Question
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 the relation in Prob. $1-8,$ determine the weight of an 80 -kg person at sea level $(z=0),$ in Denver $(z=1610 \mathrm{m})$ and on the top of Mount Everest $(z=8848 \mathrm{m})$
Step 1
However, the gravitational acceleration varies with elevation, so we need to account for this variation. The relation given in Prob. 1-8 is $g = 9.807 - 3.32 \times 10^{-6}z$, where $z$ is the elevation in meters. Show more…
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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 the relation $g=a-b z$ where $a=9.807 \mathrm{~m} / \mathrm{s}^{2}$ and $b=3.32 \times 10^{-6} \mathrm{~s}^{-2},$ determine the weight of an $80-\mathrm{kg}$ person at sea level $(z=0),$ in Denver $(z=1610 \mathrm{~m}),$ and on the top of Mount Everest $(z=8848 \mathrm{~m})$
If the variation of the acceleration of gravity, in $\mathrm{m} / \mathrm{s}^{2}$, with elevation $z$, in $\mathrm{m}$, above sea level is $g=9.81-\left(3.3 \times 10^{-6}\right) z$, determine the percent change in weight of an airliner landing from a cruising altitude of $10 \mathrm{~km}$ on a runway at sea level.
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Problems: Developing Engineering Skills
Weight $W$ is the force on an object due to the pull of gravity. On Earth, this force is given by Newton's Law of Universal Gravitation: $W=\frac{G m M}{r^{2}},$ where $m$ is the mass of the object, $M=5.974 \times 10^{24} \mathrm{~kg}$ is the mass of Earth, $r$ is the distance of the object from the center of the Earth, and $G=6.67 \times 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)$ is the universal gravitational constant. Suppose a person weighs $70 \mathrm{~kg}$ at sea level, that is, when $r=6370 \mathrm{~km}$ (the radius of Earth). Use differentials to approximate the person's weight at the top of Mount Everest, which is $8.8 \mathrm{~km}$ above sea level.
More About Derivatives
Differentials; Linear Approximations; Newton's Method
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