## Gameplay Edit

### EXACTLY HOW KEROS MULTIPLAYER SHOULD BE! Click Here!Edit

## Graphics Edit

- Blood will now cover the screen when the player is injured. This effect will eventually go away and it is still possible to see through the blood.
- The HUD is now much more sleek and easy to read than on Halo Reach

Although Keros runs on a different game engine *much like Modern Warfare 2*, the gameplay is similar to the previous two games, similar enough for players of these games to feel right at home. If players are wearing a helmet or goggles, it must have a reflective surface.

For example: *If a player comes up to another player and is observing him, the player wearing a EVA helmet has a reflective surface and the other player can see his reflection from the helmet. This goes the same for all things reflective; vehicles, Televison screens, glass, polished items, shiny items, ect.*

## Gravitational Force Edit

### Earth and Planent gravitation Edit

The gravitational force in the Keros games, or in real life, is the acceleration of any object toward the ground without significant dissipated energy, such as wind resistance. On Earth, this is (-9.81) meters per second per second. Since there is no dissipated energy to factor into Keros games, testing for g (gravitational force) is made easier. In Keros, this has been roughly calculated to be (-16 m/s/s). This, in theory, means that the g of Keros is almost twice that of Earth's.

### Space gravitation Edit

The **gravitational field** is a vector field that describes the gravitational force which would be applied on an object in any given point in space, per unit mass. It is actually equal to the gravitational acceleration at that point.

It is a generalization of the vector form, which becomes particularly useful if more than 2 objects are involved (such as a rocket between the Earth and the Moon). For 2 objects (e.g. object 2 is a rocket, object 1 the Earth), we simply write **r** instead of **r**_{12} and *m* instead of *m*_{2} and define the gravitational field **g**(**r**) as:

- $ \mathbf g(\mathbf r) = - G {m_1 \over {{\vert \mathbf{r} \vert}^2}} \, \mathbf{\hat{r}} $

so that we can write:

- $ \mathbf{F}( \mathbf r) = m \mathbf g(\mathbf r). $

This formulation is dependent on the objects causing the field. The field has units of acceleration; in SI, this is m/s^{2}.

Gravitational fields are also conservative field|conservative; that is, the work done by gravity from one position to another is path-independent. This has the consequence that there exists a gravitational potential field *V*(**r**) such that

- $ \mathbf{g}(\mathbf{r}) = - \mathbf{\nabla} V( \mathbf r). $

If *m*_{1} is a point mass or the mass of a sphere with homogeneous mass distribution, the force field **g**(**r**) outside the sphere is isotropic, i.e., depends only on the distance *r* from the center of the sphere. In that case

- $ V(r) = -G\frac{m_1}{r}. $