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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 ...


0

It's just geometry. At any given distance $r$ from a massive particle, the force from that particle is spread out over the surface of a sphere of radius $r$. The surface of a sphere scales as the square of the radius. If you assume gravitational flux[1] is conserved, then flux density, and thus force at a given point, must decrease proportional to the ...


0

What you have listed is not a formula: It's the gravitational constant itself. Getting an approximate for the gravitational constant should be fairly simple, experimentally. All you need to do is jump. Here's the equation you need: $$F = \frac{G M m}{r^2}$$ Where $F$ is the magnitude of the force exerted on each of the two objects in question, $G$ is ...


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 ...


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: $$ ...


0

This isn't an equation so there isn't anything to "solve". In terms of what the symbols mean "m3 kg-1 s-2" is just giving you the units of the gravitational constant, which in SI are meters cubed per kilogram per second squared. E^-11 is just a short hand for $\times 10^{-11}$ . The whole thing is just a number (indeed a constant!) \begin{equation} G = ...


1

It would crash into the earth because the Earth's gravitational field is not uniform and, even if said ring were to be perfectly positioned, ignoring the effects of wind, strikes from cosmic debris (not a lot that low in the atmosphere), change in mass of the ring (e.g. corrosion), change in shape of the ring (due to e.g. gravitational forces, heat ...


1

The angular momentum of the ring would stay constant. So if the ring were built on the earth and then lifted up to 100 feet and fastened together, it would would initially rotate at the same rate at the Earth's surface. The prevailing winds would probably disturb its orbit and I suspect it would eventually crash into the Earth. I think it would take extra ...


1

In the Newtonian framework, you just need to solve the integral $$\int{ \frac{ G \rho_{(r)}}{|\vec{r} - \vec{r}_o|^2} \frac{\vec{r} - \vec{r}_o}{|\vec{r} - \vec{r}_o|}}dV$$ for the volume in question, where $\vec{r}$ is the distance from an arbitrary reference point to the element of matter where density is $\rho_{(r)}$, and $\vec{r}_{o}$ the distance from ...


-1

Your question 1 is non-sensical. If the earth were sqeezed (I assume you really mean "compressed") to be much smaller, then where does your flat surface come from? One can imagine a small sphere, but there would be nothing "flat". As for question 2, there are two competing factors at work. First, the effect of gravity could be lowered because some of the ...


1

According to the shell theorem, there's no gravitational attraction inside a spherical shell of matter. At the center of the earth, its entire mass is arranged around you in spherical shells, so the gravitational acceleration due to the earth is zero. You'd still feel the gravity of the sun and moon and other external objects, as well as a (small) ...


0

I'm not a physics expert, but I do know the basics. Gravity is created by mass so all things with mass have gravity. The gravity is negligible in an object unless it is something large like the earth or moon. In the middle of earth, all of the mass should be all around you so you should feel the force of gravity pulling in all directions away from you.


1

It'll be exactly the acceleration due to gravity, because no other force is acting on it. Edit answering comment: You're thinking about velocity. Acceleration isn't something an object has. Objects accelerate because there's an external force acting on them. When B is towing A, it exerts a force on A which causes A to accelerate. When B lets go, that ...


1

True point masses and other singularities can wreak all kinds of havoc in Newtonian physics. A couple of examples: Particles can attain infinite velocity in finite time: Saari, D., and Zhihong J. (1995), "Off to infinity in finite time." Notices of the AMS 42:5. Particles can exhibit non-deterministic behavior. See this question, Norton's dome and its ...


0

Assuming two mass particles, there would be a miniumum r>0 represented by the particles themselves therefore: F= G*(m1*m2)/[r(m1)+r(m2)]^2.


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

No. the whole weight will not act on the base of the container 2. If the whole weight had acted on the base of container 2, then the pressure on the base of container 2 would be equal to that of container 1 i.e., mg/A, where mg is the whole weight of the fluid and A is area of the base. But as you know the pressure at a depth depends on the height of the ...


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 ...


1

As Rahulgarg mentioned, the pressure does not depend on the shape but on the depth. However, the direction of the force caused by pressure can be approximated as being normal to the surface, hence the total force on the sides will depend on the shape. For a fluid at rest like the one I think you are assuming, the pressure at any depth will be $p=p_0+\rho g ...


1

Assuming the train doesn't accelerate during the ball's fall, it will land in the spot you aimed at. Think about it this way. Before you drop the ball, it is moving along with the train (i.e. it has some horizontal speed). When you drop it, the ball still has this speed, and since an object in motion tends to stay in motion unless you exert a force on it ...


-1

The bullets time in the air, having been fired horizontally, depends on it's velocity. In general that varies from about 900 fps to near 3000 fps for a rifle, so obviously the time it takes to hit the ground varies as well.


-1

The first half of this theorem is: you can substitute the total mass in the centre to get the net gravity. It probably don't stays, although it needs a little bit of integration. The second (there is no net gravity inside the shell), I think it passes, if the shell is convex. The proof is exactly the same as you can see in the wiki page.


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 ...


2

Phil Frost's argument in his answer (v4) is correct. Assuming a spherical Earth with constant density $\rho$ (and assuming for simplicity that the object for some reason can move freely$^1$ through Earth so that there is no air drag, and so that we can skip all the tunnel drilling and not worry about that the Earth's rotation could press the object up ...


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 &=& - ...


1

My question is: why are these two periods (oscillating through the earth and a LEO) the same? I am sure that there is some fundamental physical reason that I am missing here. Help. It's a result of the (flawed) assumption of a uniform density Earth. The Earth is anything but a constant density object. The Earth's core is five times more dense than ...


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 ...


2

Such a dramatic change would indeed make it harder to reach space, in a precisely defined sense. The easiest measure of how 'hard' it is to get so space is the escape velocity, which is determined by the planet's mass $M$ and radius $R$ as $$ v_\text{esc}=\sqrt{\frac{2GM}R}. $$ For a planet like the one you mention, the escape velocity goes up by a factor of ...


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 ...


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 ...


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 ...


0

In my opinion "description of trajectory" is not a topic of physics it is a topic of kinematics (geometry of motion, see http://en.wikipedia.org/wiki/Kinematics). Whereas explaining the mechanism which causes an object to follow a particular trajectory IS a matter for physics. To say that "B goes around C" is to describe a trajectory in space and time. A ...


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} ...


0

The definition of work, it is done on vectors, let's say $$ \vec F = F_x\hat i+F_y\hat j+F_x\hat k, $$ that would be the force and the displacement vector it is $$ \vec \ell=\ell_x\hat i+\ell_y \hat j + \ell_z\hat k $$ so you have the 3D. Then the definition is $$W=\int \vec F \cdot d\vec\ell$$ also it would be good to check anyways the dot product and the ...


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. ...


1

I think it's better to think of these things in terms of the shape of space and time. Special relativity told us how coordinates transformed when space-time is flat. But space-time isn't necessarily flat, and general relativity told us how curved space works. General relativity then concluded that it's mass that creates this curvature - and thus gravity. ...


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 ...


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 ...


2

The formula: $$ \Delta U = mgh $$ is an approximation that applies when the distance $h$ is small enough that changes in $g$ can be ignored. As you say, the expression for $U$ is: $$ U= -\frac{G M m}{r} $$ So the change when moving a distance $h$ upwards is: $$ \Delta U = \frac{GMm}{r} - \frac{GMm}{r + h} $$ We rearrange this to get: $$\begin{align} ...


2

Your first potential energy arises from the approximation that the graviational field is approximately constant for "small heights" , i.e. $$\frac{GMm}{r^2} \approx mg$$ The full law leads to your second formula, the approximation to the first. For heights above the earth, it is justified, as we can see by taylor expanding $\frac{1}{r^2}$ around the ...


0

In short, you are right, the pendulum has an angle with respect to the gravity vector, but your reference frame has this same angle, making then aligned. A 1-D coordinated turn is a turn such that the horizontal component of the lift is equal and opposite to the force generated by centrifugal acceleration. E.g like this: Obtained from Free online pilot ...


1

Even though your integration method seems wrong, like David Hammen pointed out, the results look do not look wrong. The problem rely lies with the way you define your ellipse. Your definition for the semi-major axis, $a$, and eccentricity, $e$, are correct. The way you define the semi-minor axis, $b$, probably is not, which is defined as: $$ b = a \sqrt{1 - ...


1

Here's your key error: var dx = Add(Mul(gv , dt*dt) , Mul(o.v, dt)); // position changes In math, that's $\Delta \vec x = \vec g \Delta t^2 + \vec v \Delta t$. That's incorrect. Assuming a constant acceleration $\vec g$, the change in position over some time interval $\Delta t$ is given by $\Delta x = \frac 1 2 \vec g \Delta t^2 + \vec v \Delta t$. That ...


1

I think that a disc of orbiting material is something that arises naturally when you have enough objects to do statistics with, but not necessarily when you have only a handful of masses as in your example. I happen to be playing with an orbit simulator lately, so I put two Jupiter-mass objects in perpendicular circular orbits 1 AU from a solar-mass central ...


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

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 ...


1

I believe that in Physics, when introduced to a law, we must first ask what observation does it predict. In the case of the gravitational two body problem, we have a Hamiltonian $H = \frac{p_r^2}{2\mu} + \frac{L^2}{2\mu r^2} - \frac{G \mu M}{r}$, where $\mu=m_1 m_2/(m_1+m_2), M=m_1+m_2$, $L$ is the total angular momentum, and we have done a transformation ...


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 ...


-2

Tidal movement is various at various geo-oceanic points on our globe. Ocean is water which is fluid, as distinct from the land exposed surfaces on earth which are of course rigid; to extent of not being possible for moon effect to draw the land surface to a bulge. Therefore, ocean depth will bulge in direct response to moon position relative to Earth. ...


0

If we draw in the acceleration and velocity vectors of an orbiting body, you'll see why it acceleration due to gravity does not cause an orbiting body to fall towards an object. First, assume the orbit is circular. In any motion of an object, the velocity vector is tangent to the object's path. Because acceleration due to gravity is straight towards the ...



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