Thrown Object & Earth's Escape Velocity

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SUMMARY

The discussion centers on the concept of escape velocity and its relationship to orbital mechanics. Escape velocity, defined as the speed required to overcome Earth's gravitational pull, is calculated using the formula ve = √2gr. An object thrown upwards or downwards at escape velocity will not achieve orbit; only speeds less than escape velocity can result in elliptical orbits. Specifically, a horizontal launch at a speed equal to escape velocity divided by the square root of two is necessary for a circular orbit.

PREREQUISITES
  • Understanding of gravitational forces and their effects on motion
  • Familiarity with the concept of escape velocity
  • Knowledge of orbital mechanics and elliptical orbits
  • Basic physics principles, particularly kinematics
NEXT STEPS
  • Study the derivation and implications of the escape velocity formula ve = √2gr
  • Explore the characteristics of elliptical orbits and their mathematical descriptions
  • Learn about the differences between circular and elliptical orbits in gravitational fields
  • Investigate the effects of launch angles on orbital trajectories
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in understanding the principles of motion in gravitational fields and orbital dynamics.

Ajit Kumar
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If an object is thrown upwards with escape velocity, will it orbit the earth?
 
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Think about it - If an object is thrown downwards with escape velocity, will it orbit the earth?
 
What does the word "escape" mean? Does that definition match the definition of "orbit"?
 
So now you have two questions, approaching the problem from two different directions, These are hints. What do you think the answer is?
 
escape velocity is the velocity by which any object overcomes the gravitational pull and flies away from planet.
ve = √2gr
 
Will a body projected with a speed less than the escape velocity may orbit the Earth in circular path?
 
Of course. All elliptic paths are closed orbits, so all speeds less than escape can take any elliptic path (of which the circular is special case)
 
Ajit Kumar said:
Will a body projected with a speed less than the escape velocity may orbit the Earth in circular path?
For a body launched from the surface of the Earth the only possibility for a circular orbit is a horizontal launch with a speed equal to escape velocity divided by the square root of two. Any other direction and the orbit will intersect with the surface of the Earth. Any higher velocity and the orbit will fail to be circular. It will rise above the surface of the earth. Any lower velocity and the orbit will fail to be circular. It will fall below the surface of the earth.
 

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