Problem 61
Question
Newton's Law of Universal Gravitation states that the magnitude \(F\) of the force exerted by a point with mass \(M\) on a point with mass \(m\) is 2 $$ F=\frac{G m M}{r^{2}} $$ where \(G\) is a constant and \(r\) is the distance between the points. Assuming that the points are moving, find a formula for the instantancous rate of change of \(F\) with respect to \(r\)
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{dF}{dr} = -\frac{2GmM}{r^3} \).
1Step 1: Understand the Problem
We need to find the instantaneous rate of change of the gravitational force \( F \) with respect to the distance \( r \). This means we must find the derivative of \( F \) with respect to \( r \).
2Step 2: Identify the Function to Differentiate
The gravitational force is defined by the formula \[ F = \frac{GmM}{r^2} \]. We need to differentiate this function with respect to \( r \).
3Step 3: Apply the Power Rule for Differentiation
The expression \( \frac{1}{r^2} \) can be rewritten as \( r^{-2} \). The derivative of \( r^{-2} \) with respect to \( r \) is \( -2r^{-3} \).
4Step 4: Differentiate the Function
Using the chain rule, differentiate \( F = GmM \cdot r^{-2} \) with respect to \( r \):\[ \frac{dF}{dr} = GmM \cdot (-2r^{-3}) = -\frac{2GmM}{r^3} \]
5Step 5: Write the Result
The instantaneous rate of change of the force \( F \) with respect to the distance \( r \) is given by the derivative:\[ \frac{dF}{dr} = -\frac{2GmM}{r^3} \]
Key Concepts
Newton's Law of Universal GravitationInstantaneous Rate of ChangePower Rule for Differentiation
Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is one of the key concepts in understanding how objects attract each other due to their masses. The magnitude of the force exerted is dictated by the equation: \[ F = \frac{G m M}{r^{2}} \] where:
- \( F \) represents the gravitational force between two masses.
- \( G \) is the gravitational constant, a value that ensures the force is computed correctly in physical units.
- \( m \) and \( M \) are the masses of the two objects attracting each other.
- \( r \) is the distance between the centers of the two masses.
Instantaneous Rate of Change
The instantaneous rate of change of a function essentially tells us how fast something is changing at a particular point. In simpler terms, it's like a snapshot of the speed of change. Mathematically, this is what derivatives do. For the gravitational force \( F \), to find out how \( F \) changes as the distance \( r \) changes, we take the derivative of \( F \) with respect to \( r \). This gives us the formula for the rate of change: \[ \frac{dF}{dr} \] This derivative represents how sensitive the gravitational force is to changes in distance. In physics, finding instantaneous rates of change helps determine how swiftly quantities like speed, acceleration, and force change over tiny intervals. Unlike average rate of change, which looks over a finite interval, instantaneous rate of change gives an immediate view of change.
Power Rule for Differentiation
The Power Rule is a fundamental technique in calculus that simplifies finding derivatives of functions like polynomials. It states that for any function \( x^n \), the derivative \( \frac{d}{dx}x^n \) is \( nx^{n-1} \). This rule is especially handy for functions where variables are raised to powers. In the context of our gravitational force exercise, the force \( F \) is given by \( \frac{GmM}{r^2} \). Rewriting the denominator \( r^2 \) as \( r^{-2} \) makes it easier to apply the power rule, resulting in: \[ \frac{d}{dr}r^{-2} = -2r^{-3} \] When you multiply by \( GmM \), the constant terms outside of the power function, the derivative becomes: \[ \frac{dF}{dr} = -\frac{2GmM}{r^3} \] Utilizing the Power Rule allows us to efficiently compute these rates of change, and it is a cornerstone for computational problems in mathematics and physics, making complex expressions far more manageable.
Other exercises in this chapter
Problem 60
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