To find the escape velocity from the surface of the Earth, we need to use the concepts of gravitational potential energy and kinetic energy. The particle must have enough kinetic energy to overcome the gravitational potential energy holding it to Earth.
The gravitational potential energy at distance [tex]\( r \)[/tex] from the center of Earth is given by:
[tex]\[
U(r) = -\frac{GMm}{r}
\][/tex]
where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( M \)[/tex] is the mass of the Earth, and [tex]\( m \)[/tex] is the mass of the particle.
The kinetic energy [tex]\( K \)[/tex] the particle needs to have at the Earth's surface (where [tex]\( r = R \)[/tex]) is:
[tex]\[
K = \frac{1}{2}mv^2
\][/tex]
At the escape velocity [tex]\( v_e \)[/tex], all the kinetic energy will convert into gravitational potential energy as the particle moves away from the earth such that at [tex]\( r = \infty \)[/tex], the total energy becomes zero. Thus:
[tex]\[
\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0
\][/tex]
Solving for [tex]\( v_e \)[/tex]:
[tex]\[
\frac{1}{2}mv_e^2 = \frac{GMm}{R}
\][/tex]
[tex]\[
v_e^2 = \frac{2GM}{R}
\][/tex]
[tex]\[
v_e = \sqrt{\frac{2GM}{R}}
\][/tex]
To find [tex]\( GM \)[/tex] in terms of [tex]\( g \)[/tex] and [tex]\( R \)[/tex], we use:
[tex]\[
g = \frac{GM}{R^2}
\][/tex]
[tex]\[
GM = gR^2
\][/tex]
Plugging this back into the escape velocity formula:
[tex]\[
v_e = \sqrt{\frac{2gR^2}{R}}
\][/tex]
[tex]\[
v_e = \sqrt{2gR}
\][/tex]
Given:
[tex]\[
R = 3960 \text{ miles}
\][/tex]
[tex]\[
g = 32.17 \text{ ft/s}^2 \text{ (at Earth's surface)}
\][/tex]
[tex]\[
\text{1 mile} = 5280 \text{ feet}
\][/tex]
Converting [tex]\( R \)[/tex] to feet:
[tex]\[
R = 3960 \text{ miles} \times 5280 \text{ feet/mile} = 20908800 \text{ feet}
\][/tex]
Now calculate the escape velocity:
[tex]\[
v_e = \sqrt{2 \cdot 32.17 \text{ ft/s}^2 \cdot 20908800 \text{ feet}}
\][/tex]
[tex]\[
v_e \approx 36677.95239650109 \text{ ft/s}
\][/tex]
So, the escape velocity from the surface of the Earth, in feet per second, is approximately [tex]\( 36677.95 \text{ ft/s} \)[/tex].
