Is the Escape Velocity of a 10,500kg Spaceship from Earth Calculated Correctly?

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SUMMARY

The escape velocity of a 10,500 kg spaceship from Earth is calculated using the formula \( v = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth (5.98 x 1024 kg), and \( R \) is the radius of the Earth (6.38 x 106 m). The calculated escape velocity is approximately 3.464 x 103 m/s, which is incorrect. The correct escape velocity is approximately 11,186 m/s. The discussion emphasizes the importance of correctly substituting values into the formula.

PREREQUISITES
  • Understanding of gravitational physics
  • Familiarity with the formula for escape velocity
  • Basic algebra for substituting values into equations
  • Knowledge of the mass and radius of Earth
NEXT STEPS
  • Review the derivation of the escape velocity formula
  • Learn about gravitational constant \( G \) and its significance
  • Explore the implications of escape velocity in space travel
  • Study the effects of mass and radius on gravitational force
USEFUL FOR

Students in physics, aerospace engineers, and anyone interested in the calculations related to space travel and gravitational forces.

kubombelar
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Homework Statement

calculate the escape velocity of a 10,500kg spaceship from earth. mass of Earth is 5.98*10kg and radius of the Earth is 6.38*10kg.

elevant equations
2GM/R

The Attempt at a Solution

Using the formula i found escape velocity to be 3.464*10m/s. Is this correct?
 
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If it was correct, you would have to be careful driving your car up hills - you could easily go fast enough to escape from the Earth!
That equation has no equal sign so it isn't clear what it is. Can you write it completely and then show the numbers you substituted in? Probably just a small error or two.
 

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