Question

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as 1.55 × 105 m. Suppose further that a second group of engineers programmed the orbiter to go to 1.55 × 105 ft. What was the difference in kilometers between the two altitudes? How low did the probe go?

Answer 1

108 km

Step-by-step 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

`h_2 = 1.55 \cdot 10^5 ft \cdot (1)/(3.28 ft/m)=4.7\cdot 10^4 m`

or in kilometers,

`h_2 = 47 km`

the first altitude in kilometers is

`h_1 = 155 km`

so the difference between the two altitudes is

`\Delta h = 155 km - 47 km = 108 km`