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 v_esc appropriate to the burnout altitude. Show that the speed v of the rocket very far from the Earth is given by v=(v₀²-v²_esc)^1/2

Homework Equations

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

 

[RPinPA]

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.