How far can the rocket go trough space?

[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!

 

[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.

 

[sony]

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

 

[arildno]

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

 

[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?

 

[sony]

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

 

[Janus]

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.)

 

[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..

 

[arildno]

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

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

 

[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?

 

[sony]

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

 

[arildno]

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?

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.

 

[sony]

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

 

[arildno]

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

 

[sony]

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

Thanks a lot for your help!

Ha en fin kveld.

 

[arildno]

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

Glad we agree on that!

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

 

[arildno]

Takk det samme!

 

[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

 

[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!