imported>Patrick at 19:53, 20 September 2011

2011-09-20T19:53:42Z

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We assume that the lunar module is at rest on the Moon's surface. The total energy of the system Moon - lunar module is equal to the gravitational energy of the system

<math>U = -{G M m \over r} = -{(6.67 \times 10^{-11}) (7.4 \times 10^{22}) m \over 1.74 \times 10^6} = -2.8 \times 10^6 m J</math>

where <math>m</math> is the mass of the module. To escape the Moon's gravity, the total energy of the system must be positive or at least equal to zero. Searching for the minimum kinetic energy to escape, we will assume that the total energy of the system is zero. Therefore, the kinetic energy of the module is <math>K = 2.8 \times 10^6 m J</math>. Using the equation for the kinetic energy

<math>\frac{1}{2} m v^2 = 2.8 \times 10^6 m</math>

<math>v = \sqrt{2 \times 2.8 \times 10^6} = 2.4 \times 10^3 m/s</math>

We found the escape velocity to be 2.4 km/s. Note that the escape velocity is independent of the mass of the module.

Satellites and Binding Energy EX 6 - Revision history

imported>Patrick at 19:53, 20 September 2011