# Launching a projectile into Orbit

1. Oct 21, 2006

### Noone1982

I completely understand how normal projectiles work on a "flat" world using the x and y components.

$$x\; =\; V_{x}t\; +\; x_{o}$$

$$y\; =\; Vyt\; +\; y_{o}\; +\; 0.5gt^{2}$$

I am confused if we place the earth on the XY plane and launch a projectile at some angle. How are the equations different?

It seems that both Y and X equations should contain a gravity component using sine or cosine to take this into account. If no gravity in the x equations exists, the projectile would never come back to earth.

2. Oct 21, 2006

### PhanthomJay

I don't quite understand your question. The earth gravity force always points in one direction only....toward the center of the earth. If you are implying that you are viewing the earth from say the moon, such that the earth lies in the XY plane, and a projectile is launched from the equator in a generally left to right direction, horizonatlly or at some angle, then the y axis can be chosen as the horizontal axis, and the x axis as the vertical axis. There is no gravity component along the x axis.

3. Oct 22, 2006

### andrevdh

These equations were set up by directing the y-axis along the straight down or up direction - that is along a radius line of the earth. So if you work on the scale of the earth these directions will change during the course of a projectile's flight. We tend to develop/use equations that describe the motion of a projectile in other types of coordinate systems. Anyway, the value of g will decrease as it goes further away from the earth so these equations are not very helpfull in this context.

4. Oct 22, 2006

### Noone1982

How would I set up this problem in cylindrical coordinates?

With R(t) r-hat and phi(t) phi-hat?

5. Oct 22, 2006

### andrevdh

It has been quite a while since I have worked on orbital theory and it can get quite involved, but one of the most usefull equations I have found is the relation between the object's angular momentum per unit mass of the object and its angular velocity about the force centre:

$$h=r^2 \omega =\ constant$$

This is due to the fact that the attractive force on the object working only along the line connecting the two. So from this it follows that

$$\dot{\theta} = \frac{h}{r^2}$$

In the radial direction things look similar

$$\ddot{r} = -\frac{GM}{r^2}$$

due to Newton's law of universal gravitational attraction.

Other usefull approaches is energy considerations. Since the object is experiencing a conservative force its energy will remain constant.

Last edited: Oct 22, 2006
6. Oct 22, 2006

### Noone1982

Hmm. I have been going about this another way. My advisor told me it was a simple matter of using essentially R = 0.5a*t^2 but with vectors. Here is what I'v been doing: