AP Physics 1 Quiz: Gravitational Force

What this quiz covers

This quiz focuses on Gravitational Force, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $3F$ (correct answer)
2. $6F$
3. $\tfrac{1}{3}F$
4. $F$, because distance is unchanged.

Explanation: 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.

Question 2

1. It decreases to $\tfrac{1}{4}$ of the original value. (correct answer)
2. It decreases to $\tfrac{1}{2}$ of the original value because only one mass is doubled.
3. It stays the same because the masses do not change.
4. It increases by a factor of 2 because the larger mass is $2m$.

Explanation: 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.

Question 3

1. Case A, because the masses are the same in both cases.
2. Case B, because halving the distance increases force by a factor of 4. (correct answer)
3. They are equal because only mass determines gravitational force.
4. Case A, because gravitational force is independent of distance.

Explanation: 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, which is the same in both cases (m × 2m). It is inversely proportional to the square of the distance, so halving the distance in Case B quadruples the force compared to Case A. Qualitatively, this inverse-square law demonstrates why closer objects experience much stronger gravitational pulls, even with identical mass products. Choice C, a distractor, says they are equal because only mass determines force, but it neglects the distance factor. For such comparisons, compute F for each as proportional to mass product over r squared and directly compare the values.

Question 4

1. $3F$
2. $F/3$
3. $F/9$ (correct answer)
4. $F$ because the masses are unchanged.

Explanation: 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.

Question 5

1. It halves.
2. It doubles.
3. It quadruples. (correct answer)
4. It is unchanged because the masses are unchanged.

Explanation: This question assesses understanding of the gravitational force between two masses as described by Newton's law of universal gravitation. The gravitational force depends directly on the product of the masses, which remains unchanged here since both masses are identical and unaltered. However, the force is inversely proportional to the square of the distance, so halving the distance increases the force by a factor of four due to the inverse-square relationship. This qualitative reasoning shows how gravitational attraction intensifies dramatically as objects get closer, as the force field concentrates over a smaller area. Choice D, a common distractor, claims the force is unchanged because masses are the same, overlooking the crucial role of distance. A useful strategy for these problems is to use the formula F ∝ 1/r² and compute the scaling factor directly from the distance change.

Question 6

1. It becomes $F/4$.
2. It becomes $F/2$.
3. It becomes $2F$. (correct answer)
4. It stays $F$ because the changes cancel.

Explanation: This question assesses understanding of the gravitational force between two masses as described by Newton's law of universal gravitation. The gravitational force depends directly on the product of the masses, so halving one mass halves the product and thus the force contribution from mass. It also follows an inverse-square law with distance, meaning halving the distance quadruples the force due to the denominator shrinking to one-fourth. Qualitatively, these combined effects show how distance reductions can overpower mass decreases in gravitational interactions. Choice D, a distractor, suggests the changes cancel to keep F the same, but actually, the quadrupling from distance outweighs the halving from mass, resulting in a doubling. For similar problems, compute the overall multiplier as (mass ratio) × (1 / (distance ratio)^2) and apply it to the original force.

Question 7

1. The force on $M$ is larger because $M>m$.
2. The force on $m$ is larger because smaller masses accelerate more.
3. The forces have equal magnitude and opposite direction. (correct answer)
4. The forces depend on local $g$, so they can differ at different heights.

Explanation: 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.

Question 8

1. $m_1$ and $m_3$, because $m_3$ is an additional mass.
2. $m_1$ and $m_2$, because their separation is smaller. (correct answer)
3. All pairs, because all masses are equal.
4. No pair, because gravitational forces cancel in a three-mass system.

Explanation: 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.

Question 9

1. It becomes $9$ times larger (correct answer)
2. It becomes $3$ times larger
3. It becomes $1/9$ as large
4. It is unchanged because the masses are unchanged

Explanation: This question tests understanding of the inverse-square relationship for gravitational force with distance. The gravitational force follows F = Gm₁m₂/r², where force is inversely proportional to the square of separation distance. When distance decreases from 3r to r (a factor of 3 decrease), the force increases by a factor of 3² = 9, because the smaller denominator r² compared to (3r)² = 9r² makes the force 9 times larger. Choice D incorrectly assumes distance changes don't affect gravitational force between objects. Remember that halving the distance quadruples the force, while tripling the distance makes the force one-ninth as strong.

Question 10

1. Unchanged
2. Twice as large
3. One-half as large (correct answer)
4. One-fourth as large

Explanation: 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.

Question 11

1. $4F$ because the force increases with distance.
2. $\tfrac{1}{4}F$ because the distance quadrupled.
3. $\tfrac{1}{16}F$. (correct answer)
4. $F$ because the masses did not change.

Explanation: This problem tests understanding of the inverse-square law for gravitational force. The gravitational force follows F = Gm²/r² for identical masses. When distance increases from r to 4r, the new force becomes F' = Gm²/(4r)² = Gm²/16r² = F/16. The distance increased by a factor of 4, so the force decreases by a factor of 4² = 16, making the new force one-sixteenth of the original. Choice B incorrectly applies only the first power of the distance change rather than squaring it. To solve distance-change problems in gravitation, always remember to square the distance ratio: if distance increases by factor n, force decreases by factor n².

Question 12

1. It increases by a factor of 3.
2. It increases by a factor of 6 because both objects pull.
3. It increases by a factor of 9. (correct answer)
4. It does not change because the masses are unchanged.

Explanation: This problem tests understanding of the inverse-square law for gravitational force. The gravitational force follows F = Gm₁m₂/r², meaning force is inversely proportional to the square of the distance. When the separation decreases from 3r to r, the distance becomes one-third of its original value. Since force varies as 1/r², the new force is F' = Gm₁m₂/r² compared to the original F = Gm₁m₂/(3r)² = Gm₁m₂/9r². Therefore, F' = 9F, showing the force increases by a factor of 9. Choice A incorrectly applies a linear relationship instead of the inverse-square relationship. When distance changes in gravitational problems, always square the distance ratio to find the force change factor.

Question 13

1. It is $2$ times as large because the distance doubled
2. It is $1/2$ as large because only one object is heavier
3. It is $1/4$ as large (correct answer)
4. It is unchanged because the same two masses interact

Explanation: 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.

Question 14

1. Three times as large
2. Nine times as large (correct answer)
3. One-ninth as large
4. Unchanged because the masses are unchanged

Explanation: This question tests understanding of the inverse-square law for gravitational force with distance. The gravitational force follows F = Gm₁m₂/r², where force is inversely proportional to the square of separation. When distance decreases from r to r/3, the denominator becomes (r/3)² = r²/9, making the force 9 times larger (since dividing by a smaller denominator yields a larger result). Choice C incorrectly inverts the relationship, suggesting force decreases when objects get closer. Remember that reducing distance by a factor increases force by that factor squared.

Question 15

1. Half as large
2. Twice as large
3. Four times as large (correct answer)
4. Unchanged because $M$ is much larger than $m$

Explanation: 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.

Question 16

1. It doubles, because each sphere’s weight doubles.
2. It stays the same, because $r$ is unchanged.
3. It quadruples, because the product of the masses becomes $4m^2$. (correct answer)
4. It halves, because the masses are larger and pull less strongly per kilogram.

Explanation: This question tests comprehension of gravitational force dependence on mass. Gravitational force increases directly with the product of the masses, so doubling each mass quadruples the product and thus the force. The inverse-square relationship with distance means unchanged distance keeps that factor constant. Replacing both spheres with twice the mass at the same separation results in a force four times larger. Distractor B incorrectly claims the force stays the same because distance is unchanged, ignoring the mass increase's effect on the product. When approaching similar questions, identify unchanged variables like distance and focus on scaling the mass product to predict force changes.

Question 17

1. It becomes three times as large. (correct answer)
2. It becomes $\tfrac{1}{3}$ as large because the heavier object falls less.
3. It stays the same because distance did not change.
4. It becomes nine times as large because both masses are effectively tripled.

Explanation: 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.

Question 18

1. It becomes $\tfrac{1}{2}$ as large.
2. It becomes twice as large.
3. It becomes four times as large. (correct answer)
4. It stays the same because the masses are unchanged.

Explanation: 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.

Question 19

1. It becomes $\tfrac{1}{4}$ as large.
2. It becomes $2$ times as large.
3. It becomes $4$ times as large. (correct answer)
4. It stays the same because the masses did not change.

Explanation: 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.

Question 20

1. It doubles because the smaller craft has less weight.
2. It becomes half as large. (correct answer)
3. It becomes one-fourth as large.
4. It stays the same because distance is unchanged.

Explanation: This question tests understanding of how gravitational force depends on the masses of interacting objects. The gravitational force is proportional to the product of the two masses: F = Gm₁m₂/r². Initially, the force is F₁ = Gm(4m)/r² = 4Gm²/r². When the 4m spacecraft is replaced with a 2m spacecraft, the new force becomes F₂ = Gm(2m)/r² = 2Gm²/r². Comparing these, F₂ = F₁/2, so the force becomes half as large. Choice A incorrectly suggests the force doubles, perhaps confusing the effect of reducing mass with some other concept. To solve mass-change problems, calculate the ratio of the new mass product to the original mass product.