Hot answers tagged

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, ...


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 ...


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 ...


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 ...


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 ...


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. ...


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 ...


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 ...


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, ...


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 ...


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 ...


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 ...


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 $...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


2

That is the case for central gravity fields. You can derive Newtonian gravitation from a spacetime metric $$ ds^2 = c^2d\tau^2 = c^2\left(1 - \frac{2GM}{rc^2}\right)dt^{2} - dr^2 - r^2d\Omega^2 $$ where the only metric element different than unity is $g_{tt} = c^2 - GM/r$. The relevant Christoffel symbol is $$ \Gamma^r_{tt} = \frac{1}{2}g^{rr}\frac{\partial ...


2

According to Newton's theory of gravity, the orbit would be an ellipse, which could take the form of a perfect circle. However, Einstein's theory of General Relativity tells us that this elliptical orbit would very gradually decay due to the emission of gravitational waves, and perhaps also precess. The Wikipedia page on the two-body problem in general ...


2

The stable orbits around a star are given by the Kepler's laws oft planetary motion. In general these are ellipses with the center star in one of the two foci. Circular orbits are the special case when there is only one focus. For a orbit with a given radius, there is only one speed which allows a circular orbit. If you have a starting condition where the ...


2

From the equations $\dot q = p/m$ and $\dot p=V'(q)$ and the definition $Q=D(q)$ you can derive by differentiation $m\dot Q=D'(q)p$ and $\ddot m^2Q=D''(q)pp-mD'(q)V'(q)$, where $m$ is a diagonal matrix of masses. 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., ...


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

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 ...


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 ...


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 ...


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 ...


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

Assuming that the gravity well acts as an attracting mass whose force follows Newton's law of gravitation, your well will suck the object from Earth's surface when the force from the well is larger than the force from the earth. That is, $GM_{earth}m/R_{earth}^2<GM_{well}m/r_{well}^2$, where $r_{well}$ is the distance between object and well. Earth ...



Only top voted, non community-wiki answers of a minimum length are eligible