Vertical Projectile: Solved, Energy, Max Height

AI Thread Summary
A vertical projectile launched with initial velocity v_0 has total energy expressed as E = (1/2)m{v_0}^2 - G(Mm/R), where R is the Earth's radius. To find the maximum height, the total energy is set equal to -G(Mm/(x+R)), allowing for the calculation of height x. Concerns about algebraic errors arise when deriving x, particularly when approximating potential energy as mgx, which eliminates the potential energy term due to the chosen reference point. The derived formula for maximum height is confirmed as x = (-R^2v^2)/(Rv^2 - 2GM), which also aligns with the escape velocity equation. The discussion emphasizes the importance of understanding energy conservation and reference points in projectile motion.
cscott
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[SOLVED] Vertical projectile

Homework Statement



Is it true that a projectile launched vertically with initial velocity v_0 has total energy E = \frac{1}{2}m{v_0}^2 - G\frac{Mm}{R} where R is the radius of the earth?

And to find the maximum height I set the total energy equal to -G\frac{Mm}{x+R} where x is height, and solve for x?
 
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Yes, and yes...but do you understand why?
 
I think I do. I was just concerned because when I work it out for x I don't get a meaningful answer so I guess I messed up my algebra.

But when we approximate PE with mgx we get no PE term for total energy. This is because in that case our reference point is the surface of the Earth and not infinity, correct?
 
Is it correct that max height = x = \frac{-R^2v^2}{Rv^2-2GM}?
 
I get the correct expression for escape velocity from the above so I think it's OK.
 
cscott said:
Is it correct that max height = x = \frac{-R^2v^2}{Rv^2-2GM}?
Yes.
 
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