The Total Energy of an Object at Escape Velocity

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SUMMARY

The discussion centers on calculating the total energy and escape velocity of a 460kg satellite in a circular orbit at an altitude of 850km. The kinetic energy (Ek) is determined to be 1.27 x 1010 J, leading to a total energy of -1.27 x 1010 J, indicating the satellite's binding energy. The escape speed is derived using the formula v(esc) = √(2GM/r), where G is the gravitational constant and M is the mass of the Earth. The confusion arises regarding the additional energy required for escape, which is indeed equal to the binding energy.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts
  • Familiarity with the formula for escape velocity: v(esc) = √(2GM/r)
  • Basic knowledge of orbital mechanics and satellite dynamics
  • Proficiency in unit conversions, particularly from kilometers to meters
NEXT STEPS
  • Study the derivation of gravitational potential energy and its implications in orbital mechanics
  • Learn about the relationship between kinetic energy and escape velocity in satellite motion
  • Explore the effects of altitude on escape velocity and energy requirements
  • Investigate real-world applications of escape velocity in space missions and satellite launches
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in satellite dynamics and orbital mechanics will benefit from this discussion.

aeromat
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Question:
A 460kg satellite is launched into a circular orbit and attains an orbital altitude of 850km above Earth's surface. Calculate the:


e) additional energy and speed required for the satellite to escape.

Attempt:
I found Ek to be 1.27*10^10 J
Thus, the total energy is -1.27*10^10 J
The binding energy is 1.27*10^10 J.

Wouldn't the additonal energy it needs be the binding energy?
The escape speed would be calculated by root of:

v(esc) = root of 2GM/r

But the answer at the back says otherwis.e... :roll:
 
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