Gravitational Force - AP Physics 1

A small mass $m$ interacts gravitationally with a much larger mass $M$ in deep space. In trial 1, they are separated by $r$ . In trial 2, the separation is reduced to $r/2$ while both masses stay the same. Compared with trial 1, the gravitational force magnitude in trial 2 is
Answer: Four times as large This question tests understanding of the inverse-square relationship between gravitational force and distance. The gravitational force follows F = Gm₁m₂/r², where force is inversely proportional to distance squared. When separation is halved from r to r/2, the denominator becomes (r/2)² = r²/4, making the force 4 times larger (since dividing by a smaller number gives a larger result). Choice D incorrectly suggests that the mass ratio affects how distance changes influence force, but the inverse-square law applies regardless of mass values. To handle distance reductions, remember that halving distance quadruples force.
Two moons interact gravitationally. Moon 1 has mass $M$ and Moon 2 has mass $M$ . At distance $r$ , the force magnitude is $F$ . If their separation becomes $2r$ , what is the new force magnitude?
Answer: $\tfrac{F}{4}$ . This question tests understanding of how gravitational force changes with distance. The gravitational force between objects follows the inverse-square law: F = GM₁M₂/r². When the separation doubles from r to 2r, the distance squared in the denominator becomes (2r)² = 4r². This makes the new force F_new = GM²/(4r²) = F/4, or one-fourth of the original force. Choice D incorrectly claims that gravitational force is independent of distance, which contradicts Newton's law of universal gravitation. To solve distance-change problems, remember that the force changes by the inverse square of the distance ratio.
Two point masses interact gravitationally: mass $m$ and mass $3m$ are separated by distance $r$ . The separation is changed to $\tfrac{3r}{2}$ while masses stay the same. How does the gravitational force magnitude change?
Answer: It becomes $\tfrac{4}{9}$ as large. This question tests understanding of how gravitational force changes with distance. The gravitational force follows F = Gm₁m₂/r², inversely proportional to the square of separation. Initially, F = Gm(3m)/r² = 3Gm²/r². When distance changes from r to 3r/2, the new force is F_new = 3Gm²/(3r/2)² = 3Gm²/(9r²/4) = 3Gm²·4/9r² = 12Gm²/9r² = 4Gm²/3r². Comparing to the original, F_new = (4/3)·(Gm²/r²) = (4/9)·(3Gm²/r²) = (4/9)F. Choice C incorrectly suggests force increases with distance, violating the inverse relationship. To find the new force, divide by the square of the distance ratio: (3/2)² = 9/4.
Two small spheres interact gravitationally: sphere A has mass $m$ and sphere B has mass $2m$. Their center-to-center distance is $r$ . If the distance becomes $2r$ , how does the gravitational force magnitude between them change?
Answer: It becomes one-fourth as large. This question assesses understanding of the gravitational force between two masses as described by Newton's law of universal gravitation. The gravitational force is directly proportional to the product of the two masses involved, meaning that increasing either mass increases the force linearly with that change. Simultaneously, the force is inversely proportional to the square of the distance between the centers of the masses, so doubling the distance reduces the force to one-fourth of its original value. This inverse-square relationship arises because gravity spreads out in all directions, diluting its strength with the square of the distance. A common distractor, like choice D, incorrectly assumes that gravitational force is independent of distance, which might stem from confusing it with constant gravitational fields near Earth's surface. To solve similar problems, always identify the factors changed and apply the proportionality: multiply by the mass ratio and divide by the square of the distance ratio.
Masses $M$ and $M$ are separated by $r$ . If one mass is replaced with $3M$ while $r$ stays constant, how does the gravitational force change?
Answer: It becomes three times as large. This problem tests understanding of how gravitational force depends on mass. The gravitational force is F = Gm₁m₂/r², showing that force is directly proportional to the product of the masses. Initially, with masses M and M at distance r, the force is F = GM²/r². When one mass is replaced with 3M, the new force becomes F' = G(M)(3M)/r² = 3GM²/r² = 3F. Since only one mass tripled while distance remained constant, the force increases by a factor of 3. Choice D incorrectly suggests the force becomes nine times larger, which would only happen if both masses were tripled. To analyze mass changes in gravitational problems, multiply the force by the factor by which the mass product changes.
Masses P and Q attract gravitationally. P has mass $m$ and Q has mass $m$ . Their separation is $r$ . Q is replaced with mass $5m$ while $r$ is unchanged. What happens to the gravitational force magnitude?
Answer: It becomes five times as large. This question assesses understanding of the gravitational force between two masses as described by Newton's law of universal gravitation. The gravitational force is directly proportional to the product of the masses, so replacing one mass with five times its value multiplies the force by five. The inverse-square dependence on distance is irrelevant here since distance remains unchanged. This qualitative view highlights how gravitational attraction scales linearly with mass when other factors are fixed. Choice C, a distractor, claims it stays the same because one mass is unchanged, ignoring the product nature of the force law. A transferable strategy is to focus on the changed variables and apply F_new = F_old × (new mass product / old mass product) × (old r / new r)^2.
Two objects in space interact gravitationally: object X has mass $2m$ and object Y has mass $3m$, separated by $r$ . They are moved so the separation becomes $\tfrac{r}{2}$ with masses unchanged. How does the gravitational force magnitude change?
Answer: It becomes $4$ times as large. This question tests understanding of gravitational force and the inverse-square law for distance. The gravitational force follows F = Gm₁m₂/r², where force is inversely proportional to the square of the separation distance. When the separation decreases from r to r/2 (halves), the denominator becomes (r/2)² = r²/4. Since we're dividing by a smaller number (r²/4 instead of r²), the force becomes 4 times larger. Choice A incorrectly applies the inverse relationship, suggesting the force decreases when objects get closer. Remember that decreasing distance increases gravitational force, and halving the distance makes the force four times stronger.
Masses $m$ and $2m$ are separated by $r$ and attract with force $F$ . The $2m$ mass is replaced by $6m$ while $r$ stays the same. What is the new force magnitude?
Answer: $3F$ This question assesses how mass changes influence gravitational force. Force is directly proportional to the mass product, so replacing one mass with three times its value triples the product and the force. Distance remains constant, preserving the inverse-square factor. Here, changing 2m to 6m triples the product from $2m^2$ to $6m^2$ , yielding 3F. Distractor D claims the force stays F because distance is unchanged, ignoring the mass replacement's effect. To tackle similar problems, compute the ratio of new mass product to old and multiply by the distance ratio squared.
A satellite of mass $m$ and a space probe of mass $m$ interact gravitationally while drifting in space. Initially their separation is $r$ . The probe's mass is doubled to $2m$ while their separation is also doubled to $2r$ . Compared with the original gravitational force magnitude, the new force magnitude is
Answer: One-half as large This question tests understanding of how gravitational force depends on both mass and distance simultaneously. The gravitational force follows F = Gm₁m₂/r², depending directly on mass product and inversely on distance squared. When the probe's mass doubles (factor of 2 increase) and distance doubles (factor of 4 decrease due to squaring), the net effect is: (2 × 1)/4 = 1/2 of the original force. Choice A incorrectly assumes the mass and distance effects cancel out completely. To analyze combined changes, multiply the mass effect by the reciprocal of the squared distance effect.
Two small spheres interact gravitationally in deep space: sphere A has mass $m$ and sphere B has mass $2m$. Their center-to-center separation is increased from $r$ to $2r$ while the masses stay the same. How does the gravitational force magnitude between the spheres change?
Answer: It becomes $\tfrac{1}{4}$ as large. This question tests understanding of how gravitational force depends on distance between objects. The gravitational force between two objects follows Newton's law of universal gravitation: F = Gm₁m₂/r², where the force is inversely proportional to the square of the distance between their centers. When the separation increases from r to 2r (doubles), the distance term in the denominator becomes (2r)² = 4r². Since the force is inversely proportional to r², the new force becomes F/4, or one-fourth as large. Choice A incorrectly suggests the force increases, which would violate the inverse relationship between force and distance. To solve problems like this, remember that doubling the distance makes the force four times weaker due to the inverse-square law.
Two objects in space interact gravitationally. At separation $r$ , masses are $m$ and $m$ with force magnitude $F$ . One mass is changed to $2m$ and the separation is changed to $2r$ . How does the new force magnitude compare to $F$ ?
Answer: It is $\tfrac{F}{2}$ . This question tests understanding of how gravitational force changes when both mass and distance change. The gravitational force is F = Gm₁m₂/r². Initially, F = Gm·m/r² = Gm²/r². After changes, one mass becomes 2m and distance becomes 2r, giving F_new = Gm(2m)/(2r)² = 2Gm²/4r² = Gm²/2r² = F/2. The mass change increases force by factor of 2, while the distance change decreases it by factor of 4, resulting in a net decrease by factor of 2. Choice C incorrectly considers only the distance effect, ignoring the mass change. When both mass and distance change, multiply their individual effects on the force.
Two point masses, $M$ and $m$ , attract each other with gravitational force magnitude $F$ when separated by $r$ . Which statement is correct about the forces on the masses?
Answer: The forces have equal magnitude and opposite direction. This question probes Newton's third law in the context of gravitational forces. Gravitational force depends on the product of both masses equally, attracting each toward the other. The inverse-square law applies symmetrically to the pair, regardless of individual mass sizes. Thus, the forces on each mass are equal in magnitude but opposite in direction, as per action-reaction pairs. Distractor A wrongly states the force on M is larger because M > m, confusing force with acceleration. A key strategy is to recall that gravitational forces are mutual and equal, then consider accelerations separately using F=ma if needed.
Three masses are arranged on a line: $m_1=m$ and $m_2=m$ are separated by $r$ ; $m_3=m$ is placed so that it is $2r$ from $m_1$ . Which pair has the greatest gravitational force magnitude?
Answer: $m_1$ and $m_2$ , because their separation is smaller. This question evaluates comparing gravitational forces in multi-object systems. Force increases with larger mass products but decreases with the square of greater distances. The inverse-square law means closer objects experience stronger attraction for equal masses. In this arrangement, the pair separated by r has the greatest force due to the smallest distance. Distractor C suggests all pairs have equal force because masses are equal, but it neglects varying separations. A general strategy is to calculate each pair's force individually using F = G m1 m2 / $r^2$ and compare magnitudes directly.
Two masses $m$ and $4m$ are separated by distance $r$ and attract gravitationally. If the distance becomes $3r$ , what is the new force magnitude in terms of the original force $F$ ?
Answer: $F/9$ This question assesses understanding of the gravitational force between two masses as described by Newton's law of universal gravitation. The gravitational force is directly proportional to the product of the masses, but here masses remain m and 4m, so that factor is constant. It is inversely proportional to the square of the distance, so tripling the distance reduces the force to one-ninth of its original value due to the inverse-square law. This qualitative inverse-square reasoning illustrates how gravity diminishes rapidly over larger separations, making distant objects attract much more weakly. Choice D, a distractor, claims the force is unchanged because masses are the same, neglecting the distance's squared impact. A transferable strategy is to express the new force as original F times (new r / old r)^{-2}, adjusting for mass changes if any.
Two spheres interact gravitationally: $m_1=2M$ and $m_2=M$ . When their center-to-center distance doubles from $r$ to $2r$ , how does the force change?
Answer: It becomes $\tfrac{1}{4}$ as large. This problem tests understanding of gravitational force. The gravitational force between two objects is given by F = Gm₁m₂/r², where G is the gravitational constant, m₁ and m₂ are the masses, and r is the distance between their centers. When the distance doubles from r to 2r, the new force becomes F' = Gm₁m₂/(2r)² = Gm₁m₂/4r² = F/4. The force depends on the inverse square of the distance, so doubling the distance makes the force one-fourth as large. Choice A incorrectly suggests only mass matters, ignoring the crucial role of distance. To solve gravitational force problems, always identify what changes (mass or distance) and apply the inverse-square relationship for distance changes.
Two small spheres interact gravitationally: sphere A has mass $2m$ and sphere B has mass $m$ . Their center-to-center distance increases from $r$ to $2r$ . How does the gravitational force magnitude change?
Answer: It decreases to $\tfrac{1}{4}$ of the original value. This question assesses understanding of the gravitational force law. Gravitational force between two objects is directly proportional to the product of their masses, meaning if the product increases, the force increases accordingly. It is also inversely proportional to the square of the distance between their centers, so doubling the distance reduces the force to one-fourth of its original value. In this scenario, the masses remain unchanged while the distance doubles, leading to a force that is one-fourth of the original due to the inverse-square relationship. A common distractor, choice B, incorrectly assumes that only one mass is doubled and suggests the force halves, but no mass change occurs here. To solve similar problems, always calculate the ratio of the new force to the old by considering changes in mass product and distance squared separately.
A mass $m$ interacts gravitationally with a mass $4m$ at separation $r$ , producing force magnitude $F$ . If the separation becomes $2r$ , what is the new force magnitude?
Answer: $\tfrac{1}{4}F$ , because force varies as $\tfrac{1}{r^2}$ . This question examines the impact of distance on gravitational force. The force is proportional to the product of masses, which remains constant here. It decreases with the square of increasing distance, so doubling the separation reduces the force to one-fourth. With masses unchanged and distance doubled, the new force is one-fourth of the original. Distractor C suggests it halves because distance doubles, but this overlooks the squared nature of the inverse relationship. For transferable skill, remember to apply the inverse-square law by squaring the distance ratio when predicting force variations.
Two spheres in deep space have masses $m$ and $m$ separated by $r$ . One sphere's mass is tripled to $3m$ while the separation stays $r$ . How does the gravitational force magnitude change?
Answer: It becomes $3$ times as large. This question tests understanding of how gravitational force depends on mass. The gravitational force is directly proportional to the product of the two masses: F = Gm₁m₂/r². Initially, with both masses equal to m, the force is F₁ = Gm·m/r² = Gm²/r². When one mass is tripled to 3m, the new force becomes F₂ = G(3m)·m/r² = 3Gm²/r². Comparing these expressions, F₂ = 3F₁, so the force becomes 3 times as large. Choice B incorrectly squares the mass change factor, perhaps confusing it with how distance changes affect force. When one mass changes by a factor, the force changes by that same factor.
Two objects interact gravitationally: object $P$ has mass $m$ , object $Q$ has mass $3m$, and their separation is $r$ . In a second situation, the same two objects are separated by $2r$ . How does the gravitational force magnitude in the second situation compare to the first?
Answer: It is $1/4$ as large This question tests understanding of how gravitational force changes with distance between objects. The gravitational force follows the inverse-square law: F = Gm₁m₂/r². When the separation doubles from r to 2r, the force becomes F = Gm₁m₂/(2r)² = Gm₁m₂/4r², which is 1/4 of the original force. Choice A incorrectly suggests force increases with distance, contradicting the fundamental principle that gravitational attraction weakens with separation. To solve distance-change problems, remember that doubling distance always reduces force to one-fourth, regardless of the specific mass values.
Two small spheres of masses $m$ and $m$ are separated by $r$ . If the separation becomes $\tfrac{r}{2}$ , how does the force magnitude change?
Answer: It becomes four times as large. This problem tests understanding of the inverse-square law for gravitational force. The gravitational force follows F = Gm₁m₂/r², where force is inversely proportional to the square of the distance. When the separation changes from r to r/2, the distance becomes half its original value. Since force varies as 1/r², the new force is F' = Gm²/(r/2)² = Gm²/(r²/4) = 4Gm²/r² = 4F. Halving the distance makes the force four times larger because of the inverse-square relationship. Choice B incorrectly suggests the force only doubles, applying a linear rather than inverse-square relationship. To handle distance changes in gravitational problems, remember that halving distance quadruples force, while doubling distance reduces force to one-fourth.

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