At What Distance Will a 10 km/s Bullet Fired Vertically from Earth Arrive?

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SUMMARY

A bullet fired vertically from the Earth's surface at an initial velocity of 10 km/s will reach a maximum height of approximately 14,373 km from the Earth's center. This calculation utilizes the escape velocity formula, which equates kinetic energy to gravitational potential energy. The relevant parameters include the Earth's radius (RT = 6360 km) and mass (MT = 5.98 x 1024 kg). The discussion highlights the importance of understanding energy conservation in solving such problems.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy
  • Familiarity with the escape velocity formula
  • Basic knowledge of physics concepts related to motion and forces
  • Ability to perform calculations involving constants such as G (gravitational constant)
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  • Study the derivation of the escape velocity formula
  • Learn about energy conservation in gravitational fields
  • Explore the effects of air resistance on projectile motion
  • Investigate the implications of launching objects from different planetary bodies
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Students of physics, educators teaching mechanics, and anyone interested in the principles of motion and energy conservation in gravitational fields.

ezio1400
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If a bullet is fired vertically from the surface of the Earth with initial velocity v = 10 km / s, ignoring air resistance, at which distance h from the center of the Earth would arrive? (The radius of the Earth is RT = 6360 km, and the mass of the Earth MT = 5.98x10^24 kg)

I used the formula for escape velocity putting the speed of the bullet instead of the escape velocity but I do not think is correct.
 
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ezio1400 said:
If a bullet is fired vertically from the surface of the Earth with initial velocity v = 10 km / s, ignoring air resistance, at which distance h from the center of the Earth would arrive? (The radius of the Earth is RT = 6360 km, and the mass of the Earth MT = 5.98x10^24 kg)

I used the formula for escape velocity putting the speed of the bullet instead of the escape velocity but I do not think is correct.
You have to show your work in order for anyone to comment on it.
 
ok.

escape velocity= ((2*G*MT)/RT)1/2⇒d=(2*G*MT)/v2=2*6,67*10-11*5,98*1024/(10*103)2=7977320m
h=7977320+6360*103=14337320m→14337km
 
You are close to the right track, but why did you use the formula for the escape velocity? It gives the velocity when the kinetic energy is equal to the gravitational potential energy on the Earth surface.
In the problem, the bullet is shot upward with 10 km/s speed from the surface of the Earth, where it has some gravitational potential energy. The total energy KE+PE is conserved, how far is the bullet when its kinetic energy becomes zero?
 
Last edited:
Ok but in my book this exercise is in a chapter that precedes the chapter on energy. Maybe that's why I can not solve it. This is strange.
 
But you have learned about the escape velocity - how was it explained to you without energy?
 
The formula for escape velocity was given to me without proof .
 
ezio1400 said:
The formula for escape velocity was given to me without proof .
Strange.
 

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