Question

The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}}\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the magnitude of the instantaneous rate of change of the force increase or decrease with the separation? Explain.

Short answer

b) What is the instantaneous rate of change of the force for two identical objects of mass 0.1 kg and separation 0.01 m? c) Does the magnitude of the instantaneous rate of change increase or decrease with the separation? a) The instantaneous rate of change of the force with respect to distance can be found by differentiating the force formula \(F(x) = -\frac{GMm}{x^2}\) with respect to \(x\), which results in \(\frac{dF}{dx} = 2\frac{-GMm}{x^3}\). b) For two identical objects with mass 0.1 kg and separation 0.01 m, the instantaneous rate of change of the force is approximately -0.134 N/m. c) The magnitude of the instantaneous rate of change decreases as the separation between the objects increases.

Step-by-step solution

Differentiate F(x) with respect to x

\(\displaystyle \frac{dF}{dx} = -\frac{G M m}{x^{2}}\) Differentiate: \(\frac{dF}{dx} = \frac{d}{dx}(-\frac{GMm}{x^2})\) Using the chain rule and the power rule of differentiation, \(\frac{dF}{dx} = 2\frac{-GMm}{x^3}\) **Step 2: Evaluate the derivative for the given mass and separation (part b)** For two identical objects with \(M=m=0.1 \mathrm{kg}\) and separation \(x=0.01 \mathrm{m}\), we will plug these values into the expression we derived for \(\frac{dF}{dx}\):

Plug in the given mass and separation values

\(\frac{dF}{dx} = 2\frac{-GMm}{x^3} = 2\frac{-6.7\times 10^{-11} N m^2 kg^{-2} \cdot 0.1 kg \cdot 0.1 kg}{(0.01m)^3}\) Now we need to calculate the magnitude of this value.

Calculate the magnitude

\(\frac{dF}{dx} = -\frac{2 \cdot 6.7\times 10^{-11} N m^2 kg^{-2} \cdot 0.1 kg \cdot 0.1 kg}{(0.01m)^3} \approx -0.134 N/m\) **Step 3: Determine whether the magnitude of the derivative increases or decreases with separation (part c)** We need to analyze the sign and behavior of the expression we found for the derivative to determine how the force changes as the separation increases.

Analyze the expression for the derivative

The expression for \(\frac{dF}{dx}\) is \(\frac{dF}{dx} = 2\frac{-GMm}{x^3}\). Since \(-GMm\) is always negative (assuming masses and gravitational constant are positive), the expression is positive for all positive values of the distance x. This means that as x increases, the gravitational force F(x) is less negative, meaning that the force becomes less attractive but never repulsive (still negative). Also, since x is cubed in the denominator, the magnitude of the derivative decrease as the separation increases. So, as the separation increases, the instantaneous rate of change in force decreases. To answer part c): The magnitude of the instantaneous rate of change of the force (defined as the derivative of the force with respect to distance) decreases with the increase in separation.

Key concepts

Understanding Instantaneous Rate of Change

When we discuss the instantaneous rate of change, we're looking at how a certain quantity changes at an exact moment or point. It's akin to the speedometer reading in your car which shows your speed at that very second, rather than your average speed over a trip. In calculus, this concept is captured by the derivative.

For the gravitational force, the instantaneous rate of change with respect to distance tells us how fast the force between two objects changes as the distance between them changes. Mathematically, this involves taking the derivative of the force function with respect to the separation distance, which provides us with a new function representing that rate of change at any given point.

Derivative of Gravitational Force

Gravitational force is one of the most fundamental forces in physics, and analyzing its behavior involves calculus. To find the derivative of gravitational force, we differentiate the force function with respect to the distance between the two masses. This derivative essentially represents the rate at which the gravitational force changes due to a change in separation.

As shown in the problem solution, applying the power rule yields a formula involving the negative cube of the distance, illustrating the inverse relationship between force and distance. The derivative informs us how the force would increase or decrease should there be an infinitesimal change in distance—vital for understanding celestial movements and the dynamics of gravitational systems.

Behavior of Force with Separation

The behavior of force with separation is a key concept in understanding how gravitational interactions work. According to Newton's law of universal gravitation, as the distance between two objects increases, the force of attraction between them decreases. This is quantified by the negative square of the distance in the gravitational force formula.

When examining the derivative, we see that the force decreases at a rate that is inversely proportional to the cube of the distance. This implies that as two objects become further apart, the force diminishes rapidly. Calculus provides the precision to predict how the force will change with separation, crucial for designing space missions, understanding orbital patterns, and studying astrophysical phenomena.