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100+ Free Galicia PAU Physics / Física 2º Bachillerato CIUG Practice Questions
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Key Facts: Galicia PAU Physics / Física 2º Bachillerato CIUG Exam
90 min
Exam duration (1.5 hours)
CIUG PAU Guidelines
0–10
Grading scale
CIUG Evaluation Criteria
4.0
Minimum Access Phase mark required
Spanish University Access Regulations
63.67 €
Ordinary registration fee
CIUG Official Fees
100
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Galicia PAU Physics (CIUG 2026) is a 90-minute examination assessing 2º Bachillerato physics competencies across 5 key domain areas under LOMLOE guidelines.
Sample Galicia PAU Physics / Física 2º Bachillerato CIUG Practice Questions
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1Two point masses m1 and m2 are separated by a distance R, exerting a gravitational force F on each other. If m1 is doubled and the distance between them is doubled to 2R, what is the new gravitational force F' in terms of F?
A.0.25 F
B.0.5 F
C.1.0 F
D.2.0 F
Explanation: According to Newton's law of universal gravitation, F = G*m1*m2/R^2. Doubling m1 to 2*m1 and R to 2*R yields F' = G*(2*m1)*m2 / (2*R)^2 = (2/4) * G*m1*m2/R^2 = 0.5 F.
2The gravitational acceleration at Earth's surface is g0 = 9.80 m/s^2. What is the magnitude of the gravitational acceleration g at an altitude h equal to Earth's radius (h = R_E)?
A.4.90 m/s^2
B.2.45 m/s^2
C.3.27 m/s^2
D.1.23 m/s^2
Explanation: The gravitational field strength varies inversely with the square of the distance from Earth's center: g = G*M_E / r^2. At altitude h = R_E, the distance 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.
3A satellite of mass m = 500 kg orbits Earth in a circular orbit at an altitude h = R_E (where R_E = 6.37 x 10^6 m and Earth's mass M_E = 5.97 x 10^24 kg, G = 6.67 x 10^-11 N m^2/kg^2). What is the orbital speed of the satellite?
A.5.59 x 10^3 m/s
B.7.91 x 10^3 m/s
C.3.95 x 10^3 m/s
D.1.12 x 10^4 m/s
Explanation: Equating gravitational force to centripetal force yields G*M_E*m / r^2 = m*v^2 / r, so v = sqrt(G*M_E / r). At altitude h = R_E, r = 2*R_E = 1.274 x 10^7 m. Substituting values gives v = sqrt(6.67 x 10^-11 * 5.97 x 10^24 / 1.274 x 10^7) = sqrt(3.125 x 10^7) = 5.59 x 10^3 m/s.
4A planet orbits a star with an orbital period T1 = 8 years at a mean orbital radius r1. A second planet orbits the same star at a mean orbital radius r2 = 4 * r1. According to Kepler's Third Law, what is the orbital period T2 of the second planet?
A.16 years
B.32 years
C.64 years
D.128 years
Explanation: Kepler's Third Law states T^2 / r^3 = constant, so (T2/T1)^2 = (r2/r1)^3 = 4^3 = 64. Taking the square root gives T2/T1 = sqrt(64) = 8. Therefore T2 = 8 * T1 = 8 * 8 = 64 years.
5A space probe of mass m = 200 kg is moved from Earth's surface (r1 = R_E = 6.37 x 10^6 m) to an altitude h = 2*R_E (r2 = 3*R_E). Taking M_E = 5.97 x 10^24 kg and G = 6.67 x 10^-11 N m^2/kg^2, what is the change in gravitational potential energy delta U of the probe?
A.8.33 x 10^9 J
B.1.25 x 10^10 J
C.4.17 x 10^9 J
D.1.87 x 10^10 J
Explanation: The change in potential energy is delta U = U2 - U1 = -G*M_E*m/r2 - (-G*M_E*m/r1) = G*M_E*m * (1/R_E - 1/(3*R_E)) = G*M_E*m * (2 / (3*R_E)). Substituting values: delta U = (6.67 x 10^-11 * 5.97 x 10^24 * 200 * 2) / (3 * 6.37 x 10^6) = 8.33 x 10^9 J.
6The mass of Mars is 6.42 x 10^23 kg and its radius is 3.39 x 10^6 m. Using G = 6.67 x 10^-11 N m^2/kg^2, what is the escape velocity v_esc from the surface of Mars?
A.3.55 x 10^3 m/s
B.5.03 x 10^3 m/s
C.7.11 x 10^3 m/s
D.1.12 x 10^4 m/s
Explanation: Escape velocity is derived from energy conservation (1/2*m*v^2 - G*M*m/R = 0), yielding v_esc = sqrt(2*G*M / R). Substituting Mars parameters: v_esc = sqrt(2 * 6.67 x 10^-11 * 6.42 x 10^23 / 3.39 x 10^6) = sqrt(2.527 x 10^7) = 5.03 x 10^3 m/s.
7A 100 kg satellite moves along an equipotential surface of Earth's gravitational field from point A to point B separated by a arc distance of 5000 km. What is the work done W by the gravitational field during this displacement?
A.0 J
B.4.9 x 10^8 J
C.-4.9 x 10^8 J
D.9.8 x 10^8 J
Explanation: The work done by a conservative field is W = -m * delta V. Since points A and B lie on the same equipotential surface, V_A = V_B and delta V = 0, meaning the gravitational work done is 0 J.
8What is the ratio of the escape velocity v_esc from a given circular orbit to the circular orbital speed v_orb at that same orbital radius r?
A.0.50
B.0.71
C.1.41
D.2.00
Explanation: Orbital speed is v_orb = sqrt(G*M/r) and escape velocity is v_esc = sqrt(2*G*M/r). Taking the ratio v_esc / v_orb = sqrt(2*G*M/r) / sqrt(G*M/r) = sqrt(2) ≈ 1.41.
9A communications satellite of mass m = 1200 kg is in a stable circular orbit of radius r = 4.22 x 10^7 m around Earth (M_E = 5.97 x 10^24 kg, G = 6.67 x 10^-11 N m^2/kg^2). What is the total mechanical energy E of the satellite?
A.-5.67 x 10^9 J
B.-1.13 x 10^10 J
C.5.67 x 10^9 J
D.-2.26 x 10^10 J
Explanation: For a circular orbit, kinetic energy is E_k = G*M*m / (2r) and potential energy is U = -G*M*m / r. Total mechanical energy is E = E_k + U = -G*M*m / (2r). Substituting: E = -(6.67 x 10^-11 * 5.97 x 10^24 * 1200) / (2 * 4.22 x 10^7) = -5.67 x 10^9 J.
10The distance between Earth (M_E = 5.97 x 10^24 kg) and the Moon (M_M = 7.35 x 10^22 kg) is d = 3.84 x 10^8 m. At what distance x from the center of Earth along the line connecting their centers is the net gravitational field intensity zero?
A.3.46 x 10^8 m
B.3.84 x 10^7 m
C.1.92 x 10^8 m
D.3.80 x 10^8 m
Explanation: Setting g_Earth = g_Moon gives G*M_E / x^2 = G*M_M / (d - x)^2. Taking the square root: sqrt(M_E) / x = sqrt(M_M) / (d - x). Solving for x: x = d / (1 + sqrt(M_M / M_E)) = 3.84 x 10^8 / (1 + sqrt(7.35 x 10^22 / 5.97 x 10^24)) = 3.84 x 10^8 / (1 + 0.111) = 3.46 x 10^8 m.
About the Galicia PAU Physics / Física 2º Bachillerato CIUG Practice Questions
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