Question

a satellites escape velocity is 6.5 mi/sec, the radius of the earth is 3940 mi, and the earths gravitational constant is 32.5 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 gravitational constant \(g = 32.5\space ft/sec^{2}\) to \(mi/sec^{2}\). Since \(1\space mi=5280\space ft\), then \(g=\frac{32.5}{5280}\space mi/sec^{2}\). The escape - velocity formula is \(v = \sqrt{\frac{2gR^{2}}{r}}\), where \(v\) is the escape velocity, \(g\) is the gravitational constant, \(R\) is the radius of the earth, and \(r\) is the distance from the center of the earth to the satellite.

We know that \(v = 6.5\space mi/sec\) and \(R = 3940\space mi\). Rearranging the formula \(v=\sqrt{\frac{2gR^{2}}{r}}\) for \(r\), we get \(v^{2}=\frac{2gR^{2}}{r}\), then \(r=\frac{2gR^{2}}{v^{2}}\).

Substitute \(g=\frac{32.5}{5280}\space mi/sec^{2}\), \(R = 3940\space mi\), and \(v = 6.5\space mi/sec\) into the formula:

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$$

Step2: Calculate \(r\)

First, calculate \(3940^{2}=3940\times3940 = 15523600\)

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$$

\(65\times15523600 = 1008034000\)

\(r=\frac{1008034000}{223440}\approx4511.4\space mi\)

Step3: Find the distance from the surface

The distance from the surface of the earth \(d=r - R\). Substitute \(r\approx4511.4\space mi\) and \(R = 3940\space mi\)

\(d=4511.4−3940=571.4\approx571\space mi\)

Answer:

\(571\space mi\)