# Space station-nature of orbit?

1. Jan 18, 2012

### humanist rho

1. The problem statement, all variables and given/known data

A space station moving in a circular orbit around the earth goes into a new bound orbit by firing its engine radially outaward.This orbit is.......

Choices are;
(a) A larger circle
(b)A smaller circle
(c)An ellipse
(d)a parabola

2. Relevant equations

3. The attempt at a solution

I think it'll be a larger circle or parabola.
But donno how to work it out.

Last edited: Jan 18, 2012
2. Jan 18, 2012

### Simon Bridge

What is the free-body diagram of an object in a circular orbit?
The engine applies thrust radially outward - what does that do to the free body diagram?
Do the engines fire continuously or for a short while?

The change in velocity is important to this.
What can you say about the kind of velocity needed for a circular orbit?
(velocity is a vector - talk in terms of tangential and radial components)
What can you say about the kind of velocity needed for a parabolic orbit?
... an elliptical one?

3. Jan 18, 2012

### humanist rho

mv2/r = GMm/r2

v = Sqrt[GM/r] for circular orbit.

How can i modify these equations for elliptical orbit.

4. Jan 18, 2012

### Simon Bridge

Hint:
1. velocity is a vector - represent as radial and tangential.
2. free body diagram

the v in your equation is purely tangential.
when the thrust is applied for a short time it changes the velocity vector ...
what happens to the radial component
what happens to the tangential component

is it possible for an object in a circular orbit to have a radial component to it's velocity?

5. Jan 18, 2012

### Simon Bridge

btw: you equation gets modified as follows...

$$\underbrace{ma = F_g+F_{thrust}}_{\text{from FBD}} \Rightarrow \frac{mv_\perp^2}{R} = mg_R -\frac{m}{T}\Delta v_r$$... the thrust is applied for a fixed time T, so creates a change in speed Δv. You'll have heard of thrust being referred to as a "delta-vee"? R is the radius of the orbit so gR is the local acceleration of gravity at R. The minus sign is because the thrust is radially outwards, making the positive direction to be radially inwards. The instantanious velocity after the thrust has been applied is given by
$\vec{v}=v_r\hat{r}+v_\perp\hat{\small \perp}$
...which has magnitude $v =\sqrt{v_r^2 + v_\perp^2}$
...and the angle this makes to the radius is $\tan^{-1}(v_\perp / v_r)$.