Find the amount of work done against gravity in moving a 150kg rocket

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The discussion centers on calculating the work done against gravity when moving a 150kg rocket. The initial energy at Earth's surface is calculated as -0.9 x 10^10 J, while the energy at the orbit radius is -3.4 x 10^9 J, leading to a total work done of 5600 x 10^6 J. The textbook answer of 6 x 10^5 J is questioned, suggesting a possible error in the textbook. Participants confirm the calculations are correct, indicating the textbook may contain a typo. The conclusion affirms the accuracy of the user's calculations.
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Homework Statement
Find the amount of work done against gravity in moving a 150kg rocket from the surface of the Earth to 18000 km above the center of the Earth.
Relevant Equations
E = -Gmm/r
I started by calculating the energy at the Earths surface, which is just -G(150)(6 x 10^24)/(6400 x 10^3) = -0.9 x 10^10, and calculating the energy at the orbit radius, E = -G(150)(6 x 10^24)/(18000 x 10^3) = -3.4 x 10^9, then doing some subtraction we have -3.4 x 10^9 - (-0.9 x 10^10) = 5600 x 10^6 J. The answer provided in my textbook is 6 x 10^5 J which is clearly not what I got. Any tips?
 
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sdfsfasdfasf said:
Any tips?
Get a better textbook?
 
Possible typo in the textbook. The answers section.
 
So my answer is correct?
 
sdfsfasdfasf said:
So my answer is correct?
Yes.
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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