Unit 4 - Problem 11 ==> Escape Velocity

Definition and Givens

• Escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.
• It is the speed needed to "break free" from a gravitational field without further propulsion.
• According to Wikipedia, escape velocity equals the square root of 2 * G * m divided by r, where:
• G is the gravitational constant of 6.674 * 10^-11
• m is the mass of the body we are escaping from
• Mass of the earth is M=5.9736×10²⁴ kg
• r the distance from the center of gravity
• Radius of the earth is 6400 kilometers or 6.4 * 10^6 meters
• $$V_{escape} = \sqrt{\frac{(2 G m)}{r}}$$
• On the surface of the Earth, the escape velocity is about 11.2 kilometers per second (~6.96 mi/s), which is approximately 34 times the speed of sound (Mach 34) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s).

Deriving the escape velocity formula

• Conservation of energy provides us that the escape energies total zero.
• $$E_{initial-potential} + E_{initial-kinetic} = E_{escape-potential} + E_{escape-kinetic-energy}$$
• $$E_{initial-potential} = -\frac{G M m}{r}$$
• $$E_{initial-kinetic} = \frac{1}{2}m V^2$$
• $$-\frac{G M m}{r} + \frac{1}{2}m V_{escape}^2 = 0 + 0$$
• $$\frac{1}{2}m V^2 = \frac{G M m}{r}$$
• $$V_escape^2 = 2\frac{G M}{r}$$
• $$V_{escape} = \sqrt{\frac{(2 G M)}{r}}$$  

Calculate the escape velocity value

• $$V_{escape} = \sqrt{\frac{(2 G M)}{r}}$$
• $$V_{escape} = \sqrt{\frac{(2* (6.674 * 10^-11) ( 5.97 * 10^{24}))}{6.4 * 10^6}}$$
• $$V_{escape} = \sqrt{12.45118125 * 10^7}$$
• $$V_{escape} = 11158.486120437665\frac{m}{s} = 24,960.8\;{mph}$$

MIT Professor Lewin goes thru this same derivation.