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92

There is no tidal bulge. This was one of Newton's few mistakes. Newton did get the tidal forcing function correct, but the response to that forcing in the oceans: completely wrong. Newton's equilibrium theory of the tides with its two tidal bulges is falsified by observation. If this hypothesis was correct, high tide would occur when the Moon is at zenith ...


39

Intuitive explanation Suppose you drill two, perpendicular holes through the center of the Earth. You drop an object through one, then drop an object through the other at precisely the time the first object passes through the center. What you have now are two objects oscillating in just one dimension, but they do so in quadrature. That is, if we were to ...


27

This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked. To calculate ...


18

The picture of high tides on opposite sides of the Earth with a period of about 12 hours (actually 12 hours 25 minutes, due to the rotation of the Earth) is an oversimplification. It's just a starting point. Tides would behave this way in the limit of an all-water Earth with ocean depth so great that it had no effect on the surface wave. But the Earth has ...


10

Let's simplify. Let's eliminate the Moon. Let's get rid of the Sun temporarily. Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way. We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical ...


8

The mistake you're making is that you're looking at the full acceleration when you should look at the relative one. At distance $R=1\mathrm{au}$ from the sun, the gravitational acceleration is given by $$ a_0 = \frac{GM_\odot}{R^2} $$ Assuming a spherical cow earth (in vacuum), at midday at the equator, we're one earth-radius $r$ closer to the sun, ie $$ ...


8

Yes. Escape velocity does not depend on the flying object's mass, but only on that of the earth. The precise formula is given by $$v_e=\sqrt{\frac{2GM}{R}},$$ where $G$ is the gravitational constant, $M$ the mass of the earth and $R$ the object's distance from the center of the latter. This is of course only true if one ignores the interaction between ...


8

Phil's answer, while beautifully illustrated, is a little incomplete. It relies on the fact that in the case of the tunnel you're solving the one dimensional projection of the low earth orbit satellite, but doesn't prove this. I do this below. The force applied on the object, for a sphere of uniform density, is actually : \begin{eqnarray} F &=& - ...


6

I think the first few sentences of Landau's Mechanics puts it elegantly: One of the fundamental concepts of mechanics is that of a particle. By this we mean a body whose dimensions may be neglected in describing its motion. The possibility of so doing depends, of course, on the conditions of the problem concerned. For example, the planets may be ...


6

Yes your weight will change. The moon will have a bigger impact than the sun, so you may be heavier during the day (when there is a full moon). The effect has been measured: This figure is on page 93 of "Practical Physics" by Gordon Squires (a classical book, and one that I highly recommend). The method used is a beautiful example of careful experimental ...


5

Two terminology issues first. First, cosmology is the study of how the universe began and what will ultimately happen to it. Pieces of rock in our solar system have nothing to do with cosmology. Second, a meteoroid is a small solid body in the solar system. A meteor is what we call such a body while it is falling through our atmosphere, usually glowing hot ...


5

As of right now we have no way of deflecting an asteroid on its way to hitting the Earth. However there are lots of organisations tasked with looking into the issue. The Wikipedia article on Asteroid impact avoidance would be a good place to start. Also see the NASA Near Earth Orbit site for lots more background. You are quite correct than the kinetic ...


5

I would say, that your two first statements are not wrong, i.e. you can defend them in a discussion if you explain what you mean by extension and special case. Your conclusion, as you suspected anyway, is wrong however. Special relativity and Newtons law of gravitation are not the same thing. In fact they deal with different aspects of a physical theory. ...


5

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse. A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is ...


5

No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied ...


4

Assuming no air resistance, friction, etc., the answer is usually. @FredericBrunner is pretty much correct in his analysis, if we are assuming the object being talked about is the only object in the equations. However, if there are 2 or more objects cooperating or competing in getting away from Earth's gravitational pull, things get slightly more ...


4

In a completely ideal world, where air resistance was not present, the bomb would continue to move forward and the same horizontal speed as the airplane it was dropped from. Gravity only acts in the perpendicular direction, thus has no effect on the horizontal component of the bomb's velocity. In reality though, air friction reduced the speed of the bomb as ...


4

Let $\mathbf x(t)$ be the path of a particle. Let $\mathbf F(t)$ be a force acting on the particle as a function of time, then the work done by the force from time $t_a$ to time $t_b$ is \begin{align} W(t_b, t_a) = \int_{t_a}^{t_b} \mathbf F(t)\cdot \frac{d\mathbf x}{dt}(t)\, dt. \end{align} where the center dot denotes dot product; \begin{align} ...


4

Following the same orbit at greater velocity would mean that the satellite has greater acceleration (its velocity is changing at a greater rate than if it follows the same route more slowly). But the only force acting on it is earth's gravity, which creates a particular acceleration at any given altitude above the earth. It can't go any faster or slower ...


4

It never becomes impossible per se, but at some point there could be so much gravity that construction of a working rocket would be beyond our current ability to engineer something that could work. That is, it might take impracticably huge quantities of fuel, or require materials stronger than we can construct. There are just a couple of amazingly simple ...


3

The flaw is that you are trying to mix classical with relativistic concepts. Gravitational lensing (this is the phenomenon you are referring to) is best described in terms of general relativity. Massive bodies bend spacetime, inducing a curvature, which is described by Einstein's equations: $$G_{\mu\nu}=8\pi T_{\mu\nu},$$ where on the left hand side is ...


3

Unlike sea tide, which is quite complex, as other answers explain, the solid (not-so-solid for this part) Earth tide tends to be simple and the first-order picture can be reasonably approximated by the "bulges" metaphor mentioned in the question. Solid earth tide has an amplitude of ~1 ft typically and it can be safely ignored in most situations, including ...


3

You're confused. The number you're referring to, $6.6738410\times 10^{-11}\ \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$, is Newton's gravitational constant, $G$. This is not a formula or equation. The 'variables' are actually units: meters (m), kilograms (kg), and seconds (s). The formula where this constant most commonly appears is for Newtonian gravity: $$ ...


3

Classical mechanics has only theoretical point masses. In the simplest case of taking the particle's dimensions to a point, the following argument would hold. The mass of a particle would be given by its mass density times its volume. Take the gravitational field: $$ {\bf g}({\bf r}) = -G\frac{m_1}{|{\bf r}|^2}{\bf \hat{r}}, $$ $m_1$ will be proportional ...


3

In Newtonian gravity the acceleration due to gravity is independant of the mass of the object - a falling elephant accelerates downwards at the same speed as an accelerating gnat (ignoring air resistance). That means the orbit of an object does not depend on the object's mass (provided the object is much lighter than the star). The deflection of an object ...


3

I would say Newtonian gravity is the limit of general relativity as gravitation becomes weak. See it this way: the limit when the temperature-amplitude is small of metal extension can be written as: $$ L = L_0 (1+\alpha \delta t) $$ If $\delta t$ is quite large this equation is no longer applied. Special Relativity doesn't care about gravity, like the ...


3

You can easily convince yourself that this is not true. Just imagine a test particle on one side of a spherical shell. If the density is uniform, the theorem will apply. Now, however, decrease the density of that half of the shell until it's 0. Obviously this other half would attract the test particle. Alternatively, if changing density is not an option and ...


3

Assuming that $m_1$ and $m_2$ take up a finite amount of space (e.g., two spheres of mass with radius $r_0$), that equation isn't even valid for $r < r_0$, so there's no inconsistency. The derivation follows from Gauss' law; it is analogous to the application of Gauss' law in electrostatics; the $m_1$ and $m_2$ are the mass enclosed at some distance $r$. ...


2

The proof is most easiest if we use the vector notation. We have $$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$ where I have used the assumption that near the earth $\vec g$ is constant. Now according to the definition of center of mass we have, $${{\vec r}_{cm}} = \frac{1}{M}\int ...


2

The symbols $\text{m}$, $\text{kg}$, $\text{s}$ aren't variables that depend on a particular situation. They are the units. You can write $G = 6.67384\times10^{-11} \frac{\text{m}^3}{\text{kg}\cdot\text{s}^2}$, which means that $G$ is measured in cubic meters per kilogram-squared second, exactly the same way that you can say the Earth's radius is $6370 ...



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