How Fast to Launch an Object from Mercury for a Final Speed of 3000 m/s?

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SUMMARY

The discussion focuses on calculating the initial speed required to launch an object from Mercury to achieve a final speed of 3000 m/s when it is far from the planet. The relevant equation used is vi² = vf² + GM/ri, where G is the gravitational constant, M is the mass of Mercury (0.3x10^24 kg), and ri is the radius of Mercury (2440 km). The initial speed calculated by one participant was 4151.8315 m/s, which was identified as incorrect, indicating a need for further analysis of the formula and the kinetic energy component.

PREREQUISITES
  • Understanding of gravitational physics and orbital mechanics
  • Familiarity with the equation vi² = vf² + GM/ri
  • Knowledge of kinetic energy calculations (1/2 mv²)
  • Basic concepts of planetary mass and radius
NEXT STEPS
  • Review the derivation of the equation vi² = vf² + GM/ri
  • Study the impact of gravitational potential energy on launch speed
  • Explore the concept of escape velocity from planetary bodies
  • Investigate the role of kinetic energy in motion equations
USEFUL FOR

Students in physics, aerospace engineers, and anyone interested in celestial mechanics and the dynamics of launching objects from planetary surfaces.

ohheytai
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The radius of Mercury (from the center to just above the atmosphere) is 2440 km (2440x10^3 m), and its mass is 0.3x10^24 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 3000 m/s?


Homework Equations


vi² = vf² + GM/ri


The Attempt at a Solution


i got 4151.8315 m/s
and its wrong help me please!
 
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ohheytai said:

Homework Equations


vi² = vf² + GM/ri
The kinetic energy of an object is (1/2)mv2. Do you see what is missing in your formula?

The Attempt at a Solution


i got 4151.8315 m/s
and its wrong help me please!
 

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