# How far can the rocket go trough space?

1. Nov 13, 2004

### sony

Hi, I'm new to these boards. I hope I posted in the right category, (I believe collage is equivalent with gymnasium in Norway)

Here's the problem:
A rocket is fired of from the earth. The rocket's engine stops
85km above the earth's surface, then the rocket has a vertical velocity
of 3,8km/s. How far can the rocket go? (how high)

I'm uncertain about the criterias here. The rocket keeps going til the
velocity is 0, right? Does the rocket only have potential energi at this point?
If I try and figure it out with this, I only get rubbish. In the answer the book hints that you should use Ep = -yMm/r. But you don't know the mass of the rocket... So I guess you should find another expression equaling Ep. Anyways, I'm stuck.

Thanks for the help!

2. Nov 13, 2004

### arildno

You don't NEED the mass of the rocket, since the kinetic energy is also proportional to that mass.
Hence "m" is a common factor you can remove.

3. Nov 13, 2004

### sony

Thanks, but I still don't get it...

4. Nov 13, 2004

### arildno

Sett opp energi-regnskapet.
Vær spesielt forsiktig med uttrykkene for potensiell energi.

5. Nov 13, 2004

### sony

Takk skal du ha, men jeg er fremdeles forvirret.
Hva er kriteriene her?

Ved 85km høyde har raketten E= Ek+Ep
Og ved maksimal høyde har raketten E=Ep?

Stemmer det så langt?

6. Nov 13, 2004

### sony

Og hvis jeg setter uttrykkene lik hverandre strykes jo omtrent alt?

7. Nov 13, 2004

### Janus

Staff Emeritus
Try determining the rocket's total energy at 85km. (KE + GPE)
That will be equal to the total energy it will have at the top of its flight when its velocity is zero. write out the formula for both situations (with the appropiate unknowns) equate them and solve. (as noted already, the mass of the rocket will drop out.)

8. Nov 13, 2004

### arildno

Riktig; pass også på ENHETENE dine; lengdemålet du har fått oppgitt tall i er i km (eller km/s), gjør om disse til meter.
Altså, vi setter opp balanse av mekanisk energi (delt på massen m):
$$\frac{1}{2}v_{0}^{2}-\frac{\gamma{M}}{R_{0}}=-\frac{\gamma{M}}{R_{1}}$$
Husk at avstandene skal måles fra jordsenteret..

Note:
Let's take this in English from now on..

Last edited: Nov 13, 2004
9. Nov 13, 2004

### arildno

No, since the two distances are unequal, the potential energy terms do NOT cancel.
Your unkown is $$R_{1}$$

10. Nov 13, 2004

### sony

Ok, igjen, takk skal du ha!

In our textbook Ek = 1,5 * yMm / r
and Ep = -yMm / r

Sorry for being an imbecile, but I don't understand how to get your expression.
0,5 * yMm/r - yMm/r = - 0,5 yMm/r
How is V0 incorporated?

11. Nov 13, 2004

### sony

I forgot:
0,5 * yMm/r - yMm/r = - 0,5 yMm/x

12. Nov 13, 2004

### arildno

What kind of textbook is this??

The kinetic energy is ALWAYS $$\frac{1}{2}mv^{2}$$ where m is the mass and v the velocity.
I believe you must have confused the expression for kinetic energy with its value of it in the solution of a particular problem in your book.

13. Nov 13, 2004

### sony

Sorry! I see it now, the general formula for E = 0.5mv^2 - yMm/r

14. Nov 13, 2004

### arildno

Are you sure you are not confusing yourself with the expression for the ESCAPE VELOCITY?

15. Nov 13, 2004

### sony

I looked at the wrong expression, I think it's all clear now!

Ha en fin kveld.

16. Nov 13, 2004

### arildno

Now, take particular care with your units in solving the problem!

17. Nov 13, 2004

### arildno

Takk det samme!

18. Nov 13, 2004

### sony

Beklager at jeg maser nå, men dette forumet virket som et ganske flott sted. Postet jeg forresten i riktig kategori?

Uansett, det er kjekt å slippe å måtte vente til neste fysikktime for å få hjelp!

Sorry for nagging, shall write in english from now on.

Cheers! :D

19. Nov 14, 2004

### arildno

If you hadn't posted it in the right place, the mentors would have been alerted and moved it to its proper place.
Since you did post it where it should be, that didn't happen..

Velkommen til PF!