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

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Homework Help Overview

The discussion revolves around a physics problem involving a bullet fired vertically from the Earth's surface at a speed of 10 km/s. Participants are exploring the implications of this scenario while considering gravitational forces and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of escape velocity to analyze the bullet's trajectory and question whether this approach is appropriate given the problem's context. Some express uncertainty about their understanding of energy conservation in relation to the problem.

Discussion Status

The conversation is ongoing, with participants sharing their calculations and questioning the relevance of escape velocity in this context. There is a recognition that the problem may not align with their current understanding of energy concepts, and some participants are seeking clarification on foundational principles.

Contextual Notes

Some participants note that the problem appears in a chapter preceding the one covering energy concepts, which may contribute to their confusion. There is also mention of the escape velocity formula being presented without proof, leading to further questions about its application.

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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