# Tag Info

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P.S.2 - Done on Earth. One cannot watch the skies! If you insist on this, it cannot be done. Your condition is equivalent to us living on a permanently clouded world. In that case, we would experience alternating light and dark periods and would have no way of hypothesizing what was happening. Understanding the universe would have to wait for the ...

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There is! It corresponds just to the tides due to the Sun. Let us suppose that the Earth is not rotating around the Sun, that is, we are not in free fall towards the Sun. In this case the liquid from oceans would accumulate nearest the Sun. The effect would be only one daily tide. On the other hand when we fall towards the Sun there are accumulation of ...

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The most straightforward observation to show that the Earth moves is stellar parallax. If you take photographs of a groups of stars over a period of six months (half an orbit), some of the stars will seem to shift in position compared to the others. These stars are much closer to Earth and so seem to move more. This is similar to how, when you are riding in ...

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"Presumably, the usual method for speeds v≪c is to consider a Newtonian n-body approach and simply integrate the equations of motion with the gravitational inverse square law, using a numerical scheme such as Runge-Kutta or, something more sophisticated, like a symplectic integrator?" It might work for non-relativistic but for more than two bodies it is ...

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I am certain that mathematical analysis of tidal locking has been done many times but I have failed to find such an analysis where the mechanism for the transfer of angular momentum to spin angular momentum is included which is the question which has been asked. Perhaps someone is able to produce a reference or an analysis? Having experienced on a number ...

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It's called the virial theorem: in a bound system, the average of the potential energy of a $\frac{1}{r}$-potential V is related to the average of the kinetic energy T like $\frac{\langle V \rangle}{2} = - \langle T\rangle$.

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The tangential velocity is related to the radius of a circular orbit and the angular velocity of the orbiting object. Specifically: $$\overrightarrow{v}=\overrightarrow{\omega}\times\overrightarrow{r}\to v=\omega r$$ For the satellite, the centripetal acceleration is equal to $r\omega^2$, which is the same (according to the above relation) as the tangential ...

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The relation is indeed correct. Let me show you why. This type of problem is solved by equating forces. Other classical mechanics problems are by equating energies or, in rare cases, momentum. The gravitational potential indeed is $$V(r) = G \frac Mr \,.$$ This is not directly useful for this problem. Here you rather want to look at the gravitational ...

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You ask: could the gravitational force of an average asteroid be extrapolated to the rest of the asteroids in its family? But the question is what properties the members of a family have in common. You could argue that all the asteroids in a family will be made up from the same material and will have the same density. That means as long as you can ...

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The cross product of two vectors is the area of a parallelogram defined by those vectors. If the angle is 90 degrees then the cross product is simply the magnitude of vector 1 multiplied by the magnitude of vector 2 (i.e are of a rectangle). another way of writing the cross product is |v||u|sin(theta). If theta is 0 (both pointing in same direction) or 180 ...

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For orbit, try thinking of it this way... First simplify from three dimensional space to a 2 dimensional plane, say a sheet of paper. Also for simplification, we will consider the case of one large object and one small object. We will consider the large object to be 1) not moving and 2) so much larger than the small object that the large object is ...

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The classic thought experiment (which I think is in Feynman's lectures) is to imagine a cannon on top of a hill shooting cannon balls. Ignore air friction for this. Typically the ball curves in an ellipse (we often say parabola, but this is an approximation for a uniform gravitational field, though the two are basically identical for this case) and hits ...

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While the motion of airplanes is very different from the motion of moons, planets etc. motion is the reason aircraft stay aloft (with the exception of lighter than air cart - hot air balloons etc). To counter gravitational pull plane's wings generate lift by their motion through the air. If an airplane moves through the air too slowly it will stall and ...

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Planets (take for example the solar system) revolve around a huge massive star under the influence of gravity exerted by the massive object on the planet if the planet is under the field of influence of the star. They are actually falling towards the star, but each time missing it from colliding. Consider the moon. It is constantly falling to earth under ...

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What you are asking for is not simple at all. Retrograde motion occurs when the line joining two planets rotates with respect to a fixed coordinate system (or the fixed stars) in the opposite direction as the planets. With both planets in motion in orbits that are not nicely aligned with each other, the times between retrograde motion will only be described ...

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At it's simplest level, satellite tracking is a 3D version of triangulation. Suppose one were able to have 3 observers at the vertices of a triangle on the ground all simultaneously and instantaneously measure the distance from themselves to the satellite. Then one could use the known coordinates of the 3 observers along with the measurements they each ...

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But you can see the motion of the earth in it's orbit; it's called aberration of starlight, first measured by James Bradley in 1729. Even earlier, parallax of stars had been detected, by 1680. But you have to take detailed observations at widely separated times for the effects to be even the least bit obvious.

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We can detect it, but it takes very precise measurements. Stellar parallax, i.e. the relative displacement of close-by stars against the background of far away ones can be detected, but it's a very difficult measurement to make because the "motion" is very small (usually fractions of an arc second): Recently we have learned to build satellites that can ...

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The correct formula is actually $$M = \frac{4\pi^2 a^3}{GP^2}$$ and is a form of Kepler's third law. $M$ in this formula is the central mass which must be much larger than the mass of the orbiting body in order to apply the law. In reality the formula that should be used is $$M_1 + M_2 = \frac{4\pi^2 a^3}{GP^2},$$ where $M_1+ M_2$ is the sum of the masses ...

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Your intuition that that this equation actually gives the mass of the body being orbited is exactly right. In general, there is no way to infer the mass of a body using any measurements of its response to gravitational forces, because in its equation of motion, $$m\mathbf a=m\mathbf g,$$ where $\mathbf a$ is the body's acceleration and $\mathbf g$ is ...

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The final orbit has a smaller radius so we know the final PE must be less (more negative) and the final KE must be greater (moving faster in a smaller orbit) but the final total energy must be less (more negative), since $E_{total}=-\frac{1}{2}G\frac{Mm_{e}}{r}$. This means the satellite, or more correctly, the Earth-satellite system, has to lose energy. ...

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It is acceleration, change in direction in this case, which makes the difference as does the fact that graviton all attraction is a non-contact force. In space away from any large mass going at 1000m/hr in a straight line does not require a force to be acting on you. When you go around a corner it is a localised contact force that provides your centripetal ...

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I assume you're talking about the numerical instabilities that arise from having an infinite potential at $r=0$. Here are three common solutions: Use a soft-core potential that behaves like $1/r$ except very close to $r=0$ where it levels off to a finite value. For example, $1/\sqrt{\epsilon+r^2}$ instead of $1/r$ is common. Add hard sphere collision ...

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Take the orbit equation (using $u=\frac1r$): $$u(\theta)=A\left(1+e\cos\theta\right)$$ Where $A$ is some constant that you can work out from the differential equation if desired, and $e$ is the eccentricity of the orbit. Now, what happens when $\theta$ goes through an angle of $2\pi$? Notice that we would not have returned to the original radius if ...

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Even simpler than what's been posted: the curvature of the earth keeps slowing the orbiting body. While grabity is pulling it down the tendency to travel forward resists that pull thus keeps the "fall at a balance"

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