Josh Cryer: 1 revision

2009-01-21T11:02:30Z

1 revision

C M Edwards: /* Example */

2006-06-14T13:02:24Z

‎Example

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The kinetic energy of a moving object is its energy of motion, which is part of the body's total energy. An object's kinetic energy is equal to the work it can perform in coming to rest.

The concept of kinetic energy provides a useful relationship between mass, velocity, force, and displacement without the necessity of knowing an object's acceleration at any point.

==Mathematics==

The kinetic energy of an object is:

 

<math>T = {1 \over 2} m v^{2}</math> 

 

where 
T is the kinetic energy 
m is the mass 
v is the velocity 

The change in kinetic energy of an object is equal to the total work performed on it, regardless of its acceleration during the aplication of force.

<math>\Sigma W = \int {F \cdot ds} = {1 \over 2} m (v_{f}^2 - v_{i}^2)</math>

where 
<math>\Sigma W</math> is the total work done
F is the applied force 
ds is the derivative of the displacement 
vi and vf are the initial and final velocities, respectively. 

==Example==

For an object approaching the planet [[Mars]] on a [[Hohmann trajectory]], the object's theoretical initial velocity of approach is approximately:

 

<math>\Delta v_{2} = 2400 {m \over s}</math>

 

(This assumes that the object starts from a distance which is effectively infinite compared to the radius of the planet. In reality, the velocity of approach is greater than this value because the initial distance isn't infinite. However, since energy is conserved, the total energy inherent in the system is the same as for the case of infinite distance.)

At the escape velocity of a planet, the kinetic energy of a body is equal to its [[gravitational potential energy]].

The mass of the planet Mars is <math>M = 6.42 \cdot 10^{23} kg</math> and its radius is <math>R = 3370 km</math>. The total energy the object possesses when at rest at an infinite distance is equal to its kinetic energy at the planet's surface [[escape velocity]], which is:

 

<math>T_{esc} = {1 \over 2} m v_{s} = {G M m \over R}</math> 

 

where 
Tesc is the kinetic energy of an object at the escape velocity of Mars 
G is the Universal Gravitational constant (<math>6.67 \cdot 10^{-11} {N m^{2} \over kg^{2}}</math> 
vs is the surface escape velocity (5040 m/s for Mars) 

If the initial velocity isn't lost during the approach, then the total change in kinetic energy needed to bring an object to rest at the surface of Mars when approaching on a Hohmann trajectory is:

 

<math>W = {1 \over 2} m ( 0 - (\Delta v_{2} + v_{s}^{2})</math>

 

Effectively, the initial velocity used is the velocity the object would have if allowed to fall all the way to the surface of Mars without stopping (7400 m/s). Note that the final velocity in this case is 0 m/s - the object comes to rest.

The total work required is thus <math>\Sigma W = 27.7 {MJ \over kg}</math>.

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Kinetic energy - Revision history

Josh Cryer: 1 revision

C M Edwards: /* Example */