Gravitational Potential Energy of a rocket fired straight up

AI Thread Summary
A rocket fired straight up at half the escape velocity will reach a maximum height determined by gravitational potential energy and kinetic energy equations. The initial kinetic energy is given by 1/2mv^2, while the potential energy at the surface is -GMm/R. As the rocket ascends, its kinetic energy decreases to zero at the peak height, where the potential energy is -GMm/(R + h). To find the maximum height, one must equate the initial total energy to the potential energy at the peak. The discussion emphasizes the relationship between initial velocity and height, highlighting the need to apply gravitational equations correctly.
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Homework Statement


A rocket is fired straight up from the surface of the Earth at half the escape velocity. How high will it go relative to the surface of the earth?
Neglect dissipative forces.


Homework Equations



Uinitial + Kinitial = Uf

Kf is zero

G*m*M/radius^2 + 1/2mv^2 = -G*m*M/(2R)

The Attempt at a Solution



I used the above equations and I can solve for the initial velocity. I am stuck on how to find the height. Is it just 2R?

Many thanks...physics is a struggle for me.
 
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On the surface of the earth, PE = -GMm/R and KE = 1/2*m*v^2
At a certain height, KE = 0, and PE = -GMm/(R + h)
What is the expression for the escape velocity?
From that find the initial velocity.
 

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