# Sped of launchin rocket

1. Homework Statement
A rocket is launched straight up from the earth's surface at a speed of 15,000 m/s. What is its speed when it is very far away from earth?

2. Homework Equations
K=.5mv^2
U=-GMm/R

3. The Attempt at a Solution
Is this correct: I used conservation of energy and assumed very far was where U=0

(.5)(m)(15000m/s)^2 + -[(6.67*10^-11)(5.98*10^24)m]/[6.37*10^6m)=.5mv^2
v=9988 m/s

Also could someone explain to me why potential energy decreases with increasing height in this case as opposed to increasing U by lifting an object above the ground on earth.

Thanks

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alphysicist
Homework Helper
Your equation looks okay to me.

Also could someone explain to me why potential energy decreases with increasing height in this case as opposed to increasing U by lifting an object above the ground on earth.
It doesn't; the potential energy is a negative number that get closer to zero with increasing height (i.e., distance away from the earth), which means it is increasing.

Oh yeah. I see exactly what you are saying. Thanks!