College

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as [tex]1.55 \times 10^5[/tex] m. Suppose further that a second group of engineers programmed the orbiter to go to [tex]1.55 \times 10^5[/tex] ft.

What was the difference in kilometers between the two altitudes?

How low did the probe go?

Answer :

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

[tex]h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m[/tex]

or in kilometers,

[tex]h_2 = 47 km[/tex]

the first altitude in kilometers is

[tex]h_1 = 155 km[/tex]

so the difference between the two altitudes is

[tex]\Delta h = 155 km - 47 km = 108 km[/tex]

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