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

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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
 
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Answers and Replies

  • #2
phyzguy
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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.
 
  • #3
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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?
 
  • #4
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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 wierd?

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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  • #5
CWatters
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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
 
  • #6
jbriggs444
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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.
 
  • #7
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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" [Broken]
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!
 
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  • #8
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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" [Broken]
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:
  • #9
Redbelly98
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(Boldface added by me for emphasis.)
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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