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Solved: In 1999, NASA lost a $94 million orbiter because one group of engineers used
Chapter 2, Problem 123P(choose chapter or problem)
In 1999, NASA lost a \($94\) million orbiter because one group of engineers used metric units in their calculations while another group used English units. Consequently, the orbiter descended too far into the Martian atmosphere and burned up. Suppose that the orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as \(1.55 \times 10^{5} \mathrm{~m}\). Suppose further that a second group of engineers programmed the orbiter to go to \(1.55 \times 10^{5} \mathrm{ft}\). What was the difference in kilometers between the two altitudes? How low did the probe go?
Questions & Answers
QUESTION:
In 1999, NASA lost a \($94\) million orbiter because one group of engineers used metric units in their calculations while another group used English units. Consequently, the orbiter descended too far into the Martian atmosphere and burned up. Suppose that the orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as \(1.55 \times 10^{5} \mathrm{~m}\). Suppose further that a second group of engineers programmed the orbiter to go to \(1.55 \times 10^{5} \mathrm{ft}\). What was the difference in kilometers between the two altitudes? How low did the probe go?
Step 1 of 2
From the given,
Distance between the two altitudes = \(1.55 \times 10^{5} \mathrm{ft}\)
Let’s convert 1ft to m
\(\begin{aligned}1ft&=0.3048\mathrm{\ m}\\ &=1.55\times10^5\mathrm{ft}\times\frac{0.3048\mathrm{\ m}}{1\ \mathrm{ft}}\\ &=4.7244\times10^4\mathrm{\ m}=47.244\mathrm{\ km}\end{aligned}\)