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2026 Statistics
Key Facts: Madrid PAU Physics Exam
PAU Organising Commission / UCM
Exam Body
Community of Madrid University Admissions
90 min
Duration
Madrid PAU Regulations
0–10
Scoring
PAU Spain Marking Scheme
2º Bach.
Target Level
Madrid Education Department
€93.02
Registration Fee
UCM PAU Fee Schedule
Sample Madrid PAU Physics Practice Questions
Try these sample questions to test your Madrid PAU Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1According to Newton's law of universal gravitation, how does the gravitational force between two point masses change if the distance between their centers is tripled?
A.It decreases to 1/9 of its original value.
B.It decreases to 1/3 of its original value.
C.It increases by a factor of 9.
D.It decreases to 1/6 of its original value.
Explanation: Newton's law of universal gravitation states F = G*(m1*m2)/r^2. When separation distance r is multiplied by 3, the force becomes F' = G*(m1*m2)/(3r)^2 = (1/9)*F. Thus, force decreases to 1/9 of its original value.
2At an altitude above Earth's surface equal to Earth's radius (h = R_E), what is the magnitude of the gravitational acceleration g? (Assume surface acceleration g_0 = 9.80 m/s²).
A.2.45 m/s²
B.4.90 m/s²
C.9.80 m/s²
D.1.23 m/s²
Explanation: Gravitational acceleration at radial distance r from Earth's center is g = G*M_E / r^2. At Earth's surface r = R_E, g_0 = 9.80 m/s². At altitude h = R_E, total radial distance is r = R_E + R_E = 2 R_E. Thus g = G*M_E / (2 R_E)^2 = g_0 / 4 = 9.80 / 4 = 2.45 m/s².
3A satellite orbits Earth in a circular path at an altitude equal to Earth's radius (h = R_E = 6.37 × 10⁶ m). Given Earth's mass M_E = 5.97 × 10²⁴ kg and G = 6.674 × 10⁻¹¹ N m²/kg², what is the orbital speed of the satellite?
A.5.59 × 10³ m/s
B.7.91 × 10³ m/s
C.3.95 × 10³ m/s
D.1.12 × 10⁴ m/s
Explanation: For a circular orbit, gravitational attraction provides centripetal force: G*M_E*m / r^2 = m*v^2 / r, leading to v = sqrt(G*M_E / r). The orbital radius is r = R_E + h = 2 R_E = 1.274 × 10⁷ m. Evaluating v = sqrt((6.674 × 10⁻¹¹ × 5.97 × 10²⁴) / 1.274 × 10⁷) = sqrt(3.127 × 10⁷) = 5.59 × 10³ m/s.
4Two planets orbit a central star in circular orbits. Planet A has orbital radius R, and Planet B has orbital radius 4R. If Planet A's orbital period is T_A = 2.0 years, what is Planet B's orbital period T_B?
A.16.0 years
B.8.0 years
C.32.0 years
D.4.0 years
Explanation: By Kepler's Third Law, T^2 / r^3 is constant for orbits around the same mass. Thus (T_B / T_A)^2 = (r_B / r_A)^3 = (4R / R)^3 = 64. Taking the square root gives T_B / T_A = 8, so T_B = 8 × 2.0 = 16.0 years.
5What is the escape velocity from the surface of a planet that has twice the mass of Earth (M_P = 2 M_E) and half the radius of Earth (R_P = 0.5 R_E)? (Earth's escape velocity v_e,Earth = 11.2 km/s).
A.22.4 km/s
B.11.2 km/s
C.44.8 km/s
D.15.8 km/s
Explanation: Escape velocity is v_e = sqrt(2 G M / R). For the new planet: v_e,P = sqrt(2 G (2 M_E) / (0.5 R_E)) = sqrt(4 * (2 G M_E / R_E)) = 2 * v_e,Earth = 2 * 11.2 km/s = 22.4 km/s.
6How is the kinetic energy E_k of a satellite in a circular orbit related to its gravitational potential energy E_p (with standard infinity reference)?
A.E_k = -1/2 * E_p
B.E_k = E_p
C.E_k = -E_p
D.E_k = -2 * E_p
Explanation: For circular orbit, centripetal force yields E_k = 1/2 m v^2 = G M m / (2 r). Gravitational potential energy is E_p = -G M m / r. Comparing expressions shows E_k = -1/2 * E_p.
7A satellite of mass m = 500 kg is transferred from a circular orbit of radius r_1 = 2 R_E to a higher circular orbit of radius r_2 = 4 R_E. Given G*M_E = 3.986 × 10¹⁴ m³/s² and R_E = 6.37 × 10⁶ m, what is the work required by the engines?
A.3.91 × 10⁹ J
B.7.82 × 10⁹ J
C.1.56 × 10¹⁰ J
D.1.95 × 10⁹ J
Explanation: Total mechanical energy in circular orbit is E = -G M_E m / (2 r). Work required is W = Delta E = E_2 - E_1 = (G M_E m / 2) * (1/r_1 - 1/r_2). Substituting values: W = (3.986 × 10¹⁴ × 500 / 2) * (1/(1.274 × 10⁷) - 1/(2.548 × 10⁷)) = 9.965 × 10¹⁶ * 3.925 × 10⁻⁸ = 3.91 × 10⁹ J.
8Which of the following statements correctly describes a geostationary satellite orbit around Earth?
A.Its orbital plane must coincide with Earth's equatorial plane, moving west to east with a period of 24 hours.
B.Its orbit can pass directly over Earth's geographic North and South poles with a period of 12 hours.
C.Its orbital speed is equal to Earth's surface escape velocity.
D.It experiences zero net gravitational force from Earth.
Explanation: A geostationary satellite remains stationary relative to a point on Earth's surface. This requires a circular equatorial orbit, rotating west to east with an orbital period matching Earth's rotation period (~24 hours).
9Calculate the altitude h above Earth's surface for a geostationary satellite. (Earth mass M_E = 5.97 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N m²/kg², Earth radius R_E = 6.37 × 10⁶ m, period T = 86,400 s).
A.3.58 × 10⁷ m (35,800 km)
B.4.22 × 10⁷ m (42,200 km)
C.6.37 × 10⁶ m (6,370 km)
D.1.28 × 10⁷ m (12,800 km)
Explanation: From Kepler's 3rd Law T^2 = 4 pi^2 r^3 / (G M_E), orbital radius is r = (G M_E T^2 / (4 pi^2))^(1/3) = 4.22 × 10⁷ m. Altitude above Earth's surface is h = r - R_E = 4.22 × 10⁷ - 6.37 × 10⁶ = 3.58 × 10⁷ m = 35,800 km.
10What is the gravitational potential V at a point on Earth's surface? (Earth mass M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m, G = 6.674 × 10⁻¹¹ N m²/kg²).
A.-6.25 × 10⁷ J/kg
B.-9.80 J/kg
C.+6.25 × 10⁷ J/kg
D.-3.98 × 10¹⁴ J/kg
Explanation: Gravitational potential is V = -G M_E / R_E. Substituting values: V = -(6.674 × 10⁻¹¹ × 5.97 × 10²⁴) / 6.37 × 10⁶ = -6.25 × 10⁷ J/kg.
About the Madrid PAU Physics Practice Questions
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