Problem with Escape Velocity given final speed

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SUMMARY

The discussion centers on calculating the initial speed required for an object launched from Mercury to achieve a final speed of 2500 m/s when far from the planet. The relevant equation involves gravitational potential energy and kinetic energy, specifically 1/2mvfinal² - GMm/rfinal = 1/2mvinitial² - GMm/rinitial. Participants identified errors in the algebraic manipulation of the formula and emphasized the importance of including the radius of Mercury in the calculations. The mass of Mercury was clarified to be 0.3 x 1024 kg, correcting the initial misunderstanding of its value.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy equations
  • Familiarity with algebraic manipulation of physics formulas
  • Knowledge of Mercury's physical properties, including its mass and radius
  • Basic understanding of escape velocity concepts
NEXT STEPS
  • Review the derivation of escape velocity formulas in classical mechanics
  • Learn about gravitational potential energy and its applications in astrophysics
  • Study the impact of planetary mass and radius on escape velocity calculations
  • Practice algebraic manipulation of physics equations with various examples
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This discussion is beneficial for physics students, educators, and anyone interested in understanding escape velocity calculations, particularly in the context of planetary science and gravitational physics.

celestasia
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Homework Statement


The radius of Mercury (from the center to just above the atmosphere) is 2440 km (2440103 m), and its mass is 0.31024 kg. An object is launched straight up from just above the atmosphere of Mercury.
(a) What initial speed is needed so that when the object is far from Mercury its final speed is 2500 m/s?
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Homework Equations


1/2mvfinal^2-GMm/rfinal=1/2mvinitial^2-GMm/rfinal

The Attempt at a Solution



I tried to manipulate the formula algebraically. I got something like sqrt{2GM- vfinal^2}=vinitial. It's not working out. I think it has something to do with my algebra.
 
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The mass of Mercury, as given, is ridiculously small.

The formula you have obtained is wrong dimensionally. Where is the radius of the planet in it? Perhaps you could just show the entire manipulation that gets you there.
 
voko said:
The mass of Mercury, as given, is ridiculously small.

Since the poster apparently is not aware of how to format text on this forum I assume the given mass is to be read as 0.3∙1024 kg, which fits actual data.

I would suggest that the poster carefully checks the initial equation (which, apart from distracting formatting issues, also contains an error) and redo his calculations step by step. If the answer still mysteriously comes out wrong then by all means post the steps here for a hint or two.
 

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