Problem 100 - Physics for Scientists and Engineers with Modern Physics

Question

Suppose the gravitational potential energy of an object of mass \(m\) at a distance \(r\) from the center of the Earth is given by $$ U(r)=-\frac{G M m}{r} e^{-\alpha r} $$ where \(\alpha\) is a positive constant and \(e\) is the exponential function. (Newton's law of universal gravitation has \(\alpha=0\) ). \((a)\) What would be the force on the object as a function of \(r ?(b)\) What would be the object's escape velocity in terms of the Earth's radius \(R_{\mathrm{E}} ?\)

Step-by-Step Solution

Answer

Force: \( F(r) = \left( \frac{G M m}{r^2} + \alpha \frac{G M m}{r} \right) e^{-\alpha r} \); Escape velocity: \( v_{esc} = \sqrt{\frac{2G M}{R_{\mathrm{E}}} e^{-\alpha R_{\mathrm{E}}}} \).

Step 1: Find Expression for Force (Part a)

To find the force, we need to take the negative derivative of the potential energy function with respect to distance \(r\). Given the potential energy \(U(r) = -\frac{G M m}{r} e^{-\alpha r}\), compute the derivative:\[ \frac{dU}{dr} = \frac{d}{dr} \left(-\frac{G M m}{r} e^{-\alpha r}\right) \]Using the product rule for differentiation:\[ \frac{d}{dr} \left(\frac{-G M m}{r} e^{-\alpha r}\right) = \frac{-G M m}{r} \frac{d}{dr}(e^{-\alpha r}) + e^{-\alpha r} \frac{d}{dr}\left(-\frac{G M m}{r}\right) \]Where \(\frac{d}{dr} (e^{-\alpha r}) = -\alpha e^{-\alpha r}\) and \(\frac{d}{dr}\left(\frac{-G M m}{r}\right) = \frac{G M m}{r^2}\).Substitute these into the equation, the force \(F(r)\) becomes:\[ F(r) = -\left( \frac{G M m}{r^2} e^{-\alpha r} + \alpha \frac{G M m}{r} e^{-\alpha r} \right) \]\[ F(r) = \left( \frac{G M m}{r^2} + \alpha \frac{G M m}{r} \right) e^{-\alpha r} \]

Step 2: Calculate Escape Velocity (Part b)

Escape velocity is the minimum velocity required for an object to overcome the gravitational potential without returning.The escape velocity \(v_{esc}\) is given by the energy conservation principle, where kinetic energy equals potential energy at large distances (infinitely far from Earth):\[ \frac{1}{2}mv_{esc}^2 = G M m \left(\frac{1}{R_{\mathrm{E}}} e^{-\alpha R_{\mathrm{E}}} - 0 \right) \]Simplify and solve for \(v_{esc}\):\[ \frac{1}{2}mv_{esc}^2 = \frac{G M m}{R_{\mathrm{E}}} e^{-\alpha R_{\mathrm{E}}} \]\[ v_{esc}^2 = \frac{2G M}{R_{\mathrm{E}}} e^{-\alpha R_{\mathrm{E}}} \]\[ v_{esc} = \sqrt{\frac{2G M}{R_{\mathrm{E}}} e^{-\alpha R_{\mathrm{E}}}} \]

Key Concepts

Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation is a fundamental principle of physics describing the force of attraction between two masses. According to this law, every particle of matter in the universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The mathematical formulation of this law is given by: \[ F = \frac{G \, M_1 \, M_2}{r^2} \]where:

• \(F\) is the gravitational force between the two objects
• \(G\) is the gravitational constant \(6.674 \, \times \, 10^{-11} \, \text{N m}^2/\text{kg}^2\)
• \(M_1\) and \(M_2\) are the masses of the two objects
• \(r\) is the distance between the centers of the two masses

In relation to the exercise, when \(\alpha = 0\), the potential energy follows the standard gravitational form. Understanding this law is crucial as it forms the basis for calculating forces, orbits, and trajectories in physics, particularly in astrophysics and planetary sciences.

Escape Velocity

Escape velocity is a critical concept in physics that refers to the minimum speed an object must reach to break free from a celestial body's gravitational pull without any additional propulsion.To compute escape velocity, consider the balance between kinetic and gravitational potential energy. The kinetic energy necessary for escape must equal the gravitational potential energy of the object at a given distance, ensuring it can overcome gravitational attraction and reach infinity theoretically.The formula for escape velocity is:\[ v_{esc} = \sqrt{\frac{2G M}{R}} \]where

• \(v_{esc}\) is the escape velocity
• \(G\) is the gravitational constant
• \(M\) is the mass of the celestial body from which the object is escaping
• \(R\) is the radius from the center of the body to the point of escape

In the provided exercise, the presence of \(e^{-\alpha R_{E}}\) suggests a modification to the classical equation. This modification accounts for additional factors impacting escape velocity, possibly due to atmospheric or field variations. This is noteworthy as such adaptations can influence how we calculate and simulate space missions or trajectory designs.

Differentiation in Physics

Differentiation is a powerful mathematical tool used in physics to find rates at which quantities change. It plays a critical role in formulating physical laws and solving physics problems efficiently.In the context of the given exercise, differentiation helps us derive the expression for force from potential energy. The force exerted by a potential energy field can be derived using the negative gradient (or derivative) of the potential energy. Mathematically, this is expressed as:\[ F(r) = -\frac{dU}{dr} \]This formula is particularly significant because it links a spatially varying potential energy field to a force exerted on objects within the field. In the exercise, the differentiation involves applying the product rule due to the presence of two functions (namely, \(\frac{1}{r}\) and \(e^{-\alpha r}\)) multiplied together:

• The derivative of the exponential function \(e^{-\alpha r}\) is \(-\alpha e^{-\alpha r}\)
• The derivative of \(-\frac{G M m}{r}\) is \(\frac{G M m}{r^2}\)

Combining these using the product rule reveals how the force varies with distance. Understanding differentiation enhances comprehension of not just spatial and temporal changes but also predictions on how an object behaves under specific physical conditions.