Universal gravitation exercise

[ezio1400]

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.

 

[phinds]

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.

 

[ezio1400]

ok.

escape velocity= ((2*G*M_T)/R_T)^1/2⇒d=(2*G*M_T)/v²=2*6,67*10^-11*5,98*10²⁴/(10*10³)²=7977320m
h=7977320+6360*10³=14337320m→14337km

 

[ehild]

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?

 

[ezio1400]

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.

 

[ehild]

But you have learned about the escape velocity - how was it explained to you without energy?

 

[ezio1400]

The formula for escape velocity was given to me without proof .

 

[ehild]

The formula for escape velocity was given to me without proof .

Strange.