/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Express the acceleration of grav... [FREE SOLUTION] | 91Ó°ÊÓ

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Express the acceleration of gravity \(g_{h}\) at an altitude \(h\) above the surface of the earth in terms of the acceleration of gravity \(g_{0}\) at the surface of the earth, the altitude \(h,\) and the radius \(R\) of the earth. Determine the percent error if the weight that an object has on the surface of the earth is used as its weight at an altitude of \((a) 0.625 \mathrm{mi},(b) 625 \mathrm{mi} .\)

Short Answer

Expert verified
The percent errors are approximately 0.000708% for 0.625 miles and 6.155% for 625 miles.

Step by step solution

01

Understand the Formula for Gravitational Acceleration

The gravitational acceleration at a distance from the center of the Earth is given by the formula: \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the Earth's center to the object. At sea level, this distance is the Earth's radius \( R \). Therefore, \( g_0 = \frac{GM}{R^2} \).
02

Express Gravity at Altitude

The gravitational acceleration at an altitude \( h \) is given by \( g_h = \frac{GM}{(R+h)^2} \). To express \( g_h \) in terms of \( g_0 \), substitute \( GM = g_0 R^2 \) into the equation for \( g_h \):\[ g_h = \frac{g_0 R^2}{(R+h)^2} \].
03

Calculate Percent Error for Altitude

The percent error when using the Earth's surface gravity \( g_0 \) as an approximation for \( g_h \) at an altitude \( h \) is calculated by \(\text{Percent Error} = \left(1 - \frac{g_h}{g_0}\right) \times 100\%\). Substitute the expression for \( g_h \) from Step 2: \\[ \text{Percent Error} = \left(1 - \frac{R^2}{(R+h)^2}\right) \times 100\% \].
04

Calculate Percent Error for 0.625 Miles

Convert 0.625 miles to feet: \( 0.625 \text{ mi} = 0.625 \times 5280 = 3300 \text{ ft} \). With \( R = 20925854 \text{ ft} \), calculate the percent error:\[ \text{Percent Error} = \left(1 - \frac{20925854^2}{(20925854 + 3300)^2}\right) \times 100\% \approx 0.000708\% \].
05

Calculate Percent Error for 625 Miles

Convert 625 miles to feet: \( 625 \text{ mi} = 625 \times 5280 = 3,300,000 \text{ ft} \). Calculate the percent error:\[ \text{Percent Error} = \left(1 - \frac{20925854^2}{(20925854 + 3300000)^2}\right) \times 100\% \approx 6.155\% \].

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Key Concepts

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

percent error calculation
Understanding percent error is crucial when comparing two values, especially in scientific measurements. Percent error quantifies the difference between an experimental value and a theoretical value. This metric is often represented as a percentage, which makes it easy to understand the magnitude of error relative to the true value.

To calculate percent error, you use the following formula: \[ \text{Percent Error} = \left( \frac{\text{Experimental Value} - \text{Theoretical Value}}{\text{Theoretical Value}} \right) \times 100\% \] In this exercise, we are comparing the gravitational acceleration at an altitude, denoted as \( g_h \), with the gravitational acceleration at the Earth's surface, \( g_0 \).
  • \( g_0 \) is the theoretical value, representing the standard gravitational pull at sea level.
  • \( g_h \) is the experimental value, representing gravitational pull at a given altitude.
Using these variables in the percent error formula, we find that the difference due to altitude can be expressed as:\[ \text{Percent Error} = \left(1 - \frac{g_h}{g_0}\right) \times 100\% \] This comparison helps us understand how negligible or significant the change in gravitational force is when considering altitude.
earth's radius
Earth's radius is a fundamental factor in calculating gravitational forces. The average radius of the Earth is approximately 6,371 kilometers, or 20,925,854 feet as used in the calculations. This distance is crucial because it represents the baseline from which altitudes are measured.

The Earth's radius is integral to determining gravitational acceleration at any distance from its center.
  • At sea level, gravitational acceleration is often simplified in physics as 9.8 m/s², or \( g_0 \).
  • This figure is based on the gravitational constant and Earth's mass, contextualized by the radius \( R \).
When expressing gravitational acceleration at a specific altitude \( h \), you modify the distance used in the formula by adding or subtracting this altitude: \[ g_h = \frac{GM}{(R+h)^2} \] Understanding Earth's radius, therefore, is essential for accurately assessing how different altitudes affect gravitational acceleration.
altitude effect on gravity
Altitude impacts gravitational acceleration due to changes in distance from Earth's center. As an object moves away from the surface, the gravitational pull weakens slightly. This effect is calculated by considering both the Earth's radius \( R \) and the altitude \( h \).
  • At higher altitudes, the gravitational force is less because the distance \( r = R + h \) increases.
  • The inverse square law states that as the distance increases, the force of gravity decreases proportionately with the square of that distance.
The gravitational acceleration at an altitude is given by: \[ g_h = \frac{g_0 R^2}{(R+h)^2} \] This means that for small altitudes like 0.625 miles, the effect on gravity is almost imperceptible, resulting in very low percent error. However, at 625 miles, the effect becomes more notable, leading to a larger percent error. This relationship shows that while changes are minimal for low to moderate altitudes, they cannot be ignored at greater heights above Earth's surface. Understanding this concept is vital for applications in physics and engineering, such as satellite deployment and designing aerospace vehicles.

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Most popular questions from this chapter

In order to test the resistance of a chain to impact, the chain is suspended from a 240-lb rigid beam supported by two columns. A rod attached to the last link is then hit by a 60-lb block dropped from a 5-ft height. Determine the initial impulse exerted on the chain and the energy absorbed by the chain, assuming that the block does not rebound from the rod and that the columns supporting the beam are (a) perfectly rigid, (b) equivalent to two perfectly elastic springs.

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