Launched Rocket (Energy problem)

[I Like Pi]

Homework Statement

15. (4) A rocket is launched vertically from Earth's surface with a velocity of 3.4 km/s. How high does it go from Earth's centre?

Homework Equations

Ek = 1/2mv^2
Eg = -GMm/r

The Attempt at a Solution

1/2mv^2 = -GMm/r
1/2(3400)^2 = 6.673*10^-11*5.978*10^24/r
r = -6.9*10^11 m

How do i get rid of the negative?

Thanks for your time

 

[rock.freak667]

Remember Eg is the energy needed to move the mass to infinity, so the change in Eg is the change in ke.

The change in PE is positive, and the change in KE is 0.5mv²

So you will have 0.5mv²= GMm/r

 

[I Like Pi]

Remember Eg is the energy needed to move the mass to infinity, so the change in Eg is the change in ke.

The change in PE is positive, and the change in KE is 0.5mv²

So you will have 0.5mv²= GMm/r

Thanks! so I just ignore the negative sign?

 

[rock.freak667]

Thanks! so I just ignore the negative sign?

Well you don't really ignore is as it cancels out. But in essence, yes you can ignore it.

 

[I Like Pi]

Well you don't really ignore is as it cancels out. But in essence, yes you can ignore it.

How does it cancel out?

thanks for your help!

 

[Delphi51]

I think it is a little more complicated. Notice that -GMm/r is zero at r = infinity (as someone already wrote) and that it gets larger as r increases. So, quite a large negative value at the surface. A smaller negative value a few km up. It really only has meaning when you take the difference for two different radii. You'll need something like this:
0.5mv² = -GMm/r - (-GMm/r)
where one of the r's is the Earth's radius and the other the radius you are looking for.

 

[I Like Pi]

I think it is a little more complicated. Notice that -GMm/r is zero at r = infinity (as someone already wrote) and that it gets larger as r increases. So, quite a large negative value at the surface. A smaller negative value a few km up. It really only has meaning when you take the difference for two different radii. You'll need something like this:
0.5mv² = -GMm/r - (-GMm/r)
where one of the r's is the Earth's radius and the other the radius you are looking for.

Thank you ! Makes sense. Now if i want to find the rocket's distance from EARTH'S SURFACE, would I just subtract the radius of the Earth from the answer i get above?

Thanks again!

 

[Delphi51]

Right on the subtraction.
Most welcome and good luck on the next one.