What would happen to a bullet shot into space?

tackyattack

Ignoring whether or not the gun would be able to fire in space, what would happen if a bullet was shot, traveling 2,000 mph into space from 62 miles above the earth? How far would it go?

Vorde

Well the short answer is it would keep on going forever. The longer answer is that it will crash into something, probably the sun.

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.

tackyattack

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?

Vorde

It depends on the angle you are firing and the escape velocity at that altitude. I don't have the means to tell you that right now but a quick google search should get you the escape velocity equation (it's very simple).

russ_watters

No, 62 miles isn't that high so the escape velocity isn't much lower than on earth's surface and 2000 mph is well below it. It will follow a parabolic trajectory back to earth.

Jimmy

2000mph (894m/s) isn't even close to the escape velocity of the earth.

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.

tackyattack

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

Jimmy

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

The acceleration of gravity 100 km up is around 9.5 m/s². 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/s² 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.

GarageDweller

Once you start talking about large distances on the earth, you cannot ignore some of the things we usually ignore.
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

voko

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.