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- Thread starter tackyattack
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You see space is really really big. And even though 2000 mph seems like a lot, its actually exhaustingly slow even in stellar terms. Even if you pointed the gun in the exact opposite direction as the sun, the bullet is only traveling at about 1/40th of the escape velocity of the solar system (from the Earth), and so will eventually impact the sun.

I say eventually because something moving that slow will take a while to get anywhere in space.

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So the bullet wouldn't be able to leave our solar system because the sun would eventually gravitationally pull the bullet to it? But it could at least escape earth?

You see space is really really big. And even though 2000 mph seems like a lot, its actually exhaustingly slow even in stellar terms. Even if you pointed the gun in the exact opposite direction as the sun, the bullet is only traveling at about 1/40th of the escape velocity of the solar system (from the Earth), and so will eventually impact the sun.

I say eventually because something moving that slow will take a while to get anywhere in space.

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russ_watters

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At 62 miles (100km) above the surface:

[tex]v_{e} = \sqrt{\frac{2G*5.972*10^{24} kg}{6.471*10^{6} m}} = 11.1 km/s[/tex]

At the surface of the earth, V

Your bullet would eventually fall back to earth.

Edit: Russ beat me to it.

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How far could the bullet go before it started falling back?

At 62 miles (100km) above the surface:

[tex]v_{e} = \sqrt{\frac{2G*5.972*10^{24} kg}{6.471*10^{6} m}} = 11.1 km/s[/tex]

At the surface of the earth, V_{e}is 11.2 km/s.

Your bullet would eventually fall back to earth.

Edit: Russ beat me to it.

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Well, if I haven't bungled my calculation, about 42 km.

The acceleration of gravity 100 km up is around 9.5 m/s^{2}. That won't change significantly so we can assume a constant acceleration to simplify things.

The time it will take for the bullet to reach 0 velocity:

[tex]t = \frac{v}{a} = \frac{894m/s}{9.5m/s^2} = 94 s[/tex]

The distance it will travel in that time:

[tex] d = \frac{at^2}{2} = \frac{9.5m/s^2*(94s)^2}{2} = 42 km[/tex]

In reality, it will go a bit further because the acceleration of gravity will decrease as it moves further away from the earth. In this case, however, it won't decrease by much. Maybe around .1 m/s^{2} or so.

If I'm wrong, someone will correct me and I will be horse-whipped.

Edit: This is assuming you shoot the bullet straight up.

The acceleration of gravity 100 km up is around 9.5 m/s

The time it will take for the bullet to reach 0 velocity:

[tex]t = \frac{v}{a} = \frac{894m/s}{9.5m/s^2} = 94 s[/tex]

The distance it will travel in that time:

[tex] d = \frac{at^2}{2} = \frac{9.5m/s^2*(94s)^2}{2} = 42 km[/tex]

In reality, it will go a bit further because the acceleration of gravity will decrease as it moves further away from the earth. In this case, however, it won't decrease by much. Maybe around .1 m/s

If I'm wrong, someone will correct me and I will be horse-whipped.

Edit: This is assuming you shoot the bullet straight up.

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Air resistance, curvature of the earth (distances are portions of great circles), the rotatioan of the earth, the rotation of the bullet itself.

Let the rigid body theorists do that, they love complications

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The question is ambiguous because you did not specify what 2,000 mph is relative to, nor did you specify its direction.

If to the Earth, then it will follow an arc of an ellipse and collide with the Earth. What ellipse and what arc will depend on the direction of the velocity.

Note that the Earth is moving around the Sun at more than 60,000 mph, so the resultant velocity of the bullet relative to the Sun will not be much different.

Now, if 2,000 mph is relative to the Sun, then the velocity relative to the Earth will be almost equal to the velocity of the Earth relative to the Sun, so the bullet can collide with the Earth only accidentally. Barring that, it will move in an arc of an extremely narrow ellipse and hit the Sun.

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