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100+ Free PAU Physics (Cantabria) Practice Questions
Cantabria PAU Physics 2026 (Pruebas de Acceso a la Universidad, Universidad de Cantabria) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: PAU Physics (Cantabria) Exam
90 min
Exam time duration for the PAU Physics paper
University of Cantabria (UNICAN) PAU Organising Commission
0–10 scale
Scoring system (minimum 4.0 required in Access Phase)
Cantabria PAU Regulations
5 Blocks
Syllabus blocks: Gravitational Field, Electromagnetism, Waves, Optics, Modern Physics
2º Bachillerato Curriculum (LOMLOE)
100
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Master the 2026 Cantabria PAU Physics exam with 100 practice questions covering gravitational field, electromagnetic field, waves, optics, and modern physics. This English-language MCQ bank is a study adaptation of the theory and calculation knowledge behind the official written exam.
Sample PAU Physics (Cantabria) Practice Questions
Try these sample questions to test your PAU Physics (Cantabria) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A satellite orbits Earth at an altitude equal to Earth's radius $R_E = 6.37 \times 10^6\text{ m}$. If the gravitational field strength at Earth's surface is $g_0 = 9.80\text{ m/s}^2$, what is the gravitational field strength $g$ at the satellite's orbital altitude?
A.2.45 m/s²
B.4.90 m/s²
C.1.23 m/s²
D.9.80 m/s²
Explanation: Gravitational field strength decreases with distance according to $g = g_0 (R_E / r)^2$. At an altitude $h = R_E$, the radial distance from Earth's center is $r = R_E + h = 2 R_E$. Substituting $r = 2 R_E$ yields $g = 9.80 / 2^2 = 9.80 / 4 = 2.45\text{ m/s}^2$.
2According to Kepler's Third Law ($T^2 / a^3 = k$), if Planet A orbits a star at distance $r$ with orbital period $T$, what is the orbital period of Planet B orbiting the same star at a distance of $4r$?
A.16T
B.8T
C.4T
D.64T
Explanation: Kepler's Third Law states that $T^2 \propto r^3$, so $T_B = T_A (r_B / r_A)^{3/2}$. For $r_B = 4r$, $(4)^{3/2} = \sqrt{4^3} = \sqrt{64} = 8$. Thus, $T_B = 8T$.
3What is the escape velocity from the surface of a spherical planet with mass $M = 6.0 \times 10^{24}\text{ kg}$ and radius $R = 6.4 \times 10^6\text{ m}$? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
A.7.91 × 10³ m/s
B.1.58 × 10⁴ m/s
C.1.12 × 10⁴ m/s
D.2.24 × 10⁴ m/s
Explanation: Escape velocity is calculated using $v_{esc} = \sqrt{2GM/R}$. Substituting the values: $v_{esc} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.4 \times 10^6}} = \sqrt{1.2506 \times 10^8} \approx 1.12 \times 10^4\text{ m/s}$ ($11.2\text{ km/s}$).
4The gravitational potential at a point outside a planet of mass $M = 4.0 \times 10^{24}\text{ kg}$ is $V = -4.0 \times 10^7\text{ J/kg}$. What is the distance $r$ from the center of the planet to this point? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
A.3.34 × 10⁶ m
B.1.33 × 10⁷ m
C.4.00 × 10⁶ m
D.6.67 × 10⁶ m
Explanation: Gravitational potential is defined by $V = -GM/r$. Rearranging gives $r = -GM/V = \frac{6.67 \times 10^{-11} \times 4.0 \times 10^{24}}{4.0 \times 10^7} = 6.67 \times 10^6\text{ m}$.
5A space probe of mass $m = 500\text{ kg}$ moves in a circular orbit of radius $r = 1.0 \times 10^7\text{ m}$ around a body of mass $M = 5.0 \times 10^{24}\text{ kg}$. What is its orbital speed? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
A.5.77 × 10³ m/s
B.8.16 × 10³ m/s
C.3.34 × 10³ m/s
D.4.08 × 10³ m/s
Explanation: Equating gravitational force to centripetal force: $G M m / r^2 = m v^2 / r \implies v = \sqrt{GM/r}$. Substituting: $v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.0 \times 10^{24}}{1.0 \times 10^7}} = \sqrt{3.335 \times 10^7} \approx 5.77 \times 10^3\text{ m/s}$.
6If the mass of a planet is doubled while its radius remains unchanged, how does the gravitational field strength $g$ at its surface change?
A.It quadruples (increases by a factor of 4)
B.It doubles (increases by a factor of 2)
C.It increases by a factor of √2
D.It remains unchanged
Explanation: Surface gravitational field strength is directly proportional to planetary mass ($g = GM/R^2$). Doubling $M$ while keeping $R$ constant doubles $g$.
7A $200\text{ kg}$ satellite is transferred from a circular orbit of radius $r_1 = 7.0 \times 10^6\text{ m}$ to $r_2 = 1.4 \times 10^7\text{ m}$ around Earth ($M_E = 5.97 \times 10^{24}\text{ kg}$, $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$). How much energy must be supplied to perform this transfer?
A.5.69 × 10⁹ J
B.1.42 × 10⁹ J
C.2.84 × 10⁹ J
D.8.53 × 10⁹ J
Explanation: Total mechanical energy in a circular orbit is $E = -GMm/(2r)$. The energy required is $\Delta E = E_2 - E_1 = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$. Calculating: \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 200}{2} \left(\frac{1}{7.0 \times 10^6} - \frac{1}{1.4 \times 10^7}\right) = 3.982 \times 10^{16} \times 7.143 \times 10^{-8} = 2.84 \times 10^9\text{ J}$.
8A geostationary satellite remains fixed over a point on Earth's equator. Given Earth's mass $M_E = 5.97 \times 10^{24}\text{ kg}$ and rotational period $T = 86400\text{ s}$, what is the radius $r$ of its geostationary orbit? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
A.3.58 × 10⁷ m
B.6.37 × 10⁶ m
C.2.11 × 10⁷ m
D.4.22 × 10⁷ m
Explanation: Using Kepler's Third Law in circular orbit form: $r^3 = \frac{GM_E T^2}{4\pi^2}$. Substituting $T = 86400\text{ s}$: $r^3 = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2} \approx 7.53 \times 10^{22}\text{ m}^3$. Taking the cube root gives $r = 4.22 \times 10^7\text{ m}$ ($42,200\text{ km}$).
9A comet in an elliptical orbit around the Sun has a perihelion distance of $r_p = 1.0 \times 10^{11}\text{ m}$ where its speed is $v_p = 60\text{ km/s}$. What is its speed $v_a$ at aphelion where its distance is $r_a = 5.0 \times 10^{11}\text{ m}$?
A.12 km/s
B.30 km/s
C.2.4 km/s
D.150 km/s
Explanation: Conservation of angular momentum requires $L = m r_p v_p = m r_a v_a \implies v_a = v_p (r_p / r_a)$. Substituting the given values: $v_a = 60 \times \frac{1.0 \times 10^{11}}{5.0 \times 10^{11}} = 12\text{ km/s}$.
10How much work is done by the gravitational field of Earth ($M_E = 5.97 \times 10^{24}\text{ kg}$) to move a mass $m = 100\text{ kg}$ from $r_A = 2.0 \times 10^7\text{ m}$ to $r_B = 1.0 \times 10^7\text{ m}$? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
A.-1.99 × 10⁹ J
B.+1.99 × 10⁹ J
C.+3.98 × 10⁹ J
D.-3.98 × 10⁹ J
Explanation: Work done by conservative gravitational force is $W = -\Delta U = U_A - U_B = G M m \left(\frac{1}{r_B} - \frac{1}{r_A}\right)$. Calculating: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 100 \left(\frac{1}{10^7} - \frac{1}{2 \times 10^7}\right) = 3.982 \times 10^{15} \times 5.0 \times 10^{-7} = +1.99 \times 10^9\text{ J}$.
About the PAU Physics (Cantabria) Practice Questions
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