# Launch speed off of the earth

1. Jul 1, 2008

### baylorbelle

What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to 8REarth before coming to rest momentarily?

I think the formula for this problem is 1/2 mv2-G(mME/RE)=G(mME/2RE); I cancel out the m because it is in every variable; but whenever I plug in the numbers, I get a huge answer (about 13000 or 14000m/s) but my book says the answer should be 7.6m/s. Where am I going wrong in my formula?

2. Jul 1, 2008

### alphysicist

Are you sure you have copied the question and answer right? If a projectile has an initial speed of 7.6 m/s, it will have a maximum height of only several meters under free fall.

Also, the right side of your equation does not look right to me. (It looks like you are setting the initial energy equal to an expression for the total circular orbital energy, which does not apply here.)

3. Jul 1, 2008

### baylorbelle

The book says that the answer is 7.91 m/s. And I don't know, but it doesnt seem like its right to me on teh right side, but I dont know what I would change it to. Thats the formula that the teacher gave to us in class, but its definately not working.

4. Jul 1, 2008

### alphysicist

Well, an initial speed of 7.91m/s won't get a particle to an altitude of eight times the radius of the earth. (7.91 m/s isn't even 20 miles per hour.)

To set up the equation, you want to use conservation of energy. Your left side is okay (since it includes the kinetic and potential energy at the initial point). Now include the same type of terms, but for the highest point, on the right side.

When you solve for v, it should be considerably larger than 7.91 m/s.

5. Jul 1, 2008

### baylorbelle

i finally figured out where the formula was wrong. on the right side, the G is negative. that seemed to help a lot!

6. Jul 1, 2008

### DaveC426913

Is that 7.91 metres per second or miles per second?

7. Jul 1, 2008

### gamesguru

Use the work-energy principle:
$$\frac{1}{2}mv^2_i=\int_{R}^{9R}Fdr=\int_{R}^{9R}\frac{GMm}{r^2}dr$$.
You can handle it from there I'm pretty sure. Just solve for vi and you'll know the velocity needed to reach a height of 8R.
Hope that helps.

Last edited: Jul 1, 2008
8. Jul 1, 2008

### D H

Staff Emeritus
Is the 8 a typo in the OP? It seems to be.
You might want to justify why you "think" this is the right equation. It is the correct equation, but why?
Note: Here you have $2R_E$, which means to me that the satellite rises to an altitude above the Earth that is equal to the radius of the Earth.
In other words,

$$\frac 1 2 v^2 - \frac {GM_E} {R_E} = -\;\frac {GM_E}{2R_E}$$

9. Jul 1, 2008

### Dick

gamesguru, don't work out people's problems for them. We are supposed to coach. Not solve. Is there some confusion running around here? Is the altitude RE or 8RE? The problem statement says one and the solution implies the other.

10. Jul 1, 2008

### gamesguru

I thought I was in the regular physics forum. I removed it and left him with some work to do.

11. Jul 1, 2008

### Dick

S'ok. Thanks, but the cat seems to be largely out of the bag anyway. Looks like there was a typo in the problem statement and the OP seems to be convinced the problem is solved anyway. Unnecessary police action here, I guess. Sorry.

12. Jul 2, 2008

### baylorbelle

Thanks for all of your help! yes, the original problem did say 8 Earth Radii(sp), and the answer was 7.9 km/sec. the answer key in the back of the book had a typo on it, so it kept throwing me off.in the end, i figured out. thanks again for the tutoring.