Questions: The weight of bodies may change somewhat from one location to another as a result of the variation of the gravitational acceleration g with elevation. Accounting for this variation using g = a - bz, where a = 9.807 m / s^2 and b = 3.32 x 10^-6 s^-2, determine the weight of a 90-kg person at sea level (z=0), in Denver (z=1610 m), and on the top of Mount Everest (z=8848 m).

Solution Steps

The problem involves calculating the weight of a person at different elevations using the formula for gravitational acceleration that varies with height: \( g = a - bz \). The given values are:

• \( a = 9.807 \, \text{m/s}^2 \)
• \( b = 3.32 \times 10^{-6} \, \text{s}^{-2} \)
• Mass of the person, \( m = 90 \, \text{kg} \)

We need to calculate the weight at three different elevations: sea level (\( z = 0 \)), Denver (\( z = 1610 \, \text{m} \)), and Mount Everest (\( z = 8848 \, \text{m} \)).

At sea level, \( z = 0 \). The gravitational acceleration is: \[ g = a - b \cdot 0 = 9.807 \, \text{m/s}^2 \] The weight \( W \) is given by: \[ W = m \cdot g = 90 \, \text{kg} \times 9.807 \, \text{m/s}^2 = 882.63 \, \text{N} \]

In Denver, \( z = 1610 \, \text{m} \). The gravitational acceleration is: \[ g = a - b \cdot z = 9.807 - 3.32 \times 10^{-6} \times 1610 \] \[ g = 9.807 - 0.0053452 = 9.8017 \, \text{m/s}^2 \] The weight is: \[ W = m \cdot g = 90 \, \text{kg} \times 9.8017 \, \text{m/s}^2 = 882.153 \, \text{N} \]

On Mount Everest, \( z = 8848 \, \text{m} \). The gravitational acceleration is: \[ g = a - b \cdot z = 9.807 - 3.32 \times 10^{-6} \times 8848 \] \[ g = 9.807 - 0.02938496 = 9.7776 \, \text{m/s}^2 \] The weight is: \[ W = m \cdot g = 90 \, \text{kg} \times 9.7776 \, \text{m/s}^2 = 879.984 \, \text{N} \]

Final Answer