# Thread: Is there any speed where is a rock is thrown at a tangent to the surface of the earth

1. ## Is there any speed where is a rock is thrown at a tangent to the surface of the earth

Is there any speed where is a rock is thrown parallel at a point to the surface of the earth that it will escape into space? OR regardless of the speed of the rock would it always fall back to earth?

So I undertand that you need an escape velocity in the y direction to escape. 11.2 kilometers per second but since we are not throwing directly up in the air, it will only move in the y direction by the curvature of the earth which changes with distance.

So the rock would need to be thrown at a speed that would overcome 11.2 klometers per second of the earths curvature. That is that the earth would need to be 11.2 kilometers below the straight line in a second.

I think this would be the distance from a height of 11.2 kilometers above the surface to see the horizon right?

d^2 = 2Rh = 2 * 6400 * 11.2 + 11.2^2 = 143485 km
take the sqr of that = 379 km / second

which is less than 10% of the speed of light
29979.2458 km / s ... so do we still need to take in relativistic mass?

2. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Since the surface of the Earth is jagged and not smooth, your thrown rock would undoubtedly end up embedded inside of a chunk of land. Stupid lumpy planet.
But, if you go up to the highest mountain... You might try out Newtons Cannonball.

3. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by Neverfly
Since the surface of the Earth is jagged and not smooth, your thrown rock would undoubtedly end up embedded inside of a chunk of land. Stupid lumpy planet.
But, if you go up to the highest mountain... You might try out Newtons Cannonball.
yeah that is exactly what I was thinking.

However, does relativity come into play? @ 11,200 m/s is there any increase in the mass of the earth?

4. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by tom
yeah that is exactly what I was thinking.

However, does relativity come into play? 11,200 m/s is there any increase in the mass of the earth?
Shouldn't the mass of the cannonball be the one in question? Keep in mind that the Lorentzian Transform here is the Ball to the Earth.
At 11,200 m/s, the effect would be negligible. The Saturn Five rocket makes better time than that.

5. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Escape velocity is escape velocity, no matter which direction you throw the rock. Of course, atmospheric friction comes into play, more so horizontally (lithospheric friction even more), but we're not considering that, right?

6. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by grapes
Escape velocity is escape velocity, no matter which direction you throw the rock. Of course, atmospheric friction comes into play, more so horizontally (lithospheric friction even more), but we're not considering that, right?
yeah I was ignoring atmospheric friction

Was thinking more relativisticly

7. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by Neverfly
Shouldn't the mass of the cannonball be the one in question?
well if the earth is traveling at near the speed of light away from me ... would my mass change or the earth or both? I thought the earth.

8. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by tom
well if the earth is traveling at near the speed of light away from me ... would my mass change or the earth or both? I thought the earth.
Relativity can be tricky on this... From your perspective, you might say the Earth appears to be moving away from you at near Light Speed. But the math says that Lorentzian Contraction would be on you. This is because the Earth's mass and size greatly overwhelms your own. This was one of those things that threw Dingle for a loop. And still causes confusion, today. There are a lot of explanations on the Twin Paradox; one of the better I have seen was by a guy that used Lego's to make a video demonstration using a different explanation approach. Different but works just as well.

9. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by tom
yeah I was ignoring atmospheric friction

Was thinking more relativisticly
You missed my point. Escape velocity for a horizontally thrown rock is still the escape velocity.

Orbital velocity at sea level is escape velocity divided by square root of two

So, to keep it from "hitting" the earth, you'd need even less than escape velocity. Your math in the OP is off.

10. ## Re: Is there any speed where is a rock is thrown at a tangent to the surface of the e

Originally Posted by tom
Is there any speed where is a rock is thrown parallel at a point to the surface of the earth that it will escape into space? OR regardless of the speed of the rock would it always fall back to earth?

So I undertand that you need an escape velocity in the y direction to escape. 11.2 kilometers per second but since we are not throwing directly up in the air, it will only move in the y direction by the curvature of the earth which changes with distance.

So the rock would need to be thrown at a speed that would overcome 11.2 klometers per second of the earths curvature. That is that the earth would need to be 11.2 kilometers below the straight line in a second.

I think this would be the distance from a height of 11.2 kilometers above the surface to see the horizon right?

d^2 = 2Rh = 2 * 6400 * 11.2 + 11.2^2 = 143485 km
take the sqr of that = 379 km / second

which is less than 10% of the speed of light
29979.2458 km / s ... so do we still need to take in relativistic mass?
That's closer to 1%, but the actual value is much much less than that even.
Originally Posted by grapes
Orbital velocity at sea level is escape velocity divided by square root of two
To calculate orbital velocity, we just set the acceleration g equal to the centripetal acceleration of a body in a circular path ω x (ω x r) where ω is the angular velocity. (ω x r) is just velocity v, so ω x (ω x r) = v2/r

In other words, we set g (9.8m/s2, or .0098km/s2) equal to v2/r, so we get v = sqrt(g*r)

Then orbital velocity v at r, the surface of the earth, is approx sqrt(.0098 * 6400), or 7.9m/s2, which is 11.2m/s2 divided by the square root of 2

So a rock thrown horizontally at 7.9m/s would orbit the earth. A rock thrown horizontally at 11.2m/s would escape the earth.

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