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 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 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.

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.

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.

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.