Question

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.980 m/s\(^2\) if the acceleration due to gravity at the surface has magnitude 9.80 m/s\(^2\)?

Short answer

The distance \( h \) is approximately equal to 637 km above the Earth's surface.

Step-by-step solution

Use the formula for gravitational acceleration

The formula used to calculate the acceleration due to gravity at a certain distance from the center of the Earth is given by \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.

Relate the two accelerations

Since we know both the acceleration at the surface \( g_0 = 9.80 \text{ m/s}^2 \) and the acceleration at the distance \( g = 0.980 \text{ m/s}^2 \), we can write \( \frac{g}{g_0} = \left(\frac{R}{R + h}\right)^2 \), where \( R \) is the Earth's radius and \( h \) is the height above the surface.

Solve for \( h \)

Rearrange the equation to solve for \( \frac{R}{R + h} \) and then \( h \):1. \( \frac{0.980}{9.80} = \left(\frac{R}{R + h}\right)^2 \)2. \( \frac{1}{10} = \left(\frac{R}{R + h}\right)^2 \)3. \( \sqrt{\frac{1}{10}} = \frac{R}{R + h} \)4. \( \frac{1}{\sqrt{10}} = \frac{R}{R + h} \)5. Solve for \( h \), rearrange to find \( h = \frac{R}{\sqrt{10} - 1} - R \)

Key concepts

Gravitational Constant

The gravitational constant, represented by the symbol \( G \), is a key factor in the calculation of gravitational forces between two masses. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2 \). This constant is universal, meaning it does not change regardless of the masses involved or where in the universe these masses are located.

Understanding \( G \) is essential because it helps us calculate the gravitational force exerted by two bodies, such as a planet and an object. In the equation \( F = \frac{GMm}{r^2} \), \( G \) works as a proportionality factor. It ensures that the force is directly proportional to the product of the masses \( M \) and \( m \), and inversely proportional to the square of the distance \( r \) between their centers.

The gravitational constant helps maintain consistency in gravitational calculations, ensuring concepts like weight and orbit are predictable and stable. In problems like the one given, \( G \) remains constant, allowing us to solve for other variables such as distance without changing its value.

Mass of the Earth

The Earth's mass is a fundamental component in calculating the gravitational acceleration at any given point near or on its surface. This mass is symbolized by \( M \) in formulas like \( g = \frac{GM}{r^2} \). The approximate mass of Earth is \( 5.972 \times 10^{24} \, \text{kg} \).

The Earth's mass is crucial in determining the gravitational pull it has on objects, providing the base for gravity-related calculations. When considering the exercise problem, Earth's mass remains constant and is a vital determinant of the gravitational acceleration, affecting everything from the fall of an apple to the orbiting of satellites.

Because \( g \) is directly proportional to Earth's mass in the formula, understanding changes or variations in \( g \) at different distances can help one grasp how gravity diminishes with distance from the surface. Although the mass itself doesn't change, the effect of its gravity does.

Earth's Radius

Earth's radius, denoted as \( R \), is essential when determining the distance at which gravitational effects take place. The average radius of Earth is about \( 6,371 \, \text{km} \).

This measurement is critical when studying gravitational forces because it provides a baseline for calculating how gravity decreases with distance. In the formula \( g = \frac{GM}{r^2} \), \( r \) is the distance from the Earth's center, meaning \( R + h \) is used when objects are above the surface.

In our problem, the radius is used to relate surface gravity to gravity at a certain height. It acts as a baseline distance from the center, helping to calculate \( h \), the height above Earth's surface where a reduced gravity force is felt. This understanding is key to solving for \( h \) in various gravity-related queries.

Distance from the Center of the Earth

The distance from the center of the Earth, denoted as \( r \), plays a pivotal role in calculating gravitational force and acceleration. In gravitational equations, \( r \) is the complete radial distance from the Earth's core to the point of interest.

In practice, \( r \) is often expressed as the sum of the Earth's radius \( R \) and an additional height \( h \), where \( r = R + h \). This is particularly useful for determining gravitational force above the earth's surface. The formula \( g = \frac{GM}{r^2} \) illustrates how gravitational acceleration is inversely proportional to the square of \( r \).

In our exercise, understanding \( r \) is crucial to finding out how high an object must be from the Earth's surface to experience a specific gravitational pull, like 0.980 m/s\(^2\). Solving for \( h \) involves working backwards from the known changes in \( g \) to determine how \( r \) must change, illustrating the principle that gravity weakens with distance.