Scientists use the following formula to calculate the force of gravity that two objects exert on each other: In the equation $$ F=\frac{G \times M \times m}{r^2} $$ \(F\) is the force of gravity; \(G\) is a constant; \(M\) is the mass of one of the objects; \(m\) is the mass of the second object; \(r\) is the distance between the centers of the objects. If an object with a given mass \(m\) is replaced by an object of half its mass, which of the following will increase the force of gravity? A. increasing mass \(M\) and doubling the distance between the objects B. reducing mass \(M\) and doubling the distance between the objects C. reducing mass \(M\) and halving the distance between the objects D. increasing mass \(M\) and halving the distance between the objects

The formula to calculate the force of gravity between two objects is: \(F = \frac{G \times M \times m}{r^2}\). We are given that mass \(m\) is replaced by an object of half its mass, therefore the new mass is \(\frac{1}{2}m\).

We will now analyze each scenario considering the changes in mass and distance: A. increasing mass \(M\) and doubling the distance between the objects: The new force would be \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(2r)^2}\), where \(M'\) is the increased mass. B. reducing mass \(M\) and doubling the distance between the objects: The new force would be \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(2r)^2}\), where \(M'\) is the reduced mass. C. reducing mass \(M\) and halving the distance between the objects: The new force would be \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(\frac{1}{2}r)^2}\), where \(M'\) is the reduced mass. D. increasing mass \(M\) and halving the distance between the objects: The new force would be \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(\frac{1}{2}r)^2}\), where \(M'\) is the increased mass.

We will now compare the new forces in each scenario to the original force to find out which one increases the force of gravity: A. \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(2r)^2} = \frac{M'}{4M}F\). If \(M'\) is increased, then \(\frac{M'}{4M} < 1\), so the force of gravity will decrease. B. \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(2r)^2} = \frac{M'}{4M}F\). If \(M'\) is reduced, then \(\frac{M'}{4M} < 1\), so the force of gravity will decrease. C. \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(\frac{1}{2}r)^2} = 4\frac{M'}{M}F\). If \(M'\) is reduced, then \(\frac{M'}{M} < 1\), so the force of gravity will increase. D. \(F' = \frac{G \times M' \times (\frac{1}{2}m)}{(\frac{1}{2}r)^2} = 4\frac{M'}{M}F\). If \(M'\) is increased, then \(\frac{M'}{M} > 1\), so the force of gravity will increase. The force of gravity will increase in options C and D. However, since C and D are not separate answer choices, we must eliminate one: In option C, reducing mass \(M\) is counteracted by the increased gravitational force due to halving the distance between the objects. However, since \(m\) is also reduced to half its original value, this may potentially lessen the overall increase in gravitational force compared to option D, where mass \(M\) is increased. Therefore, option D is the most plausible answer: #answer#D. increasing mass \(M\) and halving the distance between the objects.