Question

a satellites escape velocity is 6.3 mi/sec, the radius of the earth is 3960 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 radius of the earth \(R = 3960\) mi to feet. Since \(1\) mi \(= 5280\) ft, then \(R=3960\times5280\) ft. The escape velocity formula is \(v = \sqrt{\frac{2gRr}{r - R}}\), where \(v = 6.3\) mi/sec. Convert \(v\) to ft/sec: \(v=6.3\times5280\) ft/sec and \(g = 32.1\) ft/sec².

Step2: Substitute into the formula

Substitute the values into the formula \(v=\sqrt{\frac{2gRr}{r - R}}\). Square both sides: \(v^{2}=\frac{2gRr}{r - R}\). Then \(v^{2}(r - R)=2gRr\). Expand: \(v^{2}r-v^{2}R = 2gRr\). Rearrange terms: \(v^{2}r-2gRr=v^{2}R\). Factor out \(r\): \(r(v^{2}-2gR)=v^{2}R\). So \(r=\frac{v^{2}R}{v^{2}-2gR}\).

Step3: Calculate \(r\)

\(v = 6.3\times5280=33264\) ft/sec, \(R = 3960\times5280 = 20908800\) ft.
\(v^{2}=(33264)^{2}=1106497696\), \(2gR=2\times32.1\times20908800 = 1340734560\).
\(r=\frac{1106497696\times20908800}{1106497696 - 1340734560}\)
\(r=\frac{1106497696\times20908800}{- 234236864}\approx - 98999999.9\) (This is wrong, we should use the correct formula \(v=\sqrt{\frac{2GM}{R}}\) for escape velocity, but another way: using the relation \(v=\sqrt{\frac{2gR^{2}}{r - R}}\) (derived from energy conservation \(mgR=\frac{1}{2}mv^{2}\frac{R}{r - R}\)).
Square both sides: \(v^{2}=\frac{2gR^{2}}{r - R}\), then \(r - R=\frac{2gR^{2}}{v^{2}}\).
\(r=R+\frac{2gR^{2}}{v^{2}}\)
Substitute \(g = 32.1\) ft/sec², \(R = 3960\times5280\) ft, \(v = 6.3\times5280\) ft/sec
\(r=3960\times5280+\frac{2\times32.1\times(3960\times5280)^{2}}{(6.3\times5280)^{2}}\)
First, simplify \(\frac{(3960\times5280)^{2}}{(6.3\times5280)^{2}}=\frac{3960^{2}}{6.3^{2}}=\frac{(3960\div6.3)^{2}}{1}\approx(628.57)^{2}\)
\(2\times32.1\times\frac{3960^{2}}{6.3^{2}}=2\times32.1\times628.57^{2}\approx2\times32.1\times395000\) (approximate calculation)
\(r\) (in feet) \(\div5280\) to get miles.
Another approach: Using the formula \(v=\sqrt{\frac{2gR^{2}}{r - R}}\) (from \(F = mg=\frac{GMm}{R^{2}}\), \(GM = gR^{2}\), and escape - velocity energy \(\frac{1}{2}mv^{2}=\frac{GMm}{R}-\frac{GMm}{r}\)).
\(v^{2}=\frac{2gR^{2}}{r - R}\), \(r - R=\frac{2gR^{2}}{v^{2}}\)
\(g = 32.1\) ft/sec², \(R = 3960\) mi \(=3960\times5280\) ft, \(v = 6.3\) mi/sec \(=6.3\times5280\) ft/sec
\(r - R=\frac{2\times32.1\times(3960\times5280)^{2}}{(6.3\times5280)^{2}}\)
\(r - R=\frac{2\times32.1\times3960^{2}}{6.3^{2}}\)
\(r - R=\frac{2\times32.1\times15681600}{39.69}\)
\(r - R=\frac{1006233600}{39.69}\approx25352300\) ft
Convert \(r - R\) to miles: \(\frac{25352300}{5280}\approx4801\) mi

Answer:

\(4801\) mi