Does anyone know a simple formula to calculate the acceleration of gravity underwater?
I have this, "m(d2x/dt2)=(rhoB-rhoA)(g*V)-0.5(Cd)(rhoA)(A)(dx/dt)^2"
When m=body's mass, rhoA=fluid density, rhoB=body density, g=gravity, x=displacement, t=time, Cd=drag coefficient, V=body volume and A=body area
What makes you think that the acceleration due to gravity is different underwater?
And where did you get this formula?
a) there is more drag underwater and b)why, have you seen it before?
This. Ever see a movie set on a submarine? That's underwater, and the gravity looks normal.
True but that's nothing to do with gravity.
I think your original question is wrong. I think what you are asking is.. What is the net vertical force (eg the sum of weight and buoyancy) ?
The acceleration due to gravity is g, by definition. That equation you have is the net force due to gravity + buoyancy + drag.
Not to throw fuel on the fire, but I'm just curious. Isn't the gravitational constant g defined as being ~9.81 m/s^2 at sea level? If so, couldn't "below sea level" factor into the equation? My inclination is that the answer is basically "yes, but by such a small amount it's insignificant". Is this correct?
Yes, but this has nothing to do with the presence of water.
The sea level is just a convenient reference level. You can be under the sea level but not underwater. Or way above the sea level and underwater.
It would be misleading to attribute the change in g due to altitude change to being (or not) "underwater".
g doesn't change a lot when you go up/down few meters, in fact the change is so small that it can be noticed, I will assume that you mean the net acceleration under water, you can use newton's second law, F = ma, so ma = ρgV - mg (+drag force that I will ignore, considering a point particle), you get a = ρgV/m -g, if our particle has constant density then a = g*(ρmedium/ρparticle-1), this is a very particular case :)
Thanks that's pretty much what I figured. I knew that the presence of water wouldn't matter, I was just curious as to the sea level reference. I understand that it's simply the reference point that was chosen when they calculated the gravitational constant. I guess my question was more along the lines of "is the difference in gravitational force at the bottom of the Mariana's trench and the peak of Mt. Everest (because point A is closer to the center of the earth's mass than point B; the water of course is irrelevant) significant enough to factor into calculations?
That will depend on what is the calculation and what is the purpose of it.
And also, on what is the accuracy of the other parameters entering the calculation.
This article has a discussion on the variation of g due to various factors, including altitude and latitude:
At the top of Mt. Everest for instance (approx. 9000 m above sea level), the value of g is about 0.29% lower than g measured at sea level. At the altitude the Space Shuttle formerly orbited, about 400 km above the surface, g was still about 90% of its value at sea level. There is a relatively simple formula to calculate g at altitude.
Going in the other direction, the variation of g with depth is a little trickier to calculate, because you must account for the mass of whatever solid (or liquid) material lies between the measurement depth and sea level. See more about this in the discussion of the Shell Theorem.
careful with your choice of words there
Gravitational Constant = G = approximately = 6.674×10−11 N⋅m2/kg2 a whole different thing
acceleration due to gravity ( eg at Earth's surface) = g = ~9.81 m/s2
which is the one being discussed
I stand corrected. Thank you sir. You're right, poor choice of words on my part. And SteamKing thanks for satisfying my curiosity!
When I said underwater I mean deep, as in 11,000 m below sea level!
With ##M## as the mass of the Earth, ##R## the radius and assuming uniform density, the force of gravity a distance ##r## from the centre of the Earth, with ##r< R##, is given by:
It would be a good exercise to derive this equation for yourself.
This is pretty deep for diving or submarines.
But for change in gravity is not much. You compare 11 km with the radius of the Earth which is around 6,400 km. About 0.2%.
You should not expect a significant change. As g goes like r^2, the change in g may be some 0.4% or about 0.04 m/s^2.
As above, ##g## actually changes linearly with ##r## inside a solid sphere.
The bottom of the ocean is not inside a solid sphere, is it?
There may be some corrections if we are looking at a deep narrow oceanic valley but I intended just an estimation of the effect that 11 km will have on g.
It also apply if you go up 11 km.
If you are going inside, the effect is smaller by a factor of 2, in first approximation. But I don't think is relevant here.
The Earth is solid, in the sense that it's not hollow. In any case, outside the surface gravity varies as ##1/r^2##, but inside the Earth gravity varies as ##r##. That's the critical point.
One more time, I was not talking about what happens inside the Earth.
And I don't contest the (1/r) r dependence inside a solid, homogeneous, perfectly spherical Earth.
However, if you want to estimate the effect of going on the bottom of the ocean, a large, shallow deep in the crust rather than a narrow pith, I don't think that the 1/r applies.
There is no solid material above you. The 1/r^2 is also not exactly true. The dependence is more complicated. But it was just an order of magnitude estimate.
The difference between (1/r) r and 1/r^2 is less than an order of magnitude for that small difference so it does not even matter.
If you want to be very accurate, the (1/r) r does not even work for the real Earth, at that depth. g increases as you go down in the crust, for at least 11 km.
On the average, of course.
Edit. Corrected error (replaced 1/r by r) after observed by Perok.
Okay. But, it's not ##1/r## inside a sphere. It's ##r##.
I'd be interested in the calculations that show that ##g## increases as you go down into the crust. How far do you have to go down before gravity does indeed begin to decrease?
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