Problem 58 - College Physics

Question

A man has a mass of \(75 \mathrm{~kg}\) on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" \(10 \%\) of his body weight?

Step-by-Step Solution

Answer

The man must be 305 km above Earth's surface.

Step 1: Understanding the Problem

We need to find the distance above Earth's surface where the man experiences a 10% reduction in his weight. The man's mass is 75 kg, and we need to determine how his weight changes with distance from the Earth.

Step 2: Weight Reduction Calculation

The initial weight of the man on Earth's surface is calculated as \( W = mg = 75 \times 9.8 = 735 \text{ N} \). A 10% reduction means his new weight is \( W' = 0.9 \times 735 = 661.5 \text{ N} \).

Step 3: Using Gravitational Formula

The weight of an object due to gravity is given by \( W = \frac{GMm}{r^2} \). On Earth's surface, this simplifies to \( mg \). We need to find the new distance \( r \) when \( W' = 661.5 \text{ N} \).

Step 4: Setting up the Equation

Let's use the weight formula: \[ 661.5 = \frac{G \cdot M \cdot m}{(R+h)^2} \], where \( G \cdot M \cdot m / R^2 = 735 \). Solve for \( R + h \), and hence \( h \).

Step 5: Solving for Distance

We calculate \( 735 / 661.5 = \frac{R^2}{(R+h)^2} \). Taking square roots, we get \( \sqrt{735 / 661.5} = R / (R+h) \). Solve the equation \( R / (R + h) = \sqrt{1.1111} = 1.054 \). So \( R + h = R / 0.954 \).

Step 6: Calculating Height Above Earth

Substitute \( R = 6371 \text{ km} \), Earth's radius, into the formula to find \( h \). \( R + h = 6675.713 \text{ km} \). Thus, \( h = 6675.713 - 6371 = 304.713 \text{ km} \).

Step 7: Final Result

The man would have to be approximately 305 km above the Earth's surface to "lose" 10% of his weight.

Key Concepts

Gravitational Force

Gravitational force is a fundamental interaction that draws objects toward each other. The formula governing this force between two masses, like Earth and a man, is expressed as\[ F = \frac{G \cdot M \cdot m}{r^2} \]where:

• \(F\) is the gravitational force.
• \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \text{ N(m/kg)}^2\).
• \(M\) is the mass of the larger object, such as Earth.
• \(m\) is the mass of the smaller object, like a man.
• \(r\) is the distance between the centers of the two masses.

As the distance \(r\) from Earth's center increases, the gravitational force decreases. This force is what causes objects to have weight. On Earth's surface, the weight is determined by the formula \(W = mg\), where \(g\), approximately \(9.8 \text{ m/s}^2\), replaces \(\frac{GM}{R^2}\), with \(R\) being Earth's radius.

Weight Reduction Calculation

When discussing weight, it is essential to consider the effect of gravitational force on the object's mass. To calculate a 10% reduction in weight, start with the initial weight,\[ W = mg = 75 \times 9.8 = 735 \text{ N} \]Then calculate the reduced weight:\[ W' = 0.9 \times 735 = 661.5 \text{ N} \]This reduced force tells us how much the gravitational force must decrease for the individual to "lose" 10% of his weight. We must change the distance from the Earth's surface to achieve this reduction. This calculation establishes a comparative weight that we achieve by altering the gravitational interaction.

Distance from Earth's Surface

Distance from Earth's surface significantly affects gravitational force and, therefore, weight. By changing the distance \(r\) in the gravitational force formula, the weight experienced by the person changes. In our example:1. Start with \( W' = \frac{G \cdot M \cdot m}{(R + h)^2} \).2. With a known reduction from the surface weight (735 N to 661.5N), equate: \[ \frac{735}{661.5} = \frac{R^2}{(R+h)^2} \]3. Solve for \( R + h \) using \[ \sqrt{1.1111} = \frac{R}{R+h} \]4. Rearrange to find \( h \) \[ R + h = \frac{R}{0.954} \]5. Use Earth's radius \(R = 6371 \text{ km}\) to calculate \( h \), resulting in \[ h = 6675.713 - 6371 = 304.713 \text{ km} \]Thus, approximately 305 km above Earth's surface is where the man's weight would reduce by 10%. This demonstrates the influence of the distance from Earth on gravitational force.