Problem 57 - University Physics with Modern Physics

Question

There are two equations from which a change in the gravitational potential energy \(U\) of the system of a mass \(m\) and the earth can be calculated. One is \(U=m g y(\) Eq. 7.2\() .\) The other is \(U=-G m_{\mathrm{E}} m / r(\mathrm{Eq} .13 .9) .\) As shown in Section \(13.3,\) the first equation is correct only if the gravitational force is a constant over the change in height \(\Delta y .\) The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in \(U\) between a mass at the earth's surface and a distance \(h\) above it using both equations, and find the value of \(h\) for which Eq. \((7.2)\) is in error by 1\(\% .\) Express this value of \(h\) as a fraction of the earth's radius, and also obtain a numerical value for it.

Step-by-Step Solution

Answer

The error is 1% when \( h = 0.01r \), approximately 63.71 km.

Step 1: Understand the two equations

The equations provided are for gravitational potential energy. The first, \( U = mgy \), is valid near the Earth's surface where gravity is approximately constant. The second, \( U = -\frac{G m_E m}{r} \), is universally applicable as it considers the distance from the Earth's center.

Step 2: Define U change equations

Calculate the change in potential energy \( \Delta U \) when moving from the Earth's surface to a height \( h \) above using both equations. \( \Delta U_1 = mg(h) \) (Eq. 7.2) and \( \Delta U_2 = -\frac{G m_E m}{r+h} + \frac{G m_E m}{r} \) (Eq. 13.9).

Step 3: Equate the percentage error condition

For Eq. (7.2) to be in 1% error compared to Eq. (13.9), set the condition: \( |\Delta U_1 - \Delta U_2| = 0.01 \times \Delta U_2 \). This translates to solving \( |mg(h) - (-\frac{G m_E m}{r+h} + \frac{G m_E m}{r})| = 0.01(-\frac{G m_E m}{r+h} + \frac{G m_E m}{r}) \).

Step 4: Simplify equations

Simplify the expressions: \( mg(h) = \frac{G m_E m (h)}{r^2} \) for small \( h \) using the approximation \( h << r \). Substitute back to get \( \frac{h}{r} \approx \frac{0.01}{1 - \frac{1}{1+h/r}} \).

Step 5: Solve for fraction h/r

For small \( h \), \( 1 + \frac{h}{r} \approx 1 \), we have \( \frac{h}{r} = 0.01 \). Simplify further to \( h = 0.01 \times r \).

Step 6: Calculate h numerically

Given the radius of the Earth \( r \approx 6,371,000 \text{ meters} \), calculate the numerical value of \( h \), so \( h = 0.01 \times 6,371,000 = 63,710 \text{ meters} = 63.71 \text{ kilometers} \).

Key Concepts

Universal Law of Gravitation

The Universal Law of Gravitation is a fundamental principle developed by Isaac Newton. It describes how every mass attracts every other mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This can be expressed mathematically as:\[F = \frac{G \, m_1 \, m_2}{r^2}\]where:

• \( F \) is the gravitational force between the masses,
• \( G \) 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 objects,
• \( r \) is the distance between the centers of the two masses.

Newton's law applies everywhere in the universe and is particularly useful for understanding the gravitational interactions between astronomical bodies like planets and stars. The law helps calculate not just the force, but also provides the basis for understanding gravitational potential energy when objects are moved within a gravitational field.

Gravitational Force Approximation

Close to the Earth's surface, gravity can often be approximated as a constant force, a simplification that allows easier calculations. This is because the gravitational force changes very little over small height differences relative to the radius of the Earth.

For calculations near the surface, we use the equation:\[F = mg\]where:

• \( F \) is the force due to gravity,
• \( m \) is the mass of the object,
• \( g \) is the gravitational acceleration, approximately \( 9.81 \, \text{m/s}^2 \).

This simplification assumes that gravity acts uniformly. However, as objects move significantly away from Earth's surface, the variation in gravitational force becomes more pronounced. In such cases, the Universal Law of Gravitation provides more accurate results, as it considers the change in distance from the Earth's center.

Error Analysis in Physics

Error analysis is crucial in physics to quantify the degree of uncertainty and reliability of a measurement or calculation. In the context of gravitational potential energy, error analysis helps understand how the approximations used in calculations affect the results.

When we compare the potential energy calculated using the gravitational force approximation against the more precise universal calculations, it is necessary to analyze the errors involved. These errors arise because the approximation assumes a constant gravitational force, valid only for small heights relative to Earth's radius.

In our exercise, we identified a 1% error margin that indicates the threshold at which the simple equation becomes inaccurate. Performing such error analysis involves setting up conditions like:\[|\Delta U_1 - \Delta U_2| = 0.01 \times \Delta U_2\]to determine at what heights these errors become significant, aiding in better understanding of when and how different gravitational equations should be applied.

Earth's Radius and Gravitational Effects

The radius of the Earth plays a significant role in gravitational calculations. It is approximately 6,371 kilometers. This large value means that over small distances near Earth's surface, the gravitational force remains nearly constant, justifying the use of simplified calculations.

However, when calculating potential energy or force beyond these distances, the curvature of the Earth and the increase in distance from its center cannot be ignored. This necessitates using the universal gravitational formula which accounts for the varying effects.

For our exercise, finding the height where the approximation rule is off by 1% involved calculating the fraction of Earth's radius. It turns out to be \( h = 0.01 \times r \), leading to an approximate height of 63.71 km. This calculation demonstrates how significant Earth's size is in maintaining the consistency of gravitational effects over short distances.