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

In summary: Where do you study?In summary, the problem involves a bullet being fired vertically from the surface of the Earth with an initial velocity of 10 km/s. The question is at what distance from the center of the Earth will the bullet arrive, assuming no air resistance and using the formula for escape velocity. However, the formula may not be applicable in this case as it involves the conservation of energy, which may not have been covered by the study material yet. Additionally, the formula for escape velocity was given without proof, making it difficult to understand and apply in this problem.
  • #1
ezio1400
4
0
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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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
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  • #5
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.
 
  • #6
But you have learned about the escape velocity - how was it explained to you without energy?
 
  • #7
The formula for escape velocity was given to me without proof .
 
  • #8
ezio1400 said:
The formula for escape velocity was given to me without proof .
Strange.
 

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

1. How does the universal gravitation exercise work?

The universal gravitation exercise is based on the principle of gravity, which states that all objects with mass attract each other. This exercise involves calculating the force of attraction between two objects based on their masses and the distance between them.

2. What is the formula for universal gravitation?

The formula for universal gravitation is F = G * (m1 * m2) / d^2, where F is the force of attraction between two objects, G is the gravitational constant, m1 and m2 are the masses of the objects, and d is the distance between them.

3. How does the distance between objects affect the force of universal gravitation?

The force of attraction between two objects decreases as the distance between them increases. This is because the gravitational force is inversely proportional to the square of the distance between the objects.

4. What is the unit of measurement for the gravitational constant?

The unit of measurement for the gravitational constant, G, is N*m^2/kg^2 (newton meter squared per kilogram squared). This unit is derived from the formula for universal gravitation and is used to calculate the force of attraction between two objects.

5. How does the mass of an object affect the force of universal gravitation?

The force of attraction between two objects increases as the masses of the objects increase. This is because the gravitational force is directly proportional to the product of the masses of the objects. Therefore, the greater the mass of an object, the greater its gravitational pull.

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