New answers tagged

1

While you are climbing in a sloped road, force $mg\sin\theta$ acts in opposite direction of your motion. But, while you are sliding down, that force acts in direction of your motion. For same power of vehicle (i.e. for same push on gas pedal), your speed in climbing will be less than sliding down.


0

Hint: By the use of coordinate geometry find the distance between the $2kg$ and other masses. Now first of all use the expression for the gravitation force $F=Gm_1m_2/r^2$ between two particles. And by the use of principle of superposition tackle the particles pair by pair. By this I mean Lets, assume the force between $4kg$ and $2kg$ is $\vec{F_1}$ and ...


1

From the equations $\dot q = p/m$ and $\dot p=V'(q)$ and the definition $Q=D(q)$ you can derive by differentiation $\dot Q=D'(q)p$ and $\ddot Q=D''(q)pp-D'(q)V'(q)$. The second equation produces an ODE for $Q$ if you can express $p$ in terms of $\dot Q$ up to terms in the null space of $D''(q)$ (i.e., translation and rotation degrees of freedom). This should ...


4

For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is ...


0

First remember that Air resistance is proportional to $v_{net}^2$. And $v_{net}^{\text{freefall(1)}}<v_{net}^{\text{throw at angle $\theta$ to normal(2)}}$ So air-resistive force acting in free fall proportional to $(v_{net}^{(1)})^2$ And air-resistive force acting in the other case proportional to $(v_{net}^{(2)})^2\cos(\phi) $ ($\phi$ is the angle ...


1

Dark matter might be matter which has no protons or neutrons, more like pure energy than the kind of matter which is familiar to us ... kind of like the "GEONs" which John Archibald Wheeler proposed, speculatively, many years ago (see his book Geons, Black Holes & Quantum Foam for details). Because it contains energy, and because energy is ...


4

Yes, the sun's mass isn't constant causing disturbances of orbits, but should you care. Mass of sun now: $\approx 2 * 10^{30}$ kg so $E \approx 1.8 *10^{47}$ J. Radiation per year is about $10^{34}$ J. You've got a lot bigger problems that affect your calculation of orbits. Especially like where Jupiter is relative to where you thought it was. Namely the ...


0

Infinite distributions of mass can give rise to some contradictions. The classical well known example is an infinite homogeneous universe. In this case, by symmetry you would say that the the force will be zero everywhere. However, if you chose two arbitrary points in space, one as the center of coordinates and apply gauss law centered at the origin, then ...


1

In fact, it is not convergent: the horizontal component of the gravitational force can depend arbitrarily on parts of the sheet which are far away, if you are sufficiently malicious in your choice of surfaces which exhaust the infinite plane. To make this explicit, let's suppose we are suspended at the cartesian point $(0,0,D)$, and place some mass of ...


0

The Newtonian Mechanics prediction for the Mercury's Precession is actually $532''$ per century. The general result If the central force is attractive, there is a circular orbit of radius $r_0$. This circular orbit is stabble if it correspond to a minimum of the effective potential, i.e. $$U_{ef}''(r_0)>0.$$ Using that $U_{ef}=L^2/2mr^2+U$ and $F(r_0)=-...


3

Half. The escape velocity for an object at a distance $D$ from an object of mass $M$ is $\sqrt{2GM/D}$. The circular orbital velocity (the Moon is on an orbit that's close enough to circular that I'll just assume this) at the same distance is $\sqrt{GM/D}$. Setting the escape velocity from the Earth with it's new reduced mass $M_{\rm new}$ equal to the ...


1

Assuming that we are making a certain amount of the earth's mass disappear instantly somehow, doing this will decrease the escape velocity of the earth. The moon is already moving at its orbital velocity in an orbit centered on the current center of mass of the earth-moon system. If we took away so much mass (instantaneously, somehow) that the escape ...


2

If we are at the equator on the surface of an atmosphere-free, non-rotating, perfectly spherical planet (no mountains or trees, etc) then the optimal angle should be zero (launch towards the horizon, tangent to the planet's surface). This assumes the initial velocity is the orbital speed at the surface of the planet, which is approximately the escape ...


2

When you jump from a height, you gather momentum. Absorbing this momentum at landing reduces the size of the maximum force, and thus the "pain". Let us assume that the distance over which a person can absorb the momentum of the fall is proportional to their height (proportional to the length of their legs). In that case, the taller person can absorb the ...


0

The arrow ideally will fly out like a drag race car with the parachute deployed! A well designed arrow should have these properties. 1- Sharp and proportionally heavy point to accept a large momentum and deliver it as kinetic energy E= mv^2/2 2- long and balanced stem to accommodate a big arch and maintain separation between the tip and the fletching. 3- an ...


3

As the question is finite versus infinite, I guess we don't need the exact result for a finite disc (though it is not difficult to calculate). The simple intuitive answer is that even though the disk mass is infinite, most of the forces from the bits of disk going out to infinity will cancel out due to symmetry, so the answer is finite. Let us assume we ...


25

If the disk has infinite diameter it is nothing but an infinite plane. For any finite thickness we can consider a layer of mass whose superficial density is $\sigma$. Moreover, if the plane is infinite it does not matter if you are one meter or one kilometer away from the plane. Wherever you look at the plane you will see the same structure. So the ...


7

You can do the integral, and will discover that the answer is "finite" - because not only does the distance to mass increase, but so does the angle. Consider an annulus at radial distance $r$: if you have mass per area $\sigma$, the total mass at that distance is $2\pi r \sigma$; if the vertical distance to the center of the disk is $h$, the vertical ...


12

In the movies, arrows shot into the air rotate so that during the descent, the arrow head hits ground first. What is the source of this angular momentum? An arrow shot on the Moon would not do that. Air and the geometry of the arrow are key. An arrow flying through the air is subject to two forces, gravity and aerodynamic drag. Gravitation will not make an ...


46

Air. Conservation of angular momentum does infact dictate that whatever rotation it starts with it should end with, provided nothing else acts on it. Air allows its forward momentum to act on it. Consider a weather vane, a wind sock, or a flag. They rotate when not facing into the wind because one side presents more wind resistance than the other. Once ...


8

As pointed out by dmckee in his comment, anyone (including myself) that has practised bow and arrow knows the arrow by weight and fletch inspection. As my English corrector points out, fletch doesn't seem to be a very commom word: it means that feather at the end of the dart/arrow. Everything resumes to how good is the fletching. When shot the arrow wants, ...


49

The same reason objects which are heavier on one side tend to fall with the heavy side down: the tip of the arrow is denser than the rest of the arrow. The center of gravity is offset from its geometrical center, so the air drag, which is based on the object's geometry, causes a torque together with gravity as seen in this very professional picture of a body ...


0

Would you decrease your impact impulse by jumping during the fall? Yes When? Soon enough that it's before impact, late enough that you don't hit the ceiling of the elavator. Beyond that I don't think it matters much. Would it help if you jump inside a free falling elevator? Probably not. Indeed I expect it would make things worse. The ...


1

Gas molecules move very fast and tend to mix more than they tend to settle due to gravity and density. Similar to what happens inside any bottle of liquor. Alcohol is lighter than water but it doesn't float on top, it stays mixed in. The water and the Alcohol mix naturally in part due to their shape and in larger part, due to their charges. Gases aren'...


4

In newtonian mechanics the angle does not matter, but in relativity it does. For example: An object close to the speed of light launched horizontally will orbit circular at a distance of 3GM/c² from the center of mass (the so called photon sphere), but it will escape if launched vertically. When you launch it at a distance just above 2GM/c² (the so called ...


6

Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take. Vague intuition: the vertical speed you estimate is about $...


2

The force straining a grain (which may soon be squashed) is equal and opposite to the weight (gravity force) supplied by the grain pile above. That's Newton's third law. If ALL the weight of the pile were held up by one single kernel, F_grain ~= Mass_of_pile * g and it would be easy for the ton of grain to smash the bottom kernel. It doesn't work ...


7

The physical processes that control the structure of conical piles are fascinating and imperfectly understood even today. However we can approach your question in approximate way. The angle that the surface of the pile makes with the ground is called the angle of repose. Predicting this theoretically is hard because it is is sensitive to the exact nature of ...


0

One different aspect to this in case we are really talking about a living planet. Because gravity really kicks in at that scale, anything the size of a planet is so smoothly round that a polished billard ball feels ashamed for its own imperfection. So if this being is really planet-sized, it better be of really low density ...


11

Thanks to Michael Seifert's answer, I found a paper he referenced: Satellite Relocation by Tether Deployment by G. A. Landis and F. J. Hrach, 1989. By extending a tether radially, a satellite can increase or decrease its orbital speed (pictures below copied from the paper): The principle then can be used to pump an eccentric orbit: Similarly a planet ...


2

Gravitational field is a vector field and is determined by negative gradient of the gravitational potential. $$\vec g=-\vec\nabla \phi$$ Frome equation above, it is obvious that $|\vec g|=|-\vec\nabla \phi|$ (magnitude of $\vec g$ is equal to magnitude of $-\vec \nabla \phi$) and we know that $|-\vec\nabla \phi|$ is a non-negative quantity. You have made a ...


2

First, notice that this happens on earth too, the tidal effect of the moon is about 2.3 times larger than that of the sun. The fact that both orbits go in opposite directions will not be qualitatively important, only quantitatively in terms of the irregular variation in tidal periods and sizes. If the two moons have the same density and angular size in ...


1

Fun question! Try this very simple answer (Newtonian as you asked).. If a planet changes its shape from a round ball into a twine spool like shape (i.e. with a thicker center and elongated thinner ends) and assuming the elongation is done exactly along the radial line to the star (Sun), then for simplicity sake, the amount of mass that gets closer to the ...


6

You can always use the process of tidal acceleration/deceleration. In nature this process might be very slow, such as for the system Earth/Moon. However, you can alway speed it up, by artificially increasing the frequency of the shape oscillations. In a natural system tidal acceleration will stop when the two objects are in tidal locking (both object ...


14

A different mechanism: On a long timescale, by increasing the surface area exposed to the sun (flattening the planet), the radiation pressure would increase, boosting to a higher orbit. Changing the albedo would be a more effective means to the same end but could allow assymetric force as well Either way it would be simpler in a tidally-locked planet. ...


67

If you allow for non-Newtonian gravity (i.e., general relativity), then an extended body can "swim" through spacetime using cyclic deformations. See the 2003 paper "Swimming in Spacetime: Motion by Cyclic Changes in Body Shape" (Science, vol. 299, p. 1865) and the 2007 paper "Extended-body effects in cosmological spacetimes" (Classical and Quantum Gravity, ...


16

This is a really nice question. Conservation of angular momentum tells us that in an isolated system, total angular momentum remains constant in both magnitude and direction. The key here is that the conserved quantity is the total angular momentum: spin+orbital angular momentum. An example: For a planet, angular momentum is distributed between the ...


4

By conservation of momentum and energy, the only possible way to change a planet's trajectory is to eject some (large) mass at high velocity in specific direction, like rockets do. But you are also correct that by increasing the moment of inertia, the rotational speed can be changed. But this cannot influence the movement of center of mass. Edit2: Other ...


0

Helium and hydrogen balloons don't inevitably burst as they rise, but they are frequently designed to burst at a certain altitude to return their payload to Earth. The atmospheric pressure decreases with altitude, and for a fixed mass of gas the volume is (approximately) related to pressure by the ideal gas law: $$ V = \frac{nRT}{P} $$ So as the pressure ...


2

Negative mass and by corollary negative energy have some strange consequences. If you have a set of masses in a region of space with a volume $V$ the density of energy is $\rho = \sum_im_ic^2/V$ This defines $T^{00} = \rho$ component of the stress energy tensor. The Hawking-Penrose energy conditions are that $T^{00} \ge 0$. Violations of this creates various ...


0

One has to differentiate between negative inertial mass, which is very unlikely, and a negative gravitational mass charge, which is what you are looking at. The latter has not been ruled out for antimatter. More generally, we have not confirmed, yet, that the equivalence principle holds for antimatter.This leaves room for such hypotheses and it requires ...


0

Actually, all of your derivation is correct. But, potential energy is the energy due to gravitational field with respect to point at infinity. By your derivation, u at infinity is zero. u at (r,0) is GMm/r . Thus, potential energy at (r,0) is U(r,0) = u(0,inf) - u(0, r) = 0 - GMm/r = -GMm/r


-1

I would start by measuring the pebble and the weighing it to dtermine it's size. Then I would ask "am I simply dropping it" like Galileo or am I trying to make an inference based upon throwing it then estimating the effects on landing. If it's the former then you should just get a number as no matter the rock size they all fall at the same rate.


0

So I think I've got an answer, but I'm not completely convinced it's the right one. A friend (thanks Ricky!) pointed me to several places online where people had solved this problem: Falling Body with Air Resistance Velocity-time graphs for falling objects Calculating time to reach certain velocity with drag force Each graph of $v(t)$ vs. $t$ looks ...


0

You are wrong, the orbit of the moon is changing, it is receding from the earth so its potential and kinetic energy is changing : The notional tidal bulges are carried ahead of the Earth–Moon axis by the continents as a result of Earth's rotation. The eccentric mass of each bulge exerts a small amount of gravitational attraction on the Moon, with the ...


-4

This is a wonderful question and filled with interesting albeit incomplete or inaccurate answers. Launching an "orbital body" concerns two issues and two issues ONLY namely mass and intertia. So before we "launch" anything we must first understand THE EARTH IS MASSIVE...which works AGAINST the "inertial reality" of a conical shaped item which albeit is ...


1

At the top of the loop if the normal reaction on the ball due to the track is $F_n$ down and the weight of the ball is $F_g$ down then using Newton's second law $$F_n + F_g = m\frac {v^2}{R}$$ where $v$ is the speed of the ball, $m$ is the mass of the ball and $R$ is the radius of the loop. This equation tells you that the faster the ball is moving the ...


0

If you use Newton (left), there should be no perihelion shift. If you use Einstein (right), there is one: The equations of motion are not so different, if you differentiate by proper time the only difference between Newton and Schwarzschild is the red term in $$\color{black}{\ddot{r}(t) = -\frac{G\cdot M}{r(t)^2} + r(t)\cdot \dot{\theta}(t)^2} \color{red}...


-2

when you throw the ball up wards' it is acted upon by kinetic energy in your hands equal to the force uplied and it then freely falls by gravity at stable equilibrium but when external forces are applied' it goes up hence anti-gravity


0

To begin with, it is important to clarify the scenario. When two masses are attracted to each other, regardless of their magnitudes, the gravitational force (and hence acceleration) vector acting on each is in the direction of the other. If two masses were sitting motionless in a vacuum, they would accelerate toward each other in a straight line until such ...



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