Question

In free fall, we assume the acceleration to be constant. Not only is air resistance ignored, but the gravitational field strength is assumed to be constant. From what height can an object fall to the Earth's surface such that the gravitational field strength changes less than \(1.000 \%\) during the fall?

Short answer

An object can fall from a height less than 31,855 meters (or approximately 31.86 km) for the gravitational field strength to change by less than 1%.

Step-by-step solution

Determine the Change in Gravitational Field Strength

The gravitational field strength decreases with height and is given by the formula \( g = \frac{G M}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the Earth's mass, and \( r \) is the distance from the Earth's center. To find the height \( h \) where the field strength changes by less than \( 1\% \), set \( g_0 \) as the gravitational field strength at the Earth's surface and \( g_h \) as the field strength at height \( h \). The relative change is \( \frac{g_0 - g_h}{g_0} < 0.0100 \).

Set Up the Mathematical Expression

Substitute the expressions for gravitational strength: \( g_h = \frac{GM}{(R+h)^2} \) and \( g_0 = \frac{GM}{R^2} \). Then, express the condition as \( 1 - \frac{R^2}{(R+h)^2} < 0.0100 \), where \( R \) is the Earth's radius.

Simplify and Solve the Inequality

Simplify the inequality \( 1 - \frac{R^2}{(R+h)^2} < 0.0100 \) to \( \frac{(R+h)^2 - R^2}{(R+h)^2} < 0.0100 \). This implies \( \frac{2Rh + h^2}{(R+h)^2} < 0.0100 \).

Approximate and Solve for Height \( h \)

Since \( h \ll R \), we can approximate \((R+h)^2 \approx R^2 + 2Rh\). This allows us to simplify the inequality to \( \frac{2Rh}{R^2} < 0.0100 \). Solving for \( h \), we get \( h < 0.0100 \times \frac{R}{2} = 0.0050R \).

Calculate Numerical Value for \( h \)

Substitute the Earth's average radius \( R \approx 6.371 \times 10^6 \) m into the inequality \( h < 0.0050R \), to get \( h < 0.0050 \times 6.371 \times 10^6 \). Hence, \( h < 31855 \) meters.

Key concepts

Free Fall

Free fall is a fascinating journey where objects are solely under the influence of gravity. This means that the only force acting on the object is the gravitational force, making it an idealized motion with no air resistance. In this free-fall scenario, the acceleration an object experiences due to gravity remains constant and is denoted by the symbol "g". On Earth, this constant acceleration is approximately 9.81 m/s².

Understanding free fall helps in many practical applications, like calculating the time an object takes to hit the ground when dropped from a certain height, assuming no other forces act upon it.

• Free fall implies constant acceleration.
• No air resistance is considered.
• Accurate predictions can be made about the object's motion.

By simplifying real-world conditions, free-fall scenarios allow us to deeply understand the effects of gravity on objects.

Mathematical Expressions

To address problems involving changes in gravitational field strength, mathematical expressions play a crucial role. Here, we use the principle that gravity decreases with height, which is mathematically represented as:

\[ g = \frac{G M}{r^2} \]where:

• \( G \) represents the gravitational constant
• \( M \) is the Earth's mass
• \( r \) represents the distance from Earth's center

These expressions allow us to calculate and articulate exactly how gravity behaves as an object changes position relative to Earth.

Mathematical expressions like these not only provide clarity but also form the basis for solving the inequality that defines conditions for negligible change in gravitational strength. They are the backbone for forming logical arguments and deriving solutions in physics.

Gravitational Constant

The gravitational constant, denoted by \( G \), is a key component in the law of universal gravitation. It is a constant value that interacts with masses to produce gravitational force. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \) Nm²/kg².

This constant plays a vital role in calculating gravitational forces not just on Earth but also throughout the universe. In the context of our exercise, it is utilized in the formula for calculating gravitational field strength at different heights.

• The gravitational constant is universal and constant.
• Its value is crucial in understanding gravitational interactions between objects.
• Helps calculate how gravitational influences change with varying distances.

Despite its small value, \( G \) significantly impacts the forces between massive objects, highlighting its importance in gravitational field calculations.

Inequality Approximation

In many physics problems, especially those involving small changes, approximations help simplify complex calculations. Our exercise employs an inequality approximation to determine the condition under which the change in gravitational strength is less than 1%.

To simplify, we start with the inequality \[ \frac{(R+h)^2 - R^2}{(R+h)^2} < 0.0100 \]Considering \( h \ll R \), we approximate \((R+h)^2 \approx R^2 + 2Rh\), enabling us to reduce complexity. This leads to the inequality:\[ \frac{2Rh}{R^2} < 0.0100 \]Such approximations allow clearer understanding and more manageable calculations. By utilizing this approach, we're able to present a solution that effectively estimates the maximum permissible height, ensuring practical and accessible physics problem-solving.

• Approximations aid in simplifying complex inequalities.
• They highlight manageable terms in mathematical expressions.
• Offer an efficient way to arrive at solutions when absolute precision isn't required.

Approximations ensure calculations are within acceptable limits without losing conceptual accuracy, crucial for quick evaluations.