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.

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.

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, so doubling both masses quadruples the product and thus the force. The inverse-square dependence on distance means the force would change if distance did, but here distance remains constant, so only the mass effect applies. Qualitatively, this shows how mass increases amplify attraction linearly per mass, combining multiplicatively for both. Choice C, a distractor, says it stays the same because distance is unchanged, but it ignores the mass doubling. To approach similar questions, determine the net scaling by multiplying individual mass factors and applying the inverse square for any distance change.

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.

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.

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.

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.

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, so tripling one mass while keeping the other constant triples the overall force. The force also depends inversely on the square of the distance, but since the distance remains unchanged here, that factor does not affect the outcome. Qualitatively, this inverse-square law means that gravity weakens rapidly with distance, but mass changes have a direct, linear impact when distance is fixed. A distractor like choice C suggests the force stays the same because one mass is unchanged, ignoring that the force depends on the product of both masses. For transferable strategy, calculate the ratio of new force to old by multiplying the mass scaling factors and dividing by the square of the distance scaling factor.

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 tripling one mass triples the force from that factor alone. It depends inversely on the square of the distance, so doubling the distance reduces the force to one-fourth. Qualitatively, this shows how mass increases and distance changes compete, with the inverse-square effect often dominating unless mass changes are large. Choice D, a distractor, says it becomes F/2 because distance dominates completely, but it overlooks the tripling from mass, resulting in 3/4 F. To solve these, multiply the mass scaling factor by the inverse of the distance scaling factor squared.

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 in Case 2, doubling one mass doubles the product compared to Case 1. It is inversely proportional to the square of the distance, meaning doubling the distance in Case 2 quarters the force. Qualitatively, this inverse-square relationship emphasizes that proximity greatly enhances gravitational pull, often more than mass increases do over larger distances. Choice D, a distractor, claims Case 2 is greater because force ignores distance, which contradicts the inverse-square law. A strategy for comparisons is to calculate force ratios by comparing mass products divided by distance squared for each case.