Checkpoint: Conservative Forces And Potential Energy

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SUMMARY

This discussion analyzes the kinetic energy of two identical objects released from different heights above the Earth's surface. In Case 1, an object is released from a height of 1 Earth radius, resulting in kinetic energy K1. In Case 2, an object is released from a height of 2 Earth radii, yielding kinetic energy K2. The conclusion drawn is that K2 equals 1/2 K1, demonstrating the inverse relationship between the height of release and the kinetic energy just before impact, as dictated by the conservation of mechanical energy.

PREREQUISITES
  • Understanding of gravitational potential energy (Ugrav)
  • Familiarity with kinetic energy equations (KE = 1/2mv^2)
  • Knowledge of conservation of energy principles
  • Basic concepts of gravitational force and distance
NEXT STEPS
  • Study gravitational potential energy calculations using the formula Ugrav = -G*me*mb/(R-earth)
  • Explore the implications of conservation of mechanical energy in different gravitational fields
  • Investigate the effects of varying heights on kinetic energy in physics
  • Learn about the relationship between distance from the center of mass and gravitational force
USEFUL FOR

Students studying classical mechanics, physics educators, and anyone interested in understanding the principles of energy conservation and gravitational effects on kinetic energy.

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Homework Statement



Consider two identical objects released from rest high above the surface of the Earth (neglect air resistance for this question).

In Case 1 we release an object from a height above the surface of the Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1.
In Case 2 we release an object from a height above the surface of the Earth equal to 2 Earth radii, and we measure its kinetic energy just before it hits the Earth to be K2.
1) Compare the kinetic energy of the two objects just before they hit the surface of the earth.


Homework Equations



Ugrav= -G*me*mb/ (R-earth)
KE= 1/2mv^2
conservation of energy- k+u=0

The Attempt at a Solution



K=-U
1/2mv^2 = Gmemb/r-earth

K1= 1/2mv^2
K2= 1/4mv^2

Wouldn't the K2= 1/2(k1) due to the conservation of mechanical energy. The radius of the Earth is inversely proportional to the kinetic energy
 
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The distances from the center of the Earth are 2R and 3R.
 

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