At what height should a gun be fired for bullet to have escape velocity

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Discussion Overview

The discussion revolves around the height from which a bullet must be fired to achieve escape velocity, exploring the implications of gravitational forces and the calculations involved. Participants examine the relationship between bullet speed, escape velocity, and the distance from Earth, while also considering the effects of gravitational pull at various altitudes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Walter presents a calculation suggesting that a bullet must be fired from a height of about 5.5 million miles to achieve escape velocity, which he finds surprising compared to satellite orbits.
  • Phyzguy agrees with Walter's calculation and explains that the escape velocity at Earth's surface is significantly higher than the bullet's speed, necessitating a much greater distance from Earth.
  • Walter expresses curiosity about the nature of gravity extending to infinity and questions the quantization of gravity, asking about the distance at which gravitational effects would become negligible.
  • A participant discusses a calculation involving the potential energy of a hydrogen atom in relation to the energy of a graviton, arriving at a distance of 6.0x1023 light years, which raises further questions about gravitational effects at extreme distances.
  • Another participant notes the distinction between firing the bullet from a platform at rest in an Earth-centered inertial frame versus a rotating frame, suggesting this could significantly affect the calculations.
  • Walter acknowledges the complexity of the problem and reflects on the need for a better understanding of dynamics, mentioning a series of lectures that could help clarify these concepts.
  • A later reply encourages comparing the bullet's speed to the moon's speed, noting that the moon's actual speed is about 1000 m/s, which is faster than the bullet, thus explaining why the bullet must be fired from a greater distance.

Areas of Agreement / Disagreement

Participants express varying levels of agreement on the calculations presented, but there is no consensus on the implications of gravitational effects, the nature of escape velocity, or the relevance of different frames of reference. Multiple competing views remain regarding the interpretation of gravitational influence at large distances.

Contextual Notes

Participants highlight limitations in their understanding of gravitational effects, the implications of firing from different frames of reference, and the assumptions underlying their calculations. The discussion remains open-ended with unresolved mathematical steps and interpretations.

Who May Find This Useful

This discussion may be of interest to those exploring gravitational physics, escape velocity concepts, and the implications of different reference frames in dynamics.

softport
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I think it is correct. Why are you surprised? Escape velocity at the surface of the Earth is about 11,000 m/sec, so for a bullet traveling only 300 m/sec to have escape velocity you need to go far enough away that the Earth's gravitational pull has fallen off by a lot.
 
Thanks phyzguy, my thoughts were that it's not far from Earth that you have things floating around in space (albeit with outward force from orbiting). I guess given that gravity extends to infinity, the surprising thing should be that there is an escape velocity at all.

Naive question: if gravity is quantized what is the distance at which it would fall below the minimum, and cease to have any effect? Is there a quantum of potential energy?
 
I tried to answer my question above, and found this paper that mentions a possible graviton mass of 1.3x10-69 kg .

Now, how far would a single hydrogen atom have to be, from earth, till it's potential energy due to Earth's gravity is below the energy of one graviton (E = mc2)? What came out was Avogadros number of light years, isn't that weird?

E(graviton) = mc2 = (1.3x10-69 kg) (3.0x108 m/s2)2 = 1.2x10-52 J

Mass of hydrogen atom: m = 1.7x10-27 kg ->[(1 gr/mole)( mole/6x1023 atoms)( kg/1000 gr)]
Earth mass: M = 6.0x1024 kg
Gravitation constant: 6.7x10-11 N(m/kg)2
Light year: 9.5x1015 m

Potential energy of gravity: E(gravity) = (m M G)/r

Equating energies and solving for r:

r = (m M G)/ E(graviton) = (1.7x10-27 kg)(6.0x1024kg)(6.7x10-11 N m2/kg2) / (1.2x10-52 J)
= 5.7x1039m
r = 6.0x1023 light years
 
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softport said:
I thought the height would turn out close to where satellites orbit

Satellites have to travel pretty fast to stay up...

http://en.wikipedia.org/wiki/Geostationary_orbit

A geostationary orbit can only be achieved at an altitude very close to 35,786 km (22,236 mi), and directly above the Equator. This equates to an orbital velocity of 3.07 km/s
 
softport said:
Hello, can you can tell me if this is correct? I thought the height would turn out
close to where satellites orbit, but it comes out to about 5.5 million miles away!
I was very surprised by this, that's about 20 times further than the moon.

Here's what I used
Escape velocity: Ve = sqrt(2GM/r)
Earth mass: 5.97x10^24 kg
Gravitation constant: 6.67x10^-11 N(m/kg)^2
(Ve=) Handgun bullet speed: 300 m/s

There is a niggling problem of interpretation. Are you supposed to fire the bullet from a platform that is at rest in an Earth-centered inertial frame or from a platform that is at rest in an Earth-centered rotating frame?

The distinction is rather significant at these kinds of altitudes.
 
Thanks for all your answers.
I did not think about rotating frames, only the effects between 2 masses in the same inertial frame. To frame the problem in a more realistic way, I would have to go back and watch this great series of dynamics lectures by prof Sarma at MIT. I have watched these at least three times over the last couple of years but every time, after only a few weeks, the understanding fades away again.

Anyways, I know that a post just to thank members for their replies, should be short,
so apologies for that.

PS "www.physics.louisville.edu/wkomp/teaching/spring2006/589/final/schumann.pdf"
Not sure what to think of this, as it suggests a possible speed way less than light. Maybe I'm
misreading it. I also realized that just searching for 'graviton' in these forums would probably have yielded way more information!
 
Last edited by a moderator:
Thanks for all your answers.
I did not think about rotating frames, only the effects between 2 masses in the same inertial frame. To frame the problem in a more realistic way, I would have to go back and watch this
great series of lectures by prof Sarma at MIT. I have watched these at least three times over the last couple of years but every time, after only a few weeks, the understanding fades away again.

Anyways, I know that a post just to thank members for their replies, should be short,
so apologies for that.

PS "www.physics.louisville.edu/wkomp/teaching/spring2006/589/final/schumann.pdf"
Not sure what to think of this, as it suggests a possible speed way less than light. Maybe I'm
misreading it. I also realized that just searching for 'graviton' in these forums would probably have yielded way more information!
 
Last edited by a moderator:
(Boldface added by me for emphasis.)
softport said:
Hello, can you can tell me if this is correct? I thought the height would turn out
close to where satellites orbit, but it comes out to about 5.5 million miles away!
I was very surprised by this, that's about 20 times further than the moon.

Here's what I used
Escape velocity: Ve = sqrt(2GM/r)
Earth mass: 5.97x10^24 kg
Gravitation constant: 6.67x10^-11 N(m/kg)^2
(Ve=) Handgun bullet speed: 300 m/s

regards,

Walter
Have you tried calculating the speed of the moon for comparison? It's about 1000 m/s, about 3 times faster than the bullet. But that is the moon's actual speed, its escape speed would be faster yet.

Considering that, it's not surprising that the slower bullet must be farther away than the moon to be at escape speed.
 

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