Why is acceleration due to gravity a constant?

I just learned of Newton's law of gravitation and that distance between two bodies is a factor in the gravitational force. My question is if that's true why is the Earth's gravitational acceleration a constant $9.8\mathrm{ms^{-2}}$ as change in force mathematically should mean change in acceleration if you refer to the equation $F=ma$.

Answer

Although @JamalS's answer covers the variation of $g$ due to latitude and terrain variations, here is a simpler reason why $g$ is taken to be constant close to the Earth's surface.

If the Earth has a radius $R$, and there is a mass $m$ at a distance $r$ above the surface, the force is given by: $$F=\frac{GMm}{(R+r)^2}$$

This formula can be expanded as a series for small values of $r$, and this gives (upto three terms): $$F=\frac{GMm}{R^2}-\frac{2GMmr}{R^3}+\frac{3GMmr^2}{R^4} + \mathcal O (r^3)$$

The first term is a constant, but the second term and so on have ratios of the form $\frac{r}{R^3}$ which are very small when $r$ is small compared to the radius. (for example if $r$ is in the order of a few metres)

For most examples used in high school physics or introductory classical mechanics, $r$ usually is very small compared to the radius of the Earth, which is 6371 kilometers.

In cases like these, the other terms are as good as nothing, and the first term is the most significant contribution. This first term happens to be $mg$, where: $$g=\frac{GM}{R^2}$$

When to use Quantum Mech.?

Is there any parameter (in terms of physical quantities such as mass, length, charge...) which can be used to decide when to treat a system quantum mechanically and not classically?

classical mechanics - How to combine these equations of constraint?

I want to model a nonholonomic system of an arbitrary rotating disk in 3D, which rolls without slipping, and doesn't have to stay vertical. (think spinning a penny on the table) I want to use the method I just learned of Lagrange multipliers with the Euler-Lagrange equations to solve the system.

I can parameterize the system in terms of $(x,y,\theta,\phi,\psi)$, and I can come up with several equations of constraint if I let two (or three, with $(x,y)$ changing as a pair) variables change at a time and hold the others constant. I'm using mathematica so I can afford to have unweildy representations and painful integrals.

I wanted $(x,y)$ representing the position of the center of the disk in the horizontal plane, $\theta$ representing the angle from $(x,y)$ to the point where the disk touches the ground, $\phi$ representing the angle from the xy plane to the actual (x,y,z) center of the disk (so, if $\phi=0$ the disk is flat, and if $\phi=\pi/2$ the disk is vertical), and $\psi$ representing the angle of the disk around the axis normal to its face. I wound up on the following linear transformation, which takes a stationary point on the disk's space into world space: $T(x,y,r \sin(\phi)) R_{xy}(\theta)R_{xz}(-\phi) R_{xy}(\phi) \vec{v}$

(where $T$ is a translation, $R_{xy}$ is a rotation in the xy plane etc)

This works perfectly and I can actually come up with the kinetic energy in terms of $x,\dot{x}, y,\dot{y},\phi, \dot{\phi}$ etc.

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So now where the nonholomic part comes in, I need to find the equations of constraint. The only constraint is rolling without slipping. I can find equations with partial derivatives (say, I let x and y vary as I change $\psi$ and hold all other variables constant), but these are just partial constraints and don't represent the true differentials governing the constraints. How can I find the true differentials? My sets of equations are:

1 Rotating the disk normal to its face (exactly like spinning a wheel)

$\frac{\partial x}{\partial \psi} =-r \sin (\theta ), \frac{ \partial y}{\partial\psi }=r \cos (\theta )$

2 Rotating $\theta$, the point where the disk touches the ground, without changing x, y, or $\phi$. $psi$ must change according to:

$\frac{\partial\psi}{\partial\theta} =-\cos (\phi )$

3 Changing the vertical angle of the disk, $\phi$, and having the point of contact stay the same (as well as $\theta$, $\psi$ constant), $x$ and $y$ must change according to:

$\frac{\partial x} {\partial\phi} =-r \cos (\theta ) \sin (\phi ), \frac{ \partial y} {\partial\phi} =-r \sin (\theta ) \sin (\phi )$

How can I combine these equations into full differentials for use in Lagrange multipliers with the Euler-Lagrange equations?

Animated Visualizations

Just to show what the parameters mean and what the constraint equations mean in case there's something technically wrong:

(the animations seem to freeze. If one isn't moving try dragging it to a new tab)

Adjusting the parameters on the transformation equation:

Applying partial constraint 1 to visualize rolling without slipping

Visualizing partial constraint 2

Visualizing constraint 3

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note: I'm pretty new to Lagrangian mechanics, on chapter two of Goldstein classical mechanics, but I don't see a reason why I can't apply everything I've learned (just what I've mentioned) to this problem.

Answer

The solution is much easier than I anticipated. I thought the simplest method wouldn't work and that it wouldn't take into account certain things, but taking a second look I see that it works.

The point of contact with the ground (staying consistent with the rotation matrix definition above) is: $$\mathbf{v_1}=(x-r \cos(\theta) \cos(\phi),y-r \sin(\theta) \cos(\phi))$$ A bit of movement of a point there can be equated to a wheel moving in the direction orthogonal to theta and "forwards", of magnitude $r d\psi$: $$d\mathbf{v_2}=(-r\sin(\theta) d\psi,r \cos(\theta) d \psi))$$ We should have $d\mathbf{v_1}=d\mathbf{v_2}$. Expanding everything gives: $$0=r \sin(\theta)d\psi+dx+r \sin(\theta) \cos(\phi) d\theta+r \cos(\theta) \sin(\phi) d\phi$$ $$0=-r \cos(\theta) d \psi +dy-r \cos(\theta) \cos(\phi) d\theta+r \sin(\theta)\sin(\phi)d \phi$$

With these I can successfully apply the methods of variational calculus and get a physical solution!

special relativity - Can Minkowski spacetime be redefined as a non-flat riemannian manifold?

Minkowski space time is defined in terms of a flat pseudo-Riemannian manifold. I have wondered if it can be redefined as Riamannian manifold and in the case what type of curvature would there appear.

Formally:

Let M be a semi-Riemannian manifold of dimension 4, corresponding to the Minkowski space, and let g be the metric tensor (non positive definite), T be the Riemann curvature tensor and P a generic point of M.

Question 1

Which (if any) of the following is true?

a. at every P there exists one system of coordinates for which the metric g becomes Riemannian (positive definite) in a ball of radius R non infinitesimal centred in P

b. there exists one system of coordinates for which g is Riemannian (positive definite) at all P of M

Comment: in words, can we, with a change of coordinates, get rid of semi-Riemannianity – either in finite region or globally?

If this is the case, how do we pay it in terms of curvature? This the next question:

Question 2

c. if previous a) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

d. if previous b) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

Thanks a lot

special relativity - Why does light travel at the same speed when measured by a moving observer?

I know the hypothesis that the light speed is constant is retained by experiments. But is there any theory explaining why the light speed is constant no matter how an observer moves relative to light?

My question is, specifically: Suppose an observer $O$ launches a light and $O$ starts to move at the same time with a uniform velocity $v$ in the same direction that light points. Then why $c$ is still the light speed that $O$ will measure rather than $c-v$?

Answer

A personal point of view is that you may consider that Lorentz transformations apply primarily on momenta, and not primarily on (infinitesimal or not) space-time coordinates.

This is, of course, a "strong" postulate.

If you assume (some additional postulates are needed there) that transformations are linear, and that there is a rotation invariance, you are going to study "boost" transformations : $\begin{pmatrix} p'_z\\E'\end{pmatrix} = A(v) \begin{pmatrix} p_z\\E\end{pmatrix}$. You may show that, because $A(v)A(-v)=1$, $det A(v)=1$. Supposing a group structure, you finally are looking at the one-dimensional subgroups of $SL(2, \mathbb R)$, which are :

$$\begin{pmatrix} \lambda&\\&\lambda^{-1}\end{pmatrix}\quad \begin{pmatrix} 1&v\\&1\end{pmatrix}\quad \begin{pmatrix} 1&\\v&1\end{pmatrix}\quad \begin{pmatrix} \cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{pmatrix}\quad \begin{pmatrix} \cosh \theta&\sinh \theta\\\sinh \theta&\cosh \theta\end{pmatrix}$$

If you add additional postulates that, in a boost transformation, energy and momentum must change, that there exist a transformation which puts the momentum to zero , and that, if the energy is positive for an observer, energy will be positive for all observers, the first $4$ one-dimensional subgroups of $SL(2, \mathbb R)$ are excluded, and the last dimensional subgoup corresponds to a Lorentz transformation.

relativity - Can an observer B move in such a way that his clock will run faster than that of a fixed observer A?

My understanding of Special and General Relativity comes from books which attempt to explain them to non-experts in these fields.

Let A and B be observers, which at some time t(0), are together at some location P fixed to the earth's surface. A and B are provided with "clocks", which at time t(0), are identical lumps of the same radioactive element-that has a very long half-life. A and A's "clock" remain fixed to the earth's surface. B and B's "clock" leave A and stay together, but eventually return to A after some interval of time T (as measured by A's "clock"). If, during this time T, B's velocity relative to A is sufficiently close to the velocity of light or B remains close to a sufficiently massive body, it seems that time can "pass more slowly" for B than for A. So that when B rejoins A, B's "clock" will contain more of the radioactive element than A's "clock"

My question is: Are there any scenarios that B together with B"s "clock" can undergo (during the time T) which will cause time to "pass more quickly" for B than for A? So that when B rejoins A, B's "clock" will contain less of the radioactive element than A's "clock". If there are no such scenarios, is there any fact or law in Relativity theory which rules them out?

forces - How high can be tower or building?

I tried to find in the internet some scientific explanation and calculation and looks like it is difficult. I found some calculation for house from standard bricks and it gives $170~\mathrm{m}$ for Ultimate tensile strength of brick $3~\mathrm{MPa}$. Formula is

$$ h = \dfrac{\sigma}{\rho g} $$

where $\sigma$ is Ultimate tensile strength.

this formula is explained here

As I see here there is kind of steel with Ultimate tensile strength $2600~\mathrm{MPa}$ and according to this formula it can be 32 km!

As I understand this formula for square or cylindrical shape, like the same width everywhere. But what if we make it in the shape of Eiffel Tower or kind of hyperbolic and etc? Or shape is does not matter and maximum height will be the same?

Or maybe if we build it from sticks with the same shape as diamond crystal structure we can build it up to 100 km?

Is there any well know way (formula) to calculate maximum height of the tower of the complicated shapes?

UPDATE: As I was told, I should use Compressive strength instead of Ultimate tensile strength. It looks reasonable. In this case calculation will be the same, only for steel I found value not 2600 MPa, but 300 MPa, but I can take another material from here with the similar value 2600. and if I take diamond with Compressive Strength 17000 MPa it will give 480 km.

UPDATE2/ANSWER: Looks like I found answer by myself with help of all your valuable comments. If I use assumptions like Total gravitational force to the basement less or equals breaking force ($\sigma$S) where $\sigma$ - compressive strength and S - area in square meters, I get this formula for cylinder

$$ h \leq \dfrac{\sigma}{\rho g} $$

some numbers for cylinder:

Steel (300 MPa): 3.75 km

Granite (300 MPa, but less density than steel): 11.5 km

Diamond (17000 MPa): 480 km

ABS Plastic (65 MPa): 6.5 km

Strongest concrete (80 MPa): 3.2 km

Carbon epoxy (up to 1500 MPa): 100 km

but for real building we have to divide it to 2 or 3 to have some "factor of safety". In this case only diamond and carbon epoxy can be used.

For cone

$$ h \leq \dfrac{1}{3} \dfrac{\sigma}{\rho g} $$

numbers will be 3 time more than for cylinder.

For other shape this condition should be met

$$ \sigma \geq \dfrac{\rho g V}{S} $$

I tried to calculate Exponential cone like this

which is Solid of revolution of this function

$$ f(x) = r e^{-α x} $$

where r is basement radius and α is kind of cone steepness. Volume can be calculated by formula from wikipedia article Solid of revolution Looks like for this exponential cone it is possible to build tower of any high from any material, but for materials with low compressive strength, if we take basement r=1 km, desired height 100 km for example, last 70% of the tower it will be very thick (like $10^{-5}$ meters). Of course this kind of needle is not possible to build and it does not make any sense to build. if we accept that final radius at maximum height 100 km equals 0.5 meter, the basement radius for different materials will be like this.

r without "factor of safety"

Steel (300 MPa): 400 km

Granite (300 MPa, but less density): 42 m

Diamond (17000 MPa): 0.56 m with "factor of safety" 5

ABS Plastic (65 MPa): 1.5 km

Strongest concrete (80 MPa): 5000 km

Carbon epoxy (up to 1500 MPa): 0.51 m

if we think about "factor of safety" equals 3, as I understand it is standard for this kind of things, we get this numbers

Steel (300 MPa): $10^{14}$ km

Granite (300 MPa, but less density): 230 km

Diamond (17000 MPa): 0.56 m with "factor of safety" 5

ABS Plastic (65 MPa): 5500 000 km

Strongest concrete (80 MPa): $10^{17}$ km

Carbon epoxy (up to 1500 MPa): 2 m

In this case we can really build a space elevator with high 100 km from 3 materials Granite, Diamond, Carbon epoxy. Even yearly production of Carbon epoxy will be enough to build it :)

This is exact picture for tower with 100 km height from carbon epoxy and factor of safety 3 (all axes in meters)

Of course I do not consider wind, and all other things. With precise engineering calculation might be it will not be possible.