# Homework Help: Universal gravitation/escape velocity question

1. Nov 16, 2013

### physicsdb

Q: a 4.6 kg rocket is launched directly upward from earth at 9 km/s .
What altitude above Earth's surface does the rocket reach?

Relevant equations: v=$\sqrt{2Gm/r}$
Eg=Gm1m2(1/r1-1/r2)

my attempt: i figured i should use the first equation which contains velocity so i did:

9000=$\sqrt{2(6.67x10^-11)(5.99x10^24/r}$
9000^2=2(6.67x10^-11)(5.99x10^24)/r
r=9,865 km

the answer is 1.21x10^4 km
please help, thanks!

2. Nov 16, 2013

### D H

Staff Emeritus
That equation is not relevant to this problem. This is not a question about escape velocity.

3. Nov 16, 2013

### physicsdb

so is this just a gravitational potential near earth question ?

4. Nov 16, 2013

### D H

Staff Emeritus
No. You cannot use mgh as gravitational potential energy for this problem because that expression assumes h is very small compared to the radius of the Earth.

5. Nov 16, 2013

### physicsdb

i just tried Fg=Gm1m2/r^2 and that doesn't work either
9.8=6.67^-11(5.99x10^24)/r^2

6. Nov 16, 2013

### D H

Staff Emeritus
Force isn't the relevant equation, either.

You won't make much progress in physics, or in any technical field, if your approach is to grab some random expression and run with it. You need to think. Ask yourself, what is it that would make this problem solvable?

In physics, one of the first places you should look is the conservation laws. The rocket's momentum isn't a conserved quantity; there's an external force (gravity) on the rocket. Conservation of angular momentum isn't going to help much; since the rocket was launched straight up, the rocket's angular momentum is identically zero from the perspective of an Earth-centered frame. That leaves conservation of energy, and it is very applicable since gravitation is a conservative force.

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