Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 1, Problem 31P

At 45° latitude: the gravitational acceleration as a function of elevation z above sea level is given by g = a b z , where a = 9.807 m / s 2 and b = 3.32 × 10 6 s 2 . Determine the height above sea level where the weight of an object will decrease by 1 percent.

Expert Solution & Answer
Check Mark
To determine

(a)

The height above the sea level where decrease in the object weight is 1%.

Answer to Problem 31P

The height above the sea level where decrease in the object weight is 1% is 29518m.

Explanation of Solution

Given Information:

The gravitation constant function is abz, constant a is 9.807m/s2 and the constant b is 3.32×106s-2.

Write the expression for the weight of the object above sea level.

  Wabovesealevel=(percentagedecrease)Wsealevel...... (I)

Since there is a 1% decrease in the weight. Substitute 0.99 for percentage decrease in Equation (1)

  Wabovesealevel=(0.99)Wsealevel...... (II)

Write the expression for the gravitational acceleration as a function of elevation.

  g=abz....... (III)

Here the constant are a and b, elevation above sea level is z.

Substitute 9.807m/s2 for a and 3.32×106s-2 for b in Equation (III)

  g=(9.807m/s2)(3.32×106s-2)z.... (IV)

Since at the sea level the elevation is 0, substitute 0 for z in Equation (IV)

  g=ab(0)g=a...... (V)

Write the expression for the weight of the object.

  W=mg....... (VI)

Here mass of the object is m and the acceleration due to gravity is g.

Substitute mg for W in Equation (II)

  (mg)abovesealevel=(0.99)(mg)sealevel(g)abovesealevel=(0.99)(g)sealevel...... (VII).

Calculation:

Substitute 9.807m/s2 for a in Equation (V)

  g=9.807m/s2

Substitute 9.807m/s2 for (g)sealevel and (9.807m/s2)(3.32×106s-2)z for (g)abovesealevel in Equation (VII).

  (9.807m/ s 2)(3.32× 10 6s -2)z=(0.99)9.807m/s2(3.32× 10 6s -2)z=(0.99)9.807m/s2(9.807m/ s 2)(3.32× 10 6s -2)z=0.098m/s2z=29518m

Conclusion:

The height above the sea level where decrease in the object weight is 1% is 29518m.

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Fluid Mechanics: Fundamentals and Applications

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