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Comment by ethbro | Hacker News Reader

> It’s worth noting that if you shoot a cannonball dead horizontally, it hits the ground at the same time as it would if dropped from the same height

Do you have a citation for this? I'd expect, as with all physics questions, there's a caveat.

In this case, "... over sufficiently small distances."

If you fire, say, the main guns of a battleship "over the horizon" (from your initial vantage point), I'd strongly suspect this doesn't hold.

In order for it to, the gravitational force vector, integrated over flight path, would have to perfectly counter the curvature of the Earth... which doesn't seem like it would line up so neatly.

In reality, air friction and maximum muzzle velocities probably render most of these concerns moot for practical purposes.

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"Do you have a citation for this?"

This is Newtonian physics.

Obviously, air, other things, friction mess it up.

But essentially that's it.

Horizontal motion of a body won't affect gravitations vertical pull.

in a completely Newtonian universe (ignoring relativity), if you shoot the ball at high enough speed it will never hit the earth. horizontal motion doesn't affect the vertical pull, but the direction of the vertical pull changes by the time the ball's traveled well over the horizon.

> in a completely Newtonian universe (ignoring relativity), if you shoot the ball at high enough speed it will never hit the earth.

Right. Assuming a perfectly spherical earth, at below orbital velocity, it hits the ground somewhere; at orbital velocity up to (but excluding) escape velocity it (assuming the cannon gets out of the way) orbits with the low point at (and opposite) the firing position, and beyond escape velocity it takes a curving path getting ever farther away.

Aka orthogonal vectors won't affect each other.

Does this hold true as an object's ballistic trajectory approaches significant fractions of a planet's diameter?

Granted, the object is constantly being accelerated towards the center of the planet.

But that force vector's direction changes with respect to the initial "horizontal" launch vector as the object continues on a straight path, until they're longer orthogonal.

The force of gravity is always orthogonal to the direction of motion when in orbit. So never loses momentum. Which is why the moon is still going around the earth a billion years later and has not 'fallen' into it.

We're talking suborbital velocities here.

If I'm wrong, I'd love to hear exactly why, but regurgitating basic physics doesn't resolve the difficulties in modelling a straight flight path around a curved surface, in relation to a dropped object.

This is only true for circular orbits. It’s pretty transparently obvious that elliptical orbits move closer to and farther from the center of mass (and lose and gain momentum accordingly).

It doesn't take into account the curvature of the earth. The commenter was assuming a simpler flat plane.

> If you fire, say, the main guns of a battleship "over the horizon"

...then you are not firing horizontally, but upwards. If you would insist on orienting those guns perpendicular to the gravity vector you would get a very big splash not too far away.

I'm not an artillerist, but as far as I know (and supported by a quick glance at Wikipedia), the range advantage of those big guns over smaller ones doesn't come from higher muzzle velocity (which is limited by the physical property of the propellant independent of gun size), but from the much higher kg/CdA value of their very big projectiles.

> there's a caveat

Air.

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