Gravitational Force - College Physics

1. An astronaut stands on the surface of a planet with radius $R$ and mass $M$. At what height $h$ above the surface must the astronaut travel so that their weight becomes exactly half of their weight on the surface?
   Answer: $h = R(\sqrt{2} - 1)$ When you encounter problems about gravitational force at different heights, remember that gravity follows an inverse square law with distance from the planet's center. On the surface, the astronaut's weight is $W = \frac{GMm}{R^2}$, where $m$ is the astronaut's mass. At height $h$ above the surface, the distance from the planet's center becomes $(R + h)$, so the weight becomes $W' = \frac{GMm}{(R + h)^2}$. Since we want the weight at height $h$ to be half the surface weight: $\frac{GMm}{(R + h)^2} = \frac{1}{2} \cdot \frac{GMm}{R^2}$ Simplifying by canceling $GMm$: $\frac{1}{(R + h)^2} = \frac{1}{2R^2}$ Cross-multiplying: $2R^2 = (R + h)^2$ Taking the square root: $R\sqrt{2} = R + h$ Solving for $h$: $h = R\sqrt{2} - R = R(\sqrt{2} - 1)$ This confirms answer D is correct. Answer A, $h = R(2 - \sqrt{2})$, comes from incorrectly manipulating the square root step. Answer B, $h = R$, would result from assuming a linear relationship between distance and gravitational force rather than the inverse square relationship. Answer C, $h = R/2$, likely comes from oversimplifying the problem and assuming that doubling the height halves the weight. Remember: gravitational problems always involve the inverse square law, and you must measure distance from the planet's center, not just the height above surface.
2. An object is placed at different distances from a massive sphere. At distance $r$ from the center, the gravitational potential energy is $U_1$. At distance $2r$, the potential energy is $U_2$. What is the relationship between $U_1$ and $U_2$?
   Answer: $U_1 = 2U_2$ When you encounter gravitational potential energy problems involving different distances from a massive object, focus on the inverse relationship between potential energy and distance. Gravitational potential energy is given by $U = -\frac{GMm}{r}$, where $G$ is the gravitational constant, $M$ is the mass of the sphere, $m$ is the mass of the object, and $r$ is the distance from the center. The key insight is that potential energy is inversely proportional to distance. At distance $r$: $U_1 = -\frac{GMm}{r}$ At distance $2r$: $U_2 = -\frac{GMm}{2r}$ To find the relationship, divide $U_1$ by $U_2$: $\frac{U_1}{U_2} = \frac{-\frac{GMm}{r}}{-\frac{GMm}{2r}} = \frac{-\frac{GMm}{r}} \times \frac{2r}{-GMm} = 2$ Therefore, $U_1 = 2U_2$, confirming answer A. Looking at the wrong answers: B) $U_1 = 4U_2$ incorrectly applies an inverse-square relationship, which applies to gravitational force, not potential energy. C) $U_1 = \frac{U_2}{2}$ gets the relationship backwards—this would mean the object has less potential energy when closer to the massive sphere. D) $U_1 = \frac{U_2}{4}$ combines both errors: wrong direction and wrong power relationship. Remember: gravitational potential energy has an inverse relationship with distance ($\propto \frac{1}{r}$), while gravitational force has an inverse-square relationship ($\propto \frac{1}{r^2}$). Don't confuse these two fundamental relationships.
3. A planet has the same density as Earth but twice Earth's radius. An object that weighs $100\text{ N}$ on Earth will weigh how much on this planet?
   Answer: $200\text{ N}$ When you encounter problems comparing gravitational effects between planets, focus on how mass and radius changes affect surface gravity through Newton's law of universal gravitation. Surface gravity depends on a planet's mass divided by the square of its radius: $g = \frac{GM}{R^2}$. Since weight equals mass times gravity ($W = mg$), you need to find how this planet's surface gravity compares to Earth's. First, determine the planet's mass. Since density equals mass divided by volume, and the planet has Earth's density but twice the radius: $\rho_{planet} = \rho_{Earth}$. The volume scales as $R^3$, so with radius doubled, volume increases by $2^3 = 8$. Therefore, the planet's mass is $8M_{Earth}$. Now calculate the gravity ratio: $\frac{g_{planet}}{g_{Earth}} = \frac{GM_{planet}/R_{planet}^2}{GM_{Earth}/R_{Earth}^2} = \frac{8M_{Earth}}{(2R_{Earth})^2} \times \frac{R_{Earth}^2}{M_{Earth}} = \frac{8}{4} = 2$ The object weighs $100 \times 2 = 200\text{ N}$, making (A) correct. (B) 400 N incorrectly assumes gravity scales directly with mass increase (factor of 8) without accounting for radius effects. (C) 50 N mistakenly thinks doubling radius halves the weight, ignoring the mass change entirely. (D) 800 N wrongly uses only the mass factor of 8 while completely ignoring the radius increase. Study tip: For planetary comparison problems, always work through both mass and radius effects systematically. Mass changes affect gravity proportionally, while radius changes affect it by the inverse square.
4. A spacecraft traveling from Earth to Mars must pass through the point where Earth's and Mars' gravitational influences are equal. If Mars has $0.11$ times Earth's mass and the Earth-Mars distance is $2.3 \times 10^8\text{ km}$, approximately how far from Earth's center is this equilibrium point?
   Answer: $1.8 \times 10^8\text{ km}$ When spacecraft travel between planets, there's a special point called the gravitational equilibrium point where the gravitational forces from both celestial bodies are equal. This is where you need to set up an equation using Newton's law of universal gravitation. At the equilibrium point, the gravitational force from Earth equals the gravitational force from Mars. Let's call the distance from Earth's center to this point $x$, so the distance from Mars is $(2.3 \times 10^8 - x)$ km. Setting the forces equal: $\frac{GM_E}{x^2} = \frac{GM_M}{(2.3 \times 10^8 - x)^2}$ Since $M_M = 0.11M_E$, this becomes: $\frac{1}{x^2} = \frac{0.11}{(2.3 \times 10^8 - x)^2}$ Taking the square root of both sides: $\frac{1}{x} = \frac{\sqrt{0.11}}{2.3 \times 10^8 - x}$ Cross-multiplying and solving: $2.3 \times 10^8 - x = x\sqrt{0.11}$ Since $\sqrt{0.11} \approx 0.33$, we get: $2.3 \times 10^8 = x(1 + 0.33) = 1.33x$ Therefore: $x = \frac{2.3 \times 10^8}{1.33} \approx 1.8 \times 10^8 \text{ km}$ This confirms answer A is correct. Answer B ($1.2 \times 10^8$) would place the point too close to Earth, underestimating Earth's stronger gravity. Answer C ($1.5 \times 10^8$) represents a common error of assuming the midpoint. Answer D ($2.0 \times 10^8$) places the point too far from Earth. Remember: gravitational equilibrium points are always closer to the less massive object, since gravity weakens with distance squared.
5. Two objects with masses $m_1 = 3.0\text{ kg}$ and $m_2 = 7.0\text{ kg}$ are separated by a distance of $2.0\text{ m}$. A third object with mass $m_3 = 2.0\text{ kg}$ is placed at a point such that the net gravitational force on $m_3$ is zero. How far from $m_1$ should $m_3$ be placed?
   Answer: $0.76\text{ m}$ When you encounter gravitational force problems involving a point where forces balance, you're dealing with the principle that gravitational forces are vector quantities that can cancel each other out. For the net gravitational force on $m_3$ to be zero, the gravitational forces from $m_1$ and $m_2$ must be equal in magnitude but opposite in direction. This means $m_3$ must lie on the line connecting $m_1$ and $m_2$, and since $m_2 > m_1$, the equilibrium point will be closer to the smaller mass $m_1$. Let $x$ be the distance from $m_1$ to $m_3$. Then the distance from $m_2$ to $m_3$ is $(2.0 - x)$. Setting the gravitational forces equal: $\frac{Gm_1m_3}{x^2} = \frac{Gm_2m_3}{(2.0-x)^2}$ The $G$ and $m_3$ terms cancel, leaving: $\frac{m_1}{x^2} = \frac{m_2}{(2.0-x)^2}$ Substituting values: $\frac{3.0}{x^2} = \frac{7.0}{(2.0-x)^2}$ Cross-multiplying and solving: $3.0(2.0-x)^2 = 7.0x^2$ This yields $x = 0.76$ m, confirming answer A. Answer B (1.24 m) would place $m_3$ closer to $m_2$, making the force from $m_2$ too strong. Answer C (1.50 m) places $m_3$ too close to $m_2$ for equilibrium. Answer D (0.50 m) makes $m_3$ too close to $m_1$, creating imbalance. Remember: in gravitational equilibrium problems, the test mass always sits closer to the smaller mass, and you solve by setting force magnitudes equal.
6. Two identical satellites orbit Earth in circular paths. Satellite A orbits at altitude $h$ above Earth's surface, while satellite B orbits at altitude $3h$ above Earth's surface. If Earth's radius is $R$, what is the ratio of the gravitational force on satellite A to the gravitational force on satellite B?
   Answer: $\frac{(R + 3h)^2}{(R + h)^2}$ When analyzing gravitational forces between objects at different distances, you need to apply Newton's law of universal gravitation: $F = \frac{GMm}{r^2}$, where the force is inversely proportional to the square of the distance between centers. Since both satellites are identical (same mass $m$), the ratio of gravitational forces depends only on their distances from Earth's center. Satellite A orbits at distance $r_A = R + h$ from Earth's center, while satellite B orbits at distance $r_B = R + 3h$. The force ratio becomes: $\frac{F_A}{F_B} = \frac{\frac{GMm}{(R+h)^2}}{\frac{GMm}{(R+3h)^2}} = \frac{(R+3h)^2}{(R+h)^2}$ This matches answer choice B. The satellite closer to Earth (A) experiences a stronger gravitational force than the more distant satellite (B). Answer A reverses this ratio, giving you the force on B relative to A instead of A relative to B. Answer C incorrectly uses distance to the first power instead of distance squared—a common error when students forget that gravitational force follows an inverse square law, not just an inverse relationship. Answer D completely ignores Earth's radius and only considers the altitude ratio, missing that what matters is the total distance from Earth's center, not just the height above surface. Remember: gravitational problems always require measuring distances from the center of the massive object (Earth's center), not from its surface. The inverse square relationship is crucial—doubling the distance quarters the force.
7. A spherical planet has uniform density $\rho$ and radius $R$. A small object is placed at a distance $\frac{R}{2}$ below the planet's surface (i.e., at distance $\frac{R}{2}$ from the center). What is the gravitational force on the object compared to the force it would experience on the planet's surface?
   Answer: The force is $\frac{1}{2}$ of the surface value When analyzing gravitational force inside a uniform sphere, you need to apply a key principle: only the mass within your radius contributes to the gravitational force you experience. The mass in the spherical shell outside your position has zero net gravitational effect. At distance $\frac{R}{2}$ from the center, only the mass within a sphere of radius $\frac{R}{2}$ affects the object. Since the planet has uniform density $\rho$, this inner mass is: $M_{inner} = \frac{4}{3}\pi\left(\frac{R}{2}\right)^3\rho = \frac{1}{8} \cdot \frac{4}{3}\pi R^3\rho = \frac{1}{8}M_{total}$ The gravitational force is $F = \frac{GM_{inner}m}{r^2}$, where $r = \frac{R}{2}$. Substituting: $F_{inside} = \frac{G \cdot \frac{1}{8}M_{total} \cdot m}{\left(\frac{R}{2}\right)^2} = \frac{G \cdot \frac{1}{8}M_{total} \cdot m}{\frac{1}{4}R^2} = \frac{1}{2} \cdot \frac{GM_{total}m}{R^2}$ Since the surface force is $F_{surface} = \frac{GM_{total}m}{R^2}$, the force inside is $\frac{1}{2}$ the surface value. Answer A ($\frac{1}{8}$) incorrectly uses only the mass ratio without accounting for the distance change. Answer B ($\frac{1}{4}$) likely confuses the radius ratio with the force ratio. Answer D (2 times) incorrectly assumes force increases as you go deeper. Study tip: For uniform spheres, gravitational force increases linearly from the center to the surface, reaching maximum at the surface. Remember that $F \propto \frac{M_{enclosed}}{r^2}$, where both mass and distance matter.
8. A binary star system consists of two stars, each of mass $M$, orbiting their common center of mass in circular orbits of radius $R$. If one star suddenly loses half its mass (while maintaining its orbital speed), what happens to the gravitational force between the stars immediately after the mass loss?
   Answer: The force decreases to half its original value When analyzing gravitational force problems, always start with Newton's law of universal gravitation: $F = G\frac{m_1 m_2}{r^2}$. This tells you that gravitational force depends on both masses and the distance between them. Initially, the force between the two stars is $F_0 = G\frac{M \cdot M}{(2R)^2} = G\frac{M^2}{4R^2}$. Note that the distance between stars is $2R$ since each orbits at radius $R$ from their center of mass. When one star loses half its mass, it becomes mass $\frac{M}{2}$ while the other remains mass $M$. The problem states the orbital speed is maintained, so immediately after mass loss, the distance is still $2R$. The new force becomes: $F_{new} = G\frac{M \cdot \frac{M}{2}}{(2R)^2} = G\frac{M^2/2}{4R^2} = \frac{1}{2} \cdot G\frac{M^2}{4R^2} = \frac{F_0}{2}$ The force decreases to half its original value. Choice A incorrectly assumes force depends only on distance. While distance is unchanged, the reduced mass product affects the force. Choice B applies the mass reduction twice ($\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$), which would only be correct if both masses were halved. Choice D incorrectly applies energy or momentum considerations rather than the direct force calculation. Study tip: In gravitational force problems, always identify what changes (masses, distances) and what stays constant. Apply Newton's law of gravitation directly, being careful about the distance between centers of mass, not orbital radii.
9. On planet X, an object weighs $60\text{ N}$. On planet Y, the same object weighs $90\text{ N}$. If planet X has twice the radius of planet Y, what is the ratio of planet X's mass to planet Y's mass?
   Answer: $\frac{8}{3}$ When you encounter planetary weight problems, you're dealing with gravitational force relationships. The key insight is that weight equals mass times gravitational acceleration, and gravity depends on both planetary mass and radius. The gravitational acceleration on any planet is $g = \frac{GM}{R^2}$, where G is the gravitational constant, M is the planet's mass, and R is its radius. Since weight $W = mg$, we can write: For planet X: $60 = m \cdot \frac{GM_X}{R_X^2}$ For planet Y: $90 = m \cdot \frac{GM_Y}{R_Y^2}$ Taking the ratio of these equations eliminates the object's mass: $\frac{60}{90} = \frac{GM_X/R_X^2}{GM_Y/R_Y^2} = \frac{M_X}{M_Y} \cdot \frac{R_Y^2}{R_X^2}$ Since $R_X = 2R_Y$, we have: $\frac{2}{3} = \frac{M_X}{M_Y} \cdot \frac{R_Y^2}{(2R_Y)^2} = \frac{M_X}{M_Y} \cdot \frac{1}{4}$ Solving for the mass ratio: $\frac{M_X}{M_Y} = \frac{2}{3} \times 4 = \frac{8}{3}$ Answer A is correct. Answer B ($\frac{3}{2}$) ignores the radius relationship entirely. Answer C ($\frac{2}{3}$) represents the weight ratio but forgets that gravity also depends on radius squared. Answer D ($\frac{4}{3}$) incorrectly applies the radius factor as linear rather than squared. Remember: gravitational problems always involve the inverse square relationship with distance. When comparing planetary gravities, both mass and radius matter, and radius effects are squared.
10. A uniform solid sphere of mass $M$ and radius $R$ has a small cavity of radius $\frac{R}{4}$ carved out from its center. What is the gravitational field at a point on the surface of the original sphere, directly above the center of the cavity?
    Answer: $\frac{63GM}{64R^2}$ When you encounter a gravitational field problem involving a cavity, use the superposition principle: treat the cavity as the original solid sphere minus a smaller sphere of material that was removed. The original uniform sphere would create a gravitational field of $\frac{GM}{R^2}$ at its surface. However, we need to account for the missing mass in the cavity. The cavity has radius $\frac{R}{4}$, so its volume is $\frac{1}{64}$ of the original sphere's volume, meaning it contained $\frac{M}{64}$ of mass. The distance from the cavity's center to the point on the surface is $R$ (the full radius). The "missing mass" $\frac{M}{64}$ would have contributed a field of $\frac{G(M/64)}{R^2} = \frac{GM}{64R^2}$ at this point. Using superposition, the actual field equals the original field minus the contribution from the removed mass: $\frac{GM}{R^2} - \frac{GM}{64R^2} = \frac{64GM - GM}{64R^2} = \frac{63GM}{64R^2}$ Answer A is correct. Answer B ($\frac{GM}{R^2}$) ignores the cavity entirely. Answer C ($\frac{15GM}{16R^2}$) likely results from incorrectly calculating the cavity's mass fraction as $\frac{1}{16}$ instead of $\frac{1}{64}$. Answer D ($\frac{7GM}{8R^2}$) suggests using $\frac{1}{8}$ as the mass fraction, confusing linear and volume ratios. Study tip: For cavity problems, always remember that volume scales as the cube of linear dimensions, so a cavity with radius $\frac{R}{4}$ has volume fraction $(\frac{1}{4})^3 = \frac{1}{64}$.
11. A geosynchronous satellite orbits Earth at a radius where the orbital period equals Earth's rotation period (24 hours). If a satellite orbits at half this radius, what will be its orbital period?
    Answer: $8.5\text{ hours}$ This question tests your understanding of Kepler's Third Law, which relates orbital period to orbital radius. When you see problems about satellite orbits at different distances, immediately think about how period scales with radius. Kepler's Third Law states that the square of the orbital period is proportional to the cube of the orbital radius: $T^2 \propto r^3$. For two satellites, this gives us $\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$. Let's call the geosynchronous period $T_1 = 24$ hours at radius $r_1$. The new satellite orbits at $r_2 = \frac{r_1}{2}$. Substituting into Kepler's Third Law: $\frac{T_2^2}{24^2} = \frac{(\frac{r_1}{2})^3}{r_1^3} = \frac{1}{8}$ Therefore: $T_2^2 = \frac{24^2}{8} = 72$, so $T_2 = \sqrt{72} = 8.5$ hours. Looking at the wrong answers: Choice A (6.0 hours) likely comes from incorrectly assuming period is inversely proportional to radius ($T \propto \frac{1}{r}$), giving $24 \times \frac{1}{2} = 12$, then making an arithmetic error. Choice C (12 hours) results from the common misconception that period is directly proportional to radius. Choice D (17 hours) might come from incorrectly applying $T \propto \sqrt{r}$, giving $24 \times \frac{1}{\sqrt{2}} \approx 17$. The answer is B (8.5 hours). Remember: For orbital mechanics problems, always start with Kepler's Third Law ($T^2 \propto r^3$). The period decreases dramatically as you move closer to the central body—cutting the radius in half reduces the period by a factor of $2^{3/2} = 2.83$.
12. The gravitational field strength at the surface of a planet is $12\text{ m/s}^2$. At what altitude above the surface will the gravitational field strength be reduced to $3\text{ m/s}^2$? Express your answer in terms of the planet's radius $R$.
    Answer: $R$ When you encounter gravitational field problems involving altitude changes, remember that gravitational field strength follows an inverse square relationship with distance from the planet's center. The gravitational field strength is given by $g = \frac{GM}{r^2}$, where $r$ is the distance from the planet's center. At the surface, $r = R$ and $g = 12 \text{ m/s}^2$. At altitude $h$, the distance becomes $r = R + h$ and $g = 3 \text{ m/s}^2$. Setting up the ratio: $\frac{g_{\text{altitude}}}{g_{\text{surface}}} = \frac{R^2}{(R+h)^2}$ Substituting values: $\frac{3}{12} = \frac{R^2}{(R+h)^2}$ This simplifies to: $\frac{1}{4} = \frac{R^2}{(R+h)^2}$ Taking the square root: $\frac{1}{2} = \frac{R}{R+h}$ Cross-multiplying: $R + h = 2R$, so $h = R$ Choice A ($R$) is correct. Choice B ($2R$) represents a common error where students confuse the altitude with the total distance from center. At altitude $2R$, you'd be at distance $3R$ from center, giving field strength $\frac{12}{9} = 1.33 \text{ m/s}^2$. Choice C ($3R$) would put you at distance $4R$ from center, yielding field strength $\frac{12}{16} = 0.75 \text{ m/s}^2$. Choice D ($0.5R$) gives distance $1.5R$ from center, producing field strength $\frac{12}{2.25} = 5.33 \text{ m/s}^2$. Strategy tip: Always distinguish between altitude above surface and total distance from planet center. Set up ratios using the inverse square law to avoid calculation errors.
13. A binary star system consists of two stars of equal mass $M$ separated by distance $2d$. A small test mass $m$ is placed at the midpoint between the stars. If the test mass is displaced slightly toward one star along the line connecting the stars, what happens?
    Answer: The test mass experiences a net force toward the star it was displaced toward, making the equilibrium unstable When analyzing equilibrium in gravitational systems, you need to examine what happens to forces when the system is slightly perturbed. This reveals whether the equilibrium is stable (restoring forces) or unstable (amplifying forces). At the midpoint between two equal masses, the gravitational forces from each star exactly cancel, creating equilibrium. However, when you displace the test mass slightly toward one star by distance $x$, the force balance changes dramatically. The displaced test mass is now closer to one star (distance $d-x$) and farther from the other (distance $d+x$). Since gravitational force follows $F = \frac{GMm}{r^2}$, the force from the nearer star becomes $\frac{GMm}{(d-x)^2}$ while the force from the farther star becomes $\frac{GMm}{(d+x)^2}$. The inverse-square relationship means the stronger force from the nearer star dominates, creating a net force that pulls the test mass further toward that star, away from the original equilibrium position. Option B incorrectly suggests stable equilibrium with restoring forces back toward the midpoint. Option C misunderstands that gravitational forces depend on distance—positions are definitely not equivalent. Option D assumes the restoring force needed for harmonic motion exists, but we've shown the opposite occurs. The correct answer is A: the test mass experiences amplifying forces that make the equilibrium unstable. Study tip: For gravitational equilibrium problems, always check the stability by examining how forces change with small displacements. Unstable equilibria are common in gravitational systems, unlike spring systems where equilibria are typically stable.
14. A pendulum clock that keeps perfect time on Earth is taken to a planet where the gravitational acceleration is $\frac{g}{4}$. How does the period of the pendulum change?
    Answer: The period doubles When you encounter pendulum problems involving different gravitational environments, you need to recall that the period of a simple pendulum depends on both length and gravitational acceleration through the formula $T = 2\pi\sqrt{\frac{L}{g}}$. Since the pendulum clock is simply moved to a different planet, its physical length $L$ remains unchanged. The only variable that changes is the gravitational acceleration, which decreases from $g$ to $\frac{g}{4}$. Let's compare the periods. On Earth: $T_{Earth} = 2\pi\sqrt{\frac{L}{g}}$. On the new planet: $T_{planet} = 2\pi\sqrt{\frac{L}{g/4}} = 2\pi\sqrt{\frac{4L}{g}} = 2\pi \cdot 2\sqrt{\frac{L}{g}} = 2T_{Earth}$ The period doubles, confirming answer A is correct. Now for the wrong answers: B suggests the period halves, which would occur if gravity increased by a factor of 4, not decreased. C claims the period increases by a factor of 4 – this misconception likely comes from directly using the gravity ratio without considering the square root relationship. D suggests an increase by $\sqrt{2}$, which doesn't match any reasonable manipulation of our gravity change. Remember this key pattern: in the pendulum period formula, gravitational acceleration appears in the denominator under a square root. When gravity decreases by a factor of 4, you take the square root of the inverse ratio ($\sqrt{4} = 2$) to find how the period increases. This inverse square root relationship is crucial for pendulum problems across different gravitational environments.
15. The gravitational acceleration on the surface of a neutron star is approximately $10^{12}\text{ m/s}^2$. If a neutron star has a radius of $10\text{ km}$, what is the approximate escape velocity from its surface?
    Answer: $1.4 \times 10^8\text{ m/s}$ When you encounter escape velocity problems, you're dealing with the minimum speed needed for an object to break free from a gravitational field. The escape velocity formula is $v_e = \sqrt{2gR}$, where $g$ is surface gravity and $R$ is the radius. With the given values: $g = 10^{12} \text{ m/s}^2$ and $R = 10 \text{ km} = 10^4 \text{ m}$, you can substitute directly: $v_e = \sqrt{2 \times 10^{12} \times 10^4} = \sqrt{2 \times 10^{16}} = \sqrt{2} \times 10^8$ Since $\sqrt{2} \approx 1.41$, this gives $v_e \approx 1.4 \times 10^8 \text{ m/s}$, confirming answer (A). Looking at the wrong answers: (B) $1.0 \times 10^8 \text{ m/s}$ would result from forgetting the factor of 2 in the escape velocity formula, using $v_e = \sqrt{gR}$ instead. (C) $2.0 \times 10^8 \text{ m/s}$ comes from incorrectly using 2 instead of $\sqrt{2}$ as the coefficient. (D) $7.1 \times 10^7 \text{ m/s}$ appears to involve calculation errors, possibly confusing the relationship between the given parameters. Remember that escape velocity problems on physics exams often test whether you know the correct formula and can handle the square root properly. The factor of 2 under the square root is crucial—it comes from energy conservation between kinetic and gravitational potential energy. Always double-check your formula before plugging in numbers.
16. Consider a uniform ring of mass $M$ and radius $R$. A point mass $m$ is located on the axis of the ring at distance $h$ from the center. What is the magnitude of the gravitational force on the point mass?
    Answer: $\frac{GMmh}{(h^2 + R^2)^{3/2}}$ When you encounter gravitational force problems involving extended objects like rings or disks, the key is using symmetry to simplify the vector addition of forces from all mass elements. To find the gravitational force, consider how each small mass element $dm$ on the ring exerts a force on the point mass. Every element is at the same distance $\sqrt{h^2 + R^2}$ from the point mass (using the Pythagorean theorem). The force from each element has magnitude $\frac{G \cdot dm \cdot m}{h^2 + R^2}$. Here's the crucial insight: due to the ring's symmetry, all horizontal force components cancel out. Only the vertical components (along the axis) survive. Each force makes an angle with the axis where $\cos\theta = \frac{h}{\sqrt{h^2 + R^2}}$. The net force is therefore: $F = \int \frac{G \cdot dm \cdot m}{h^2 + R^2} \cdot \frac{h}{\sqrt{h^2 + R^2}} = \frac{GMmh}{(h^2 + R^2)^{3/2}}$ This confirms answer A is correct. Answer B, $\frac{GMm}{h^2 + R^2}$, ignores the angular factor and assumes all force acts along the axis. Answer C, $\frac{GMm}{(h^2 + R^2)^2}$, incorrectly squares the distance term. Answer D, $\frac{GMmR}{(h^2 + R^2)^{3/2}}$, mistakenly uses $R$ instead of $h$ in the numerator, confusing the geometry. Remember: in symmetric mass distribution problems, always identify which force components cancel due to symmetry, then apply the appropriate geometric factor to the surviving components.
17. A satellite orbits Earth at a distance of $2R$ from Earth's center, where $R$ is Earth's radius. If the satellite is moved to orbit at distance $3R$ from Earth's center, by what factor does the gravitational force on the satellite change?
    Answer: The force decreases by a factor of $\frac{4}{9}$ The gravitational force follows an inverse square law: $F = \frac{GMm}{r^2}$. Initially at $r_1 = 2R$: $F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$. Finally at $r_2 = 3R$: $F_2 = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}$. The ratio is $\frac{F_2}{F_1} = \frac{\frac{GMm}{9R^2}}{\frac{GMm}{4R^2}} = \frac{4}{9}$. Choice B uses linear scaling instead of inverse square. Choice C inverts the correct ratio. Choice D incorrectly suggests the force increases.
18. The gravitational potential energy between two masses $m_1$ and $m_2$ separated by distance $r$ is $U = -\frac{Gm_1m_2}{r}$. If the distance between the masses is tripled, by what factor does the gravitational force between them change?
    Answer: Decreases by a factor of $9$ When you encounter gravitational problems, remember that force and potential energy are related but follow different distance dependencies. The key is knowing how to derive one from the other. To find how gravitational force changes with distance, you need to use the relationship between force and potential energy: $F = -\frac{dU}{dr}$. Taking the derivative of the given potential energy $U = -\frac{Gm_1m_2}{r}$: $F = -\frac{d}{dr}\left(-\frac{Gm_1m_2}{r}\right) = -Gm_1m_2 \frac{d}{dr}(r^{-1}) = -Gm_1m_2(-r^{-2}) = \frac{Gm_1m_2}{r^2}$ This gives you Newton's law of universal gravitation, showing that gravitational force is inversely proportional to $r^2$. When distance triples ($r \rightarrow 3r$), the new force becomes $\frac{Gm_1m_2}{(3r)^2} = \frac{Gm_1m_2}{9r^2}$, which is the original force divided by 9. Choice A is correct—the force decreases by a factor of 9. Choice B incorrectly applies the inverse relationship from potential energy ($r^{-1}$) directly to force, missing the derivative step. Choice C might come from confusing this with some other physics relationship or calculation error. Choice D suggests the force increases, which violates the basic principle that gravitational force weakens with distance. Study tip: Always remember the "inverse square law" pattern in physics—gravitational force, electric force, and light intensity all follow $1/r^2$ relationships. When distance increases by factor $n$, these quantities decrease by factor $n^2$.
19. Two spherical masses $m_1 = 8.0\text{ kg}$ and $m_2 = 2.0\text{ kg}$ are initially at rest and separated by $1.0\text{ m}$. If they are released and allowed to move under their mutual gravitational attraction only, what is the ratio of distances they each travel when they meet?
    Answer: $m_1$ travels $\frac{1}{4}$ the distance that $m_2$ travels When you encounter problems involving two objects attracting each other gravitationally, the key principle is conservation of momentum. Since the system starts at rest with zero total momentum, the momentum must remain zero throughout the motion. Let's work through this systematically. If $m_1$ travels distance $d_1$ and $m_2$ travels distance $d_2$, then at any moment their momenta must be equal in magnitude but opposite in direction: $m_1 v_1 = m_2 v_2$. Since both objects start from rest and experience the same time interval, their velocities are proportional to their distances traveled: $v_1 = d_1/t$ and $v_2 = d_2/t$. Substituting: $m_1 \frac{d_1}{t} = m_2 \frac{d_2}{t}$, which simplifies to $m_1 d_1 = m_2 d_2$. Therefore: $\frac{d_1}{d_2} = \frac{m_2}{m_1} = \frac{2.0}{8.0} = \frac{1}{4}$. This means $m_1$ travels $\frac{1}{4}$ the distance that $m_2$ travels, confirming answer A. Answer B ($\frac{1}{2}$) incorrectly uses the square root of the mass ratio. Answer C (same distance) ignores the mass difference entirely—a common misconception that heavier objects somehow "fall faster" in this context. Answer D (2 times) inverts the correct ratio, perhaps confusing which mass should be in the numerator. Remember: in gravitational problems with two bodies, the lighter object always moves farther because momentum conservation requires $m_1 d_1 = m_2 d_2$. The inverse relationship between mass and distance traveled is crucial.
20. Two identical satellites orbit a planet. Satellite A orbits at radius $r$ and satellite B orbits at radius $4r$. What is the ratio of the gravitational force on satellite A to the gravitational force on satellite B?
    Answer: $16:1$ When you encounter orbital mechanics problems involving gravitational force, the key principle is Newton's law of universal gravitation: $F = \frac{GMm}{r^2}$, where the force is inversely proportional to the square of the distance. Since both satellites are identical (same mass $m$) orbiting the same planet (same $M$), you can set up a ratio. For satellite A at radius $r$: $F_A = \frac{GMm}{r^2}$. For satellite B at radius $4r$: $F_B = \frac{GMm}{(4r)^2} = \frac{GMm}{16r^2}$. Taking the ratio: $\frac{F_A}{F_B} = \frac{\frac{GMm}{r^2}}{\frac{GMm}{16r^2}} = \frac{GMm}{r^2} \times \frac{16r^2}{GMm} = 16$ Therefore, the ratio is $16:1$, making (A) correct. (B) 4:1 represents the common error of forgetting to square the radius ratio. Students often think "4r is 4 times larger, so force is 4 times smaller," but this ignores the inverse square relationship. (C) 8:1 has no clear physical basis and likely results from confused arithmetic, perhaps mixing up the radius factor with the inverse square law incorrectly. (D) 2:1 suggests taking the square root of the correct answer, which doesn't correspond to any relevant physical relationship in gravitational problems. Study tip: Always remember that gravitational force follows an inverse square law. When the distance increases by a factor of $n$, the force decreases by a factor of $n^2$. This pattern appears frequently in physics problems involving fields that spread out in three dimensions.

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