91Ó°ÊÓ

If \(M\) is the mass of the earth and \(G\) is a constant, the acceleration due to gravity, \(g,\) at a distance \(r\) from the center of the earth is given by $$g=\frac{G M}{r^{2}}$$ (a) Find \(d g / d r\) (b) What is the practical interpretation (in terms of acceleration) of \(d g / d r ?\) Why would you expect it to be negative? (c) You are told that \(M=6 \cdot 10^{24}\) and \(G=6.67 \cdot 10^{-20}\) where \(M\) is in kilograms, \(r\) in kilometers, and \(g\) in \(\mathrm{km} / \mathrm{sec}^{2} .\) What is the value of \(d \mathrm{g} / \mathrm{dr}\) at the surface of the earth \((r=6400 \mathrm{km}) ?\) Include units. (d) What does this tell you about whether or not it is reasonable to assume \(g\) is constant near the surface of the earth?

Step by step solution

01

Differentiate the gravity equation

Given the equation for acceleration due to gravity, \(g = \frac{GM}{r^2}\), we need to find \(\frac{dg}{dr}\). To do this, we'll differentiate the equation with respect to \(r\). Using the power rule and chain rule, we have:\[\frac{dg}{dr} = \frac{d}{dr} \left(\frac{GM}{r^2}\right) = -\frac{2GM}{r^3}\]

02

Interpret the derivative

The derivative \(\frac{dg}{dr}\) represents the rate of change of the gravitational acceleration \(g\) with respect to the distance \(r\) from the earth's center. A negative \(\frac{dg}{dr}\) means that as the distance \(r\) increases, the gravitational acceleration decreases. This aligns with the expectation because gravity weakens with distance.

03

Substitute values into the derivative

We know that \(M = 6 \cdot 10^{24} \, kg\) and \(G = 6.67 \cdot 10^{-20} \, km^3/kg\cdot s^2\). Given \(r = 6400\, km\), substitute these into the expression for \(\frac{dg}{dr}\):\[\frac{dg}{dr} = -\frac{2 \times 6.67 \cdot 10^{-20} \cdot 6 \cdot 10^{24}}{(6400)^3} \]Calculate:\[\frac{dg}{dr} = -\frac{2 \times 4.002 \cdot 10^{5}}{262144000000} \approx - \frac{8.004 \cdot 10^{5}}{262144000000} \approx -3.05 \times 10^{-9} \, km/s^2/km\]

04

Reasonability of \(g\) being constant near Earth's surface

Since \(\frac{dg}{dr}\) at the Earth's surface is very small \((-3.05 \times 10^{-9} \, km/s^2/km)\), it indicates that the change in gravitational acceleration with respect to radial distance is minimal over short distances. Therefore, it is reasonable to assume \(g\) is approximately constant near the Earth's surface.

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Key Concepts

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

Differentiation and Its Role in Calculating Gravitational Change

Differentiation is a mathematical technique used to calculate the rate at which one quantity changes in relation to another. It's a core part of calculus, and in this context, it helps us understand how gravity varies with distance.

When we differentiate an equation like the gravity formula \(g = \frac{GM}{r^2}\), we're looking to find \(\frac{dg}{dr}\). This tells us how quickly gravitational acceleration \(g\) changes as the distance from the Earth's center \(r\) changes.

To derive this expression, we use rules like the power rule and the chain rule, resulting in \(\frac{dg}{dr} = -\frac{2GM}{r^3}\).

• The power rule helps manage exponents during differentiation.
• The chain rule deals with the function's composition.

This derivative is essential for understanding the behavior of gravity over various distances.

Understanding Gravitational Acceleration

Gravitational acceleration \(g\) describes how fast an object speeds up as it falls due to gravity. It depends on both the mass of the Earth \(M\) and the distance \(r\) from its center.

The formula \(g = \frac{GM}{r^2}\) shows this relationship, placing emphasis on the inversely proportional effect of distance. As distance increases, because \(r\) is squared in the denominator, the gravitational force significantly decreases.

• This inverse square law implies gravity diminishes rapidly as you move away from Earth.
• The constancy of \(G\), the gravitational constant, illustrates uniform behavior of gravitational forces universally.

The derivative \(\frac{dg}{dr}\), being negative, clearly proves that gravitational acceleration decreases as you move further from the Earth, reinforcing the intuitive understanding that gravity is weaker at greater distances.

Rate of Change in Gravitational Acceleration

When discussing the rate of change in gravitational acceleration, we refer to how gravitational force varies with increasing distance. This is expressed in terms of \(\frac{dg}{dr}\).

The term \(\frac{dg}{dr} = -\frac{2GM}{r^3}\) specifically tells us how much gravitational acceleration changes per unit increase in distance \(r\). The negative sign indicates a decrease, showing gravity gets weaker as you get further away.

In practical terms, near the Earth's surface, this rate is very small, about \(-3.05 \times 10^{-9} \ km/s^2/km\).

• This small value tells us that over short ranges, such as those experienced on Earth's surface, gravity can be considered nearly constant.
• It underlines why, in daily life, we don't notice gravity changing even if we move around on Earth's surface.

Thus, assuming \(g\) constant near the surface is reasonable. It's a crucial concept in fields like physics and engineering, and in real-world applications like designing buildings or bridges.