| Intro | What is it? | How does it work? | What can I do with it? | The Future of Gravity! | Bibliography |
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What can be done with Gravity?
Escape Speed
Escape speed is the minamum speed needed for a mass to stop moving at an infinate distance from the initial mass. Because gravitational potential and kinetic energy are both path independent, It doesn't matter the direction of the velocity only the magnitude. We can use conservation of energy to calculate the Escape speed. %DELTA E = 0; (Kf - Ki) + (Uf - Ui) = 0; 1/2 mv2 = Gm1m2/R; v=(2GM/R)1/2. Again substituting in the mass and radius of the earth we find the escape velocity to be 11.2 km/s.
Orbits
Up to this points all the objects discussed were non-moving, but what would happen if the mass was moving perpindictular to the force of gravity. We already have a situation very simalar to this: circular motion. Gm1m2/R2 = m1v2/R. This looks very simalar to kenitic energy. If we simplify the R's and divide both sides by 2, we can find the energy of an orbit: Gm1m2/2R.
http://www.orbitsimulator.com/gravity/barycenter.GIF
Artificial Gravity
The objective of artificial gravity is to simulate a downward acceleration close to 9.8 m/s2 but without a secondary massive object. Imagine a person standing on the surface of the Earth. This can be obtained through circular motion. Again circular motion is when a force is acting perpindicular to an objects velocity. If a person were standing on the inside of a rotating cylinder their normal force would be this radial force and the rotation would be the tangential velovity.
http://mvl.mit.edu/AG/images/2001gravityferris.jpg
Lagrangian Points
In the movie space cowboys one of the characters describs geting to the moon as "You only need to get halfway there then the moons gravity will do the rest." While not eactly half the distance, the idea of an equilibrium point is true. The point L1 is easyly enough found as the point were "G MSun MSat / R12 = G MEarth MSat / R12". L2 and L3 are a little more complicated. The same equation needs to be solved for a point where the resultant acceleration puts the satalite in circular motion. And finaly L2 and L3 have an x and y component to the resulting acceleration.
