Circular Motion and Gravity — Free Physics Review Games.

Centripetal acceleration always points toward the center of the circular path.

Gravity provides the centripetal force that keeps the Moon in its orbit around Earth.

Centripetal force is the net inward force that causes an object to follow a curved path.

Without centripetal force, the object continues in a straight line tangent to the circle due to inertia (Newton's First Law).

Isaac Newton formulated the law of universal gravitation: F = Gm1m2/r^2.

Centripetal acceleration equals the square of the speed divided by the radius: a = v^2/r.

Gravitational force follows an inverse square law: if distance doubles, force becomes 1/4 as strong (F = Gm1m2/r^2).

Mass is a measure of the amount of matter (constant), while weight is the force of gravity on that mass (W = mg) and varies with location.

Static friction between the tires and road surface provides the centripetal force needed for a car to turn.

Orbital velocity is the speed at which an object must travel to maintain a stable orbit, balanced between gravity and inertia.

F = mv^2/r = 1000(20^2)/50 = 1000(400)/50 = 8000 N.

Astronauts and the ISS are in continuous free fall (orbit) around Earth, so everything falls at the same rate, creating the sensation of weightlessness.

Kepler's Third Law states T^2 is proportional to r^3, relating orbital period to the semi-major axis of the orbit.

Centrifugal force is not a real force but an apparent outward force experienced in a rotating (non-inertial) reference frame.

Weight = mg = m(GM/r^2). If M doubles and r stays the same, g doubles, so weight doubles.

Gravitational field strength is defined as force per unit mass, \(g = F/m\), so its unit is newtons per kilogram, \(\text{N/kg}\). The choice \text{kg/N} inverts this relationship and does not represent force divided by mass. Recognizing that field quantities are always 'something per unit source quantity' helps identify correct SI units across physics topics.

Uniform circular motion is defined by a constant speed, even though the velocity direction is continuously changing, which produces centripetal acceleration. The choice 'It oscillates periodically' describes non-uniform motion where speed varies, which is not the case here. Students should remember that in circular motion, acceleration can exist without a change in speed because acceleration depends on direction change too.

Newton's law states that gravitational force is proportional to the product of the two masses and inversely proportional to the square of the distance between them, \(F = G\frac{m_1 m_2}{r^2}\). The choice \(F = G\frac{m_1 m_2}{r}\) incorrectly uses an inverse-linear relationship rather than the inverse-square law that gravity actually follows. This inverse-square dependence is a hallmark of gravity and should be memorized precisely for exam calculations.

Because gravitational force follows an inverse-square law, tripling the distance \(r\) causes the force to scale by \(\frac{1}{3^2} = \frac{1}{9}\). The choice 'It becomes \(\frac{1}{3}\) of the original' would only apply if force were inversely proportional to distance rather than distance squared. Students must apply the squared relationship whenever distance changes in gravitation problems.

The universal gravitational constant is conventionally denoted by capital \(G\), distinct from lowercase \(g\) which represents local gravitational acceleration near Earth's surface. The choice \(g\) is a common point of confusion because it looks similar but represents a different, location-dependent quantity. Always distinguish \(G\) (universal constant) from \(g\) (acceleration due to gravity at a specific location) on exams.

The net force in uniform circular motion is the centripetal force, which always points toward the center of the circular path to continuously redirect the velocity vector. The choice 'Tangent to the circle' describes the direction of instantaneous velocity, not the direction of the net force causing the curved path. Remember that centripetal force is not a new type of force but rather the net inward force required to maintain circular motion.

Weight depends on gravitational force, which decreases with the square of the distance from Earth's center according to \(F = G\frac{Mm}{r^2}\), so weight decreases as distance increases. The choice 'Its weight stays the same' ignores the well-established inverse-square dependence of gravitational force on distance. Mass remains constant regardless of location, but weight is a force that varies with gravitational field strength.

Kepler's First Law states that planets orbit the Sun in elliptical paths, with the Sun located at one of the two foci rather than at the center. The choice 'Perfect circles centered on the Sun' is a common misconception, since circles are only a special case of ellipses with zero eccentricity. Understanding that orbits are elliptical explains why planets have varying orbital speeds at different points in their orbit.

At any instant, the velocity of an object in circular motion points tangent to the circle, perpendicular to the radius at that point. The choice 'Parallel to the centripetal force' is incorrect because centripetal force points radially inward, which is perpendicular to velocity, not parallel to it. This perpendicularity is why centripetal force changes direction but not speed in uniform circular motion.

By Kepler's Third Law, \(T^2 \propto r^3\), so a larger orbital radius corresponds to a longer orbital period for objects orbiting the same central body. The choice 'The orbital period stays the same' ignores this direct relationship between orbital radius and period confirmed by both Kepler's law and Newtonian gravity. This radius-period relationship is essential for comparing orbits of different satellites or planets around the same body.

Gravitational attraction between the Sun and planet supplies the necessary inward centripetal force that continuously bends the planet's straight-line inertial path into a curved orbit. The choice 'The planet's own inertia' actually works against curving the path, since inertia tends to keep an object moving in a straight line unless acted on by a force. This interplay between gravity and inertia is central to understanding why planets maintain stable orbits rather than flying off or falling in.

Once the string breaks, the centripetal force vanishes, so the ball obeys Newton's first law and travels in a straight line tangent to the circle at the instant of release, in the direction of its instantaneous velocity. The choice 'Directly away from the center of the circle' incorrectly assumes a radial trajectory, but there is no outward force acting on the ball, only its tangential velocity. This scenario illustrates that circular motion requires a continuous centripetal force, and its removal reveals the object's straight-line inertial motion.

Using \(F_c = \frac{mv^2}{r} = \frac{(0.5)(4)^2}{2} = \frac{0.5 \times 16}{2} = 4\ \text{N}\), the centripetal force is 4 newtons. The choice \(16\ \text{N}\) mistakenly omits dividing by the radius, using only \(mv^2\) instead of \(\frac{mv^2}{r}\). Always apply the full centripetal force formula, ensuring both mass, velocity squared, and radius are correctly incorporated.

Using Kepler's Third Law \(T^2 \propto r^3\), doubling the radius gives $T_{new}^2 = T_{old}^2 \times 2^3 = T_{old}^2 \times 8$, so $T_{new} = T_{old}\sqrt{8} \approx 24 \times 2.83 \approx 68$ hours. The choice 'About 48 hours' incorrectly assumes a linear relationship between radius and period rather than the correct \(r^3\) dependence inside the square root. Students should remember that orbital period scales with the cube root of radius cubed, not directly with radius.

Gravitational force is \(F = G\frac{m_1m_2}{r^2}\); doubling one mass multiplies force by 2, and halving distance multiplies force by \((1/0.5)^2 = 4\), giving a combined factor of \(2 \times 4 = 8\), so the new force is \(8F\). The choice \(4F\) only accounts for the distance change and neglects the doubled mass, underestimating the total effect. When multiple variables change simultaneously in the gravitation formula, multiply their individual scaling factors together.

On a frictionless banked curve, resolving the normal force into vertical and horizontal components shows that the horizontal component provides centripetal force while the vertical component balances gravity, leading to \(\tan\theta = \frac{v^2}{rg}\). The choice \(\sin\theta = \frac{v^2}{rg}\) incorrectly uses sine instead of the correct tangent relationship derived from dividing the horizontal and vertical force equations. This banked curve derivation is a classic application combining centripetal force with basic trigonometric force resolution.

The Moon's smaller mass produces a weaker gravitational field strength at its surface compared to Earth, so the gravitational force (weight) on the astronaut is smaller even though mass stays constant. The choice 'The astronaut's mass changes upon reaching the Moon' is physically incorrect, since mass is an intrinsic property that does not change with location. This distinction between mass, an invariant quantity, and weight, a location-dependent force, is fundamental to gravitation problems.

At the minimum speed for maintaining contact at the top of a loop, gravity alone supplies exactly the required centripetal force, meaning \(mg = \frac{mv^2}{r}\) and the normal force approaches zero. The choice 'The normal force must equal the centripetal force' describes a situation with excess speed where both gravity and normal force contribute, not the critical minimum condition. This minimum-speed condition is a classic multi-force analysis problem testing understanding of when normal force can vanish in circular motion.

Surface gravitational field strength is \(g = \frac{GM}{r^2}\); doubling both mass and radius gives $g_{new} = \frac{G(2M)}{(2r)^2} = \frac{2GM}{4r^2} = \frac{1}{2}\cdot\frac{GM}{r^2}$... actually this equals half, so recompute: the correct calculation shows $g_{new} = \frac{GM}{2r^2}$, which is half of Earth's \(g\), matching the answer 'It is the same as Earth's' being incorrect—so the true answer is half. The choice 'It is twice Earth's' incorrectly assumes mass and radius changes reinforce each other multiplicatively rather than mass increasing \(g\) while radius squared decreases it. Always substitute scaled variables directly into the formula to correctly track how competing changes affect the final field strength.

As the ring-shaped station rotates, the floor pushes inward on occupants with a normal force that provides the centripetal force needed for circular motion, and by Newton's third law, occupants feel this as a force pressing them 'down' onto the floor, mimicking gravity. The choice 'Centrifugal force pushes outward and is a real force pressing occupants to the floor' incorrectly treats centrifugal force as real, when it is actually a fictitious force that only appears in the rotating (non-inertial) reference frame. This rotating reference frame analysis is key to understanding artificial gravity concepts in engineering and space travel discussions.

Using Newton's second law, \(a = \frac{F}{m} = \frac{6000}{1500} = 4\ \text{m/s}^2\), which is the satellite's centripetal acceleration since gravity is the only force acting. The choice \(0.25\ \text{m/s}^2\) mistakenly inverts the division, computing mass over force instead of force over mass. Since gravity provides the sole centripetal force for orbiting satellites, Newton's second law directly links gravitational force to centripetal acceleration.

Orbital speed is given by \(v = \sqrt{\frac{GM}{r}}\), so doubling the radius scales speed by \(\frac{1}{\sqrt{2}} \approx 0.71\), meaning Satellite A moves slower than Satellite B. The choice 'Satellite A's speed is twice Satellite B's speed' incorrectly assumes a direct proportionality between radius and speed, but the true relationship involves an inverse square root. This inverse relationship explains why satellites farther from a planet travel more slowly in their orbits.

At the bottom of the loop, the seat must push upward with a normal force greater than gravity to provide both support against gravity and the additional centripetal force directed toward the loop's center, so the pilot feels heavier than normal. The choice 'The apparent weight equals actual weight exactly' ignores the extra centripetal force requirement present at the bottom of a curved path. This increased apparent weight at the bottom of loops explains why pilots experience high g-forces during such maneuvers.

Orbital speed follows \(v = \sqrt{\frac{GM}{r}}\), so halving the Sun's mass \(M\) while keeping \(r\) constant scales speed by \(\sqrt{0.5} \approx 0.71\), meaning Earth would need a lower speed to maintain that radius. The choice 'It would decrease by half' incorrectly assumes a direct linear relationship between mass and speed rather than the correct square root dependence. This relationship shows that orbital speed depends on the square root of the central mass, not on the mass directly.

At the bottom of the circle, both the tension and gravity's opposing component combine such that tension must supply the centripetal force plus support the rock's weight, giving \(T = \frac{mv^2}{r} + mg\), the maximum value in the cycle. The choice 'At the top of the circle' is incorrect because at the top, gravity assists the centripetal force, reducing the tension needed, following \(T = \frac{mv^2}{r} - mg\). This variation in tension around a vertical circle is a key multi-force analysis concept for circular motion problems.

Since \(F_c = \frac{mv^2}{r}\), increasing the tangential speed \(v\) while keeping mass and radius constant causes the required centripetal force to increase proportionally to the square of the speed. The choice 'The required centripetal force stays the same' ignores the direct dependence of centripetal force on velocity squared in the formula. This quadratic relationship means even small speed increases can substantially raise the force needed to maintain circular motion.

Since gravitational force is inversely proportional to \(r^2\), larger distances produce weaker forces, so distance \(2r\) gives the weakest force, followed by \(r\), then \(0.5r\) gives the strongest force, yielding the order B, A, C from weakest to strongest. The choice 'C, B, A' incorrectly reverses the ranking, implying larger distances produce stronger forces, which contradicts the inverse-square relationship. Ranking problems like this test whether students correctly apply the inverse-square law directionally, not just numerically.

At the top of the ferris wheel, gravity points toward the center (downward, matching the centripetal direction), so less normal force is needed to provide the same centripetal force, making the rider feel lighter as \(N = mg - \frac{mv^2}{r}\). The choice 'At the bottom, because gravity opposes the centripetal direction, increasing normal force' actually describes where the rider feels heaviest, not lightest, since normal force must overcome gravity plus provide centripetal force there. This asymmetry between the top and bottom of vertical circular paths is a recurring theme in AP-level circular motion problems.

By Newton's third law, the gravitational forces the two stars exert on each other are always equal in magnitude and opposite in direction, regardless of their masses being equal or not. The choice 'The more massive star exerts a stronger force on the other' misunderstands Newton's third law, since gravitational forces between any two masses form an action-reaction pair of exactly equal magnitude. This principle applies universally to gravitational interactions, not just to systems with equal masses.

Increasing speed beyond the circular orbital velocity at that radius means gravity can no longer provide exactly the centripetal force needed for a circle, so the spacecraft moves outward into a higher, elliptical orbit with the original point becoming its perigee. The choice 'The spacecraft will remain in the same circular orbit' ignores that circular orbits require a very specific speed for a given radius, and any deviation changes the orbital shape. This principle underlies orbital maneuvers used by real spacecraft to raise or adjust their orbits by firing thrusters.

As the cone angle increases, the radius of the circular path increases (since \(r = L\sin\theta\)), and using \(\tan\theta = \frac{v^2}{rg}\), larger angles require higher speeds to maintain the larger radius circular path with the corresponding centripetal force. The choice 'The required speed decreases' contradicts the physics, since a wider cone angle demands a faster orbital speed to generate sufficient centripetal force at the larger radius. This conical pendulum problem combines trigonometric relationships with centripetal force analysis, a synthesis often tested at the AP level.

Kepler's Third Law states \(T^2 \propto r^3\), so \(\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3\), giving \(8^2 = 64 = \left(\frac{r_2}{r_1}\right)^3\), and taking the cube root gives \(\frac{r_2}{r_1} = 4\). The choice '8' incorrectly assumes radius scales the same as period, ignoring the cube-root relationship required by Kepler's law. Working through Kepler's law ratio problems requires carefully applying both the square and cube exponents before solving for the desired ratio.

Orbital speed for a circular orbit is \(v = \sqrt{\frac{GM}{r}}\), so reducing mass to a quarter scales speed by \(\sqrt{1/4} = 1/2\), meaning the satellite would need to slow to half its original speed to remain in circular orbit at the same radius. The choice 'The satellite's speed would remain unchanged' ignores that orbital speed directly depends on the central mass through this square-root relationship, and the satellite would otherwise spiral outward if it kept its original higher speed. This kind of what-if scenario tests whether students can manipulate the orbital velocity formula for hypothetical mass changes.

The kinematic formula assumes constant acceleration, but gravitational acceleration \(g = \frac{GM}{r^2}\) changes significantly as an object falls from far away and gets closer to Earth, making \(g\) non-constant and invalidating the simple kinematics equations. The choice 'The object would not accelerate at all if released from a great distance' is false, since gravitational force, though weaker at large distances, is still nonzero and produces acceleration. This distinction reminds students that near-surface constant-\(g\) kinematics only applies over relatively small height changes where \(g\) is approximately uniform.

Since density \(\rho = \frac{M}{\frac{4}{3}\pi r^3}\) is constant, mass scales as \(M \propto r^3\), and surface gravity \(g = \frac{GM}{r^2} \propto \frac{r^3}{r^2} = r\), so doubling the radius while keeping density constant doubles the surface gravitational field strength. The choice 'It is the same as Earth's' incorrectly assumes gravity depends only on radius and ignores that mass also increases with volume when density is fixed. This combined density-mass-radius reasoning is a common synthesis question testing whether students can derive relationships rather than just plug into formulas.

At the critical speed for losing contact, the normal force drops to zero and gravity alone provides the centripetal force, so \(mg = \frac{mv^2}{r}\), which solves to \(v = \sqrt{gr}\). The choice \(v = \sqrt{2gr}\) incorrectly includes an extra factor of 2 that does not arise from correctly setting normal force to zero in the force equation. This scenario mirrors the top-of-the-loop problem and is a frequently tested synthesis of circular motion and gravity concepts.

Although drag removes total mechanical energy from the orbit, causing the radius to shrink, the required orbital speed at a smaller radius is actually higher according to \(v = \sqrt{\frac{GM}{r}}\), so counterintuitively the satellite speeds up as it spirals inward due to increasing kinetic energy despite losing total energy overall. The choice 'The satellite's orbital speed decreases as it spirals inward' misapplies the everyday intuition that losing energy should reduce speed, but for orbits, potential energy decreases faster than total energy, forcing kinetic energy (and thus speed) to increase. This paradoxical result is a classic hard-level topic connecting orbital mechanics with energy conservation in a nontrivial way.

For orbits around a common center of mass, the radii are inversely proportional to the masses, so \(\frac{r_M}{r_{4M}} = \frac{4M}{M} = 4\), meaning the lighter star orbits four times farther from the center of mass than the heavier star. The choice 'The star of mass \(4M\) orbits at four times the radius of the star of mass \(M\)' reverses this relationship, incorrectly placing the more massive star farther from the center of mass. This inverse mass-radius relationship for binary systems is essential for correctly analyzing multi-body gravitational systems.

Gravity remains the dominant force acting on orbiting objects, providing the centripetal force for their curved path, and the floating sensation occurs because both the astronaut and spacecraft are in continuous free fall together, canceling any apparent contact force. The choice 'Astronauts float because there is no air resistance in space' misidentifies the actual cause, as weightlessness is about the absence of a normal force from continuous free fall, not the presence or absence of air. This misconception about weightlessness is one of the most commonly tested conceptual errors in orbital mechanics units.

Since gravitational force follows an inverse-square law with distance, the planet experiences the strongest gravitational pull when it is closest to the star, at perihelion, where \(r\) is smallest. The choice 'The force is constant throughout the orbit' contradicts the elliptical nature of the orbit, since distance from the star varies continuously and force scales with \(1/r^2\). This varying force throughout an elliptical orbit is directly responsible for the varying orbital speed described by Kepler's Second Law.

Using \(a = \omega^2 r\), solving for \(\omega\) gives \(\omega = \sqrt{\frac{a}{r}} = \sqrt{\frac{9.8}{100}} = \sqrt{0.098} \approx 0.313\ \text{rad/s}\). The choice 'About \(0.098\ \text{rad/s}\)' mistakenly reports the value of \(a/r\) itself rather than taking the square root to solve for angular velocity. This type of rotational artificial-gravity calculation is a practical, multi-step application of the centripetal acceleration formula in rotational form.

Conservation of angular momentum, $L = mvr$, requires that as the comet's distance \(r\) from the Sun decreases at perihelion, its speed \(v\) must increase to keep \(L\) constant, making it move fastest at closest approach. The choice 'The same at both points, since gravity does no work on the comet' is incorrect because gravity does perform work on the comet as it moves along the radial direction of a non-circular elliptical path, changing its kinetic energy. This speed variation captured by Kepler's Second Law (equal areas in equal times) is a direct consequence of angular momentum conservation in gravitational orbits.

The block slides once the required centripetal force, \(\frac{mv^2}{r}\), exceeds the maximum static friction force available, \(f_{s,max} = \mu_s N\), since friction is the only horizontal force providing the centripetal force in this scenario. The choice 'When the normal force becomes zero' is irrelevant here because the normal force on a horizontal turntable simply balances gravity vertically and does not directly limit centripetal motion in the horizontal plane. This friction-limited centripetal force scenario is a standard application connecting circular motion with force-of-friction concepts.

Gravitational potential energy is given by $U = -\frac{GMm}{r}$, which increases (becomes less negative, approaching zero) as \(r\) increases, while gravitational force magnitude $F = \frac{GMm}{r^2}$ decreases as \(r\) increases, so these two quantities move in different mathematical senses but both trend toward zero at infinite distance. The choice 'Gravitational potential energy decreases while force magnitude increases' reverses both trends incorrectly, contradicting the standard formulas for gravitational potential energy and force. Understanding that potential energy is negative and increases toward zero while force always decreases with distance is crucial for correctly analyzing orbital energy problems.

In horizontal circular motion, tension alone must supply the entire centripetal force since gravity acts vertically and is balanced separately, keeping tension roughly constant, but in vertical circular motion gravity has components that add to or subtract from the required tension depending on position, causing tension to vary throughout the loop. The choice 'Tension is constant in both horizontal and vertical circular motion' ignores how gravity's changing relationship to the centripetal direction in a vertical circle causes tension to fluctuate. Recognizing this difference is essential for correctly solving vertical circular motion problems where tension (or normal force) at the top and bottom differ significantly.