Question

a satellites escape velocity is 6.3 mi/sec, the radius of the earth is 3950 mi, and the earths gravitational constant is 32.1 ft/sec². how far is the satellite from the surface of the earth?
the satellite is approximately □ mi from the surface of the earth.
(do not round until the final answer. then round to the nearest whole number as needed.)

Explanation:

Step1: Convert units

First, convert the escape velocity \(v = 6.3\) mi/sec to ft/sec. Since \(1\) mile \(= 5280\) feet, then \(v=6.3\times5280 = 33264\) ft/sec. The radius of the Earth \(R = 3950\) mi. Convert \(R\) to feet: \(R = 3950\times5280=20856000\) ft.

Step2: Use the escape - velocity formula

The formula for escape velocity is \(v=\sqrt{\frac{2gM}{R + h}}\), where \(g = 32.1\) ft/sec². Also, from the gravitational force \(F=\frac{GMm}{r^{2}}\) and near the Earth's surface \(F = mg\) (where \(r = R\)), we have \(GM = gR^{2}\). The escape - velocity formula can be rewritten as \(v=\sqrt{\frac{2gR^{2}}{R + h}}\).
Squaring both sides gives \(v^{2}=\frac{2gR^{2}}{R + h}\). Then \(R + h=\frac{2gR^{2}}{v^{2}}\).
Substitute \(g = 32.1\), \(R = 20856000\), and \(v = 33264\) into the formula:

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Step3: Find \(h\) in feet and convert back to miles

\(h=(R + h)-R\). \(h = 25203252.03-20856000=4347252.03\) ft.
Convert \(h\) to miles: \(h=\frac{4347252.03}{5280}\approx823\) mi.

Answer:

\(823\) mi