# A rocket burns out at an altitude h above the Earth's surface

Benjamin Fogiel

## Homework Statement

A rocket burns out at an altitude h above the Earth's surface. Its speed v0 at burnout exceeds the escape speed vesc appropriate to the burnout altitude. Show that the speed v of the rocket very far from the Earth is given by v=(v02-v2esc)1/2

KEf-KEi=Ui-Uf

## The Attempt at a Solution

I plugged in the values of kinetic and potential energies:

(m1v2)/2 - (m1v02)/2 = (-Gm1m2)/r1 + (Gm1m2)/r2

Simplified to get:

v2=(-2Gm2)/r1 + (2Gm2)/r2 + v02

Im not sure where to go from here

Homework Helper
What are ##r_1## and ##r_2##?

Escape velocity for mass ##m_1## to escape mass ##m_2## starting at distance ##r## is the value such that ##(1/2)m_1v_0^2 = G m_1 m_2/r##. In other words, added to the initial potential energy, you get a total energy of 0, which means that the mass ##m_1## will not stop before reaching potential energy 0 at "infinite" or "very far away" distance.

Perhaps that will help?

The intent of the phrase "very far away" refers to the zero point of potential energy. So you can just use 0 for PE.

Benjamin Fogiel
r1 = h
r2 = "very far away"

Sorry, forgot to define those values.

Yes, that helped! So Uf is essentially zero which then gives me the right answer.

Thank you.