Explaining uphill rivers scientifically?
I see this in so many maps. Normally, the water flows downhill because of gravity but here it is the opposite, rivers will go above the hills. When you mention this incoherence, the authors refuse to change it because that's how they want it. Apparently, the laws of physic does not need to apply with rivers.
Is there any phenomenon that could explain this even if it's only on a short distance?
To add to the other answers, there's also the possibility of the water appearing to run uphill due to a an optical illus …
9y ago
As user3453518 says, this could happen for a short distance by inertia. I advise you to read his answer; there are some …
9y ago
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2 answers
To add to the other answers, there's also the possibility of the water appearing to run uphill due to a an optical illusion.
I'm not aware of any river where this is the case, but since it happens with roads I see no reason why it couldn't also happen with rivers.
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As user3453518 says, this could happen for a short distance by inertia. I advise you to read his answer; there are some very good points made well in it. I'll add some science to it.
The formula for gravitational potential energy is
$$ \large E_\text{GP} = mgh $$
where $m$ is the mass of the object in question, $g$ is the gravitational constant, and $h$ is its height above the turning point. On Earth, $g = 9.81\text{ Nkg}^{-1}$.
So let's say we have 1 cubic metre of pure water, weighing in at 1000kg. It's starting down a hill of 50m vertical height (we'll assume this hill is somehow made of super-hydrophobic material so as to easily discount friction). It has:
$$ E_\text{GP} = 1000 \times 9.81 \times 50 $$ $$ E_\text{GP} = 490.5\text{ kJ} $$
At the bottom of this slope, since energy is conserved and we're discounting friction, it will also have 490.5 kJ. It'll be travelling at a velocity we can determine using the formula for kinetic energy:
$$ E_\text{K} = \frac{1}{2} mv^2 $$ $$ v = \sqrt{\frac{2E_\text{K}}{m}} $$ $$ v = 31.32\text{ ms}^{-1} $$ or about 70 mph.
Using this, we can show how high it could go if projected straight up with no loss of kinetic energy:
$$ v_f^2 - v_i^2 = 2as $$
We're looking for $s$, the displacement from the original position. $a$ is deceleration due to gravity, so just −9.81, $v_f$ is the final velocity (0) and $v_i$ the initial velocity which we just found.
$$ s = \frac{v_f^2 - v_i^2}{2a} $$ $$ s = \frac{0 - 980.9424}{2\times -9.81} $$ $$ s = 49.9971\text{m} $$
That's close to the original height of 50m (it's not exact because of my rounding)"”but here's the crucial part: it's not over it. Unless you can manipulate gravity specifically for water without affecting anything else on the planet, it's physically and scientifically impossible to go over the value of the original height. Moreover, in these calculations I have excluded several important considerations such as friction and air resistance. In reality, you'll never get anywhere near this value.
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