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Chapter 8: Problem 62

Show that the form \(\Delta U=m g \Delta r\) follows from Equation 8.5 when \(r_{1} \simeq r_{2} .\) [Hint: Write \(r_{2}=r_{1}+\Delta r\) and apply the binomial approximation (Appendix A).]

Short Answer

Expert verified
By substituting \(r_{2}=r_{1}+\Delta r\) into Equation 8.5 and applying the binomial approximation, it simplifies to the formula \(\Delta U=m g \Delta r\).

Step by step solution

01

Substitute the hint into the equation

Start by incorporating the hint \(r_{2}=r_{1}+\Delta r\) into Equation 8.5. Considering the definition of \(\Delta r\) as \(r_2 - r_1\), you have what is needed to progress to the next step.
02

Apply the binomial approximation

Now, apply the binomial approximation, which for small \(\Delta r\) is given by \((1+x)^n = 1 + nx\). This step involves using these variables in the context of Equation 8.5.
03

Simplify the equation

Following the application of the binomial approximation, simplify the equation further by cancelling identical terms on both sides. This step should lead us directly to the desired formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Understanding gravitational potential energy is fundamental to physics, as it's the energy an object possesses due to its position in a gravitational field. The Earth, for instance, exerts a gravitational pull on objects, giving them potential energy which is directly proportional to their height above the ground and their mass.

The gravitational potential energy, often denoted as U, can be expressed as: \[ U = mgh \] where m is the mass of the object, g is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\) on the Earth's surface), and h is the height above the reference point. As the object falls, this potential energy is converted into kinetic energy.
Differential Height in Physics
When dealing with gravitational potential energy, the concept of differential height becomes significant. Differential height, denoted as \(\Delta r\) or \(\Delta h\), is essentially the change in height or vertical position of an object. This small change can have considerable effects on the object鈥檚 potential energy.

It's important to note that for small differences in height, the change in potential energy can be approximated as: \[ \Delta U = mg\Delta r \] This implies that the change in potential energy is linearly related to the change in height, under the assumption that the change is small compared to the total height.
Equation 8.5 in Physics
Equation 8.5 typically relates to a fundamental physics formula explaining some aspect of gravitational interaction or energy relationships. In the context of this exercise, we interpret Equation 8.5 as the expression for the change in gravitational potential energy when an object is moved in a gravitational field.

The specific form of Equation 8.5 could be a more general equation representing the gravitational potential energy at different locations, which becomes simplified through approximations such as the binomial approximation when the differentials in the two locations are minimal, that is when \(r_1 \approx r_2\).
Work-Energy Theorem
The work-energy theorem is a principle in physics that states the work done by the net force on an object is equal to the change in its kinetic energy. Expressed as an equation, it looks like this: \[ W = \Delta KE \] where W is the work done, and \(\Delta KE\) is the change in kinetic energy of the object.

When we talk about potential energy, if an object's height changes within a gravitational field, the work done by gravity is related to the change in gravitational potential energy. Therefore, the work-energy theorem is closely related to the change in gravitational potential energy since it hints at the energy conversions happening within the system.

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Most popular questions from this chapter

Exact solutions for gravitational problems involving more than two bodies are notoriously difficult. One solvable problem involves a configuration of three equal-mass objects spaced in an equilateral triangle. Forces due to their mutual gravitation cause the configuration to rotate. Suppose three identical stars, each of mass \(M,\) form a triangle of side \(L .\) Find an expression for the period of their orbital motion.

Tidal forces are proportional to the variation in gravity with position. By differentiating Equation \(8.1,\) estimate the ratio of the tidal forces due to the Sun and the Moon. Compare your answer with the ratio of the gravitational forces that the Sun and Moon exert on Earth. Use data from Appendix E.

Neglecting Earth's rotation, show that the energy needed to launch a satellite of mass \(m\) into circular orbit at altitude \(h\) is $$\left(\frac{G M_{\mathrm{E}} m}{R_{\mathrm{E}}}\right)\left(\frac{R_{\mathrm{E}}+2 h}{2\left(R_{\mathrm{E}}+h\right)}\right).$$

The force of gravity on an object is proportional to the object's mass, yet all objects fall with the same gravitational acceleration. Why?

Given the Moon's orbital radius of \(384,400 \mathrm{km}\) and period of 27.3 days, calculate its acceleration in its circular orbit, and compare with the acceleration of gravity at Earth's surface. Show that the Moon's acceleration is lower by the ratio of the square of Earth's radius to the square of the Moon's orbital radius, thus confirming the inverse-square law for the gravitational force.
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