.76 hours.
6If the gravitational potential at Earth's surface is V0 = -6.25 x 10^7 J/kg, what is the gravitational potential V at a distance r = 2 R_E from Earth's center?
A.-3.125 x 10^7 J/kg
B.-1.56 x 10^7 J/kg
C.-12.5 x 10^7 J/kg
D.0 J/kg
Explanation: Gravitational potential due to a point mass is V(r) = -G M_E / r. At r = 2 R_E, V = -G M_E / (2 R_E) = V0 / 2 = (-6.25 x 10^7 J/kg) / 2 = -3.125 x 10^7 J/kg.
7How much work is done by the gravitational field when a mass m = 100 kg is moved from r1 = R_E to r2 = 2 R_E above Earth? (Use g0 = 9.80 m/s^2, R_E = 6.37 x 10^6 m).
A.+3.12 x 10^9 J
B.-3.12 x 10^9 J
C.-6.24 x 10^9 J
D.+6.24 x 10^9 J
Explanation: Work done by conservative gravitational force is W_field = -delta U = -(U2 - U1) = G M_E m (1/r2 - 1/r1). Since G M_E = g0 R_E^2, W_field = g0 R_E^2 m (1/(2 R_E) - 1/R_E) = -0.5 m g0 R_E = -0.5 * 100 * 9.80 * 6.37 x 10^6 = -3.12 x 10^9 J. The negative sign indicates gravity opposes outward displacement.
8Calculate the altitude h above Earth's surface of a geostationary satellite with orbital period T = 24 hours (86,400 s). (Use M_E = 5.97 x 10^24 kg, R_E = 6,370 km, G = 6.67 x 10^-11 N m^2/kg^2).
A.35,830 km
B.42,200 km
C.20,200 km
D.6,370 km
Explanation: From Kepler's Third Law, r^3 = G M_E T^2 / (4 pi^2) = (6.67 x 10^-11 * 5.97 x 10^24 * 86400^2) / (4 pi^2) = 7.537 x 10^22 m^3, yielding r = 42,240 km. The altitude above Earth's surface is h = r - R_E = 42,240 km - 6,370 km = 35,870 km (approx. 35,830 km with exact values).
9According to Kepler's Third Law (T^2 / r^3 = constant), if a planet's mean orbital radius around a star is increased by a factor of 4, by what factor does its orbital period increase?
A.4
B.8
C.16
D.64
Explanation: Kepler's Third Law states T^2 proportional to r^3, so T proportional to r^(3/2). If r2 = 4 r1, then T2 / T1 = (4)^(3/2) = (sqrt(4))^3 = 2^3 = 8.
10What is the total mechanical energy E_tot of a satellite of mass m in a circular orbit of radius r around a planet of mass M?
A.-G M m / (2 r)
B.-G M m / r
C.+G M m / (2 r)
D.0
Explanation: The kinetic energy is K = 0.5 m v^2 = G M m / (2 r) and potential energy is U = -G M m / r. Total mechanical energy is E_tot = K + U = G M m / (2 r) - G M m / r = -G M m / (2 r). The negative energy indicates a bound orbital state.
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