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When exploring the universe or even something as simple as lifting an object from the ground, we continuously interact with gravitational forces. Newton's Law of Gravitation provides us with a fundamental equation to describe how two masses attract each other. This law is elegantly simple yet profound: the gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is calculated as:\[F = \frac{Gm_1m_2}{r^2}\]where:

• \( G \) is the gravitational constant, approximately equal to \( 6.674 \times 10^{-11} \text{Nm}^2/ ext{kg}^2 \), which quantifies the strength of gravity.
• \( m_1 \) and \( m_2 \) are the masses involved.
• \( r \) is the distance between the centers of the two masses.

This equation tells us that the gravitational force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them. This is why the force becomes significantly weaker as distance increases.

The problem we usually face in physics is how to optimize certain quantities under given conditions. Maximizing force often involves placing objects in positions that enhance the gravitational pull between them. This exercise specifically asks us to concentrate on a setup involving a mass that can be subdivided. For this scenario, envision splitting a mass \( M \) into two parts: a smaller mass \( m \) and the remaining mass \( M-m \). The aim is to maximize the gravitational force between these two mass parts when they are separated by a fixed distance. The question becomes where to cut or what size \( m \) should be to achieve the maximum force.This scenario hinges on understanding that as you adjust the sizes of \( m \) and \( M - m \), you're effectively playing with the components that define gravitational force in the equation \( F = \frac{Gm(M-m)}{(3.0 \times 10^{-3})^2} \). The task is to mathematically optimize this setup by considering the variables and constants in play.

To solve problems like maximizing the gravitational force, differentiation is a key tool in physics. Differentiation allows us to find how changing one quantity affects another. In this particular problem, the task is to differentiate the force function \( F = Gm(M-m) \) with respect to the variable \( m \).Here's a simple breakdown:

• You start with a function expressing the force: \( Gm(M-m) \).
• Then, differentiate this function with respect to \( m \), which results in the derivative \( \frac{d}{dm}[Gm(M-m)] = G(M - 2m) \).
• The next step is to set the derivative equal to zero to find the critical points: \( G(M - 2m) = 0 \).

Through this process, we determine where the force is at its greatest or smallest—or in some cases, where it may not change at all. Finding that \( M = 2m \) provides a critical point, reliant on being able to successfully carry out this differentiation properly.

In calculus, critical points help us identify where a function reaches its maximum, minimum, or saddle point. For our physics problem, solving the derivative equation \( M - 2m = 0 \) gave us a critical point at \( m = \frac{M}{2} \).To decide if this point is a maximum or minimum, it's essential to evaluate the second derivative, as it tells us about the concavity of the function.

• A second derivative \( \frac{d^2}{dm^2}[Gm(M-m)] = -2G \) that is negative implies the function is concave down at \( m = \frac{M}{2} \).
• Therefore, given the negative value of the second derivative, the critical point is a maximum point.

Thus, the exercise confirms that when \( m = \frac{M}{2} \), the gravitational force is maximized. Understanding critical points through derivatives lets us not just solve mathematical issues, but help visualize and optimize real-world problems.