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1

If you want to actually simulate the behavior of the planet as it experiences the (vector) force as it moves around, then you need to find a stepping method and write your velocity vectors and position vector in terms of coordinates. I recommend a Verlet velocity method. Others at this site have their favorites, too. Euler's method is not good enough for ...


0

Consider an arbitrary point P and a point close to it, let's say A. $\underline{g}_P = -\frac{GM}{r^2}$ Whereas $\underline{g}_A = -\frac{GM}{(r+a)^2}$ Taylor expansion on $\underline{g}_A = -\frac{GM}{(r)^2}(1-\frac{2a}{r} +...)$ Finding the difference between $\underline{g}_P$ and $\underline{g}_A$ yields $\frac{2aGM}{r^3}$ note that this answer ...


1

Limitations of $g \propto \frac{1}{r^2}$: The relationship $g(r) = G \frac{M}{r^2} \rightarrow g \propto \frac{1}{r^2}$ (where $g$ is the acceleration due to gravity, $G$ is the universal gravitational constant, and $r$ is the distance between the massive object and the accelerating object) is just fine in Newtonian physics—it doesn't need to be fixed. ...


1

If you have a spherical body of radius $R$ with mass $M$, the gravitational field at any point at a radial distance $r$ is given by: $$\phi=\frac{GM(r)}{r^2}$$ where $M(r)$ is the mass enclosed inside a spherical shell of radius $r$. This is the only mass that matters in this case. (because of the Shell Theorem: https://en.wikipedia.org/wiki/Shell_theorem) ...


0

Since proportionality is NOT the same as equality, you cannot say that v proportional to mT is incorrect or false solely based on dimensional analysis. But when we deal with equalities, the situation is quite different, as both members of the equation, to be equal, must be of the same dimensions. Consider the equation 3x = 9x where x is length (a ...


3

Dimensional analysis should never take the place of common sense - it should be used to inform it. Let's take Newton's example. He knew that objects of different mass fell at the same speed - which had to mean they experienced a gravitational force proportional to their mass. Hence $$F \propto m$$ If gravitational attraction between two bodies is ...


0

Imagine that you had a bunch of atoms all equally spaced on the x axis from $x=0 \to x=L$. The force you feel would be the sum of the forces from each one. And the potential would be the sum of the potentials from each one. That sum is what you approximate by the integral. (This is where early physics classes and early math classes can be opposite, in an ...


0

The acceleration they experience locally is different due to the different masses, whilst the acceleration they experience one to the other is of course equal (or at least synchronizing during any approach), otherwise A would hit B (when the planets crash together) later or sooner than B would hit A.


7

Your misconception has nothing to do with gravity - you're just getting a little mixed up about acceleration vs. relative acceleration. Let's dispense with gravity, since it's a red herring here. Say there are two cars. Car A accelerates at $+3 ~\rm m/s/s$ (to the right). Car B accelerates at $-5 ~\rm m/s/s$ (that is, to the left). So far, so good, right? ...


2

It doesn't have to be modified, it's fine as it is. There's no paradox at all. The force that attracts both to each other is indeed $F=G\frac{M_A M_B}{r^2}$. But the acceleration they experience is not the same (at least provided $M_A \neq M_B$) because their inertias (masses) are not the same. For one $F=M_A a_A$, for the other $F=M_B a_B$. There's no ...


1

If a human is physically incredibly strong, and can push up off the ground no matter the gravitational pull, then the maximum amount of gravitational force the person can push off in would make him walk the fastest. My reasoning behind this is if he/she has no trouble pushing off the ground, and all that matters is minimizing the duration the persons foot is ...


0

The force of Earth's gravity is inversely proportional to the square of the distance to the Earth.


2

The gravity g from a body of mass m at d distance from the center of the body can be found by the equation g = (m × G) / (d2), where G is the universal gravitational constant. Usually, g is in meters per second2, m is in kilograms, d is in meters, and G is in meters3 per kilogram per second2. If you want to apply this to the Earth, the Earth's mass is ...


5

How “large” is a Lagrange point? L1, L2 and L3 are essentially zero size cause they're never stable. They're still useful cause an orbital near L1, L2 or L3 doesn't require a lot of energy to stay in that general area. so we can use L1, 2 or 3 for not quite stable orbits that don't require much energy adjustment. L1 Halo orbits are used too, not ...


0

One way to think about this is to realise that the time interval is proportional to the speed $v$ of the particle when it rebounds, in fact it's just $$ T=\frac{2v}{g}. $$ On each successive rebound the particle's upwards velocity $v_n$ is $e\cdot v_{n-1}$ so the time intervals must also be related by $T_n$ = $e\cdot T_{n-1}$


0

If friction is constant, then the path followed doesn't matter. This is true because the only force involved is gravity, which is a conservative field. And that means (or is the definition of :-) ) that the energy expended to go from point A to point B is path-independent. Now, if you're interested in like the required initial velocity, you'll see that ...


0

In this system: Around half circle #1, gravity is a component of centripetal force. The normal force between coaster and path is due to centrifugal force. Around half circle #2, gravity is also a component of centripetal force, via the normal force between coaster and path. The normal force is a combination of gravity and centrifugal force. Around half ...


1

The approximation of 9.81 m/s^2 is a generalisation. The exact value is most likely different at a specific location, due to the distance from the centre of the earth to the point being evaluated. The reference to "surface of the earth" is also a relative since the earth is known not to be perfectly round due to centrifugal forces making the radius ...


1

Physically, a negative total energy is a necessary and sufficient condition for avoiding all particles flying off to infinity separately. However, it is always possible for some particles to be given enough energy to escape a system. In fact, this tends to happen in real life. Planets get ejected from their solar systems over millions of years, and stars ...


0

I think this formula would work better F=MxA Since you are hovering you need to consider the acceleration of gravity in your calculations. So you would get F= M x 9.832 meters per second squared. Use kilograms for the mass in this case


1

I would choose highly symmetric initial conditions which are easily shown to be periodic. Then I would perturb the system by small steps. For example, four identical particles (of mass $m$) moving on a circle and interacting through Newton gravitation alone is a solution of the equations of motion. $$ \vec{x}_i = R \left(\begin{array}{c}\cos\left(\omega t ...


2

There is a confusion of what that $v$ means. You are thinking about the velocity of the drone, which is stationary this $v_{drone}=0$. But in your equation, to calculate the power needed by the wings, you have to consider the velocity of the motor providing the thrust (propelling air downwards at a certain rate) to keep the drone floating which is making ...


0

Power equals work/time. Work is force times distance, so you can simplify to force times speed, when the force is constant and the force is causing the speed. In your situation there is a force due to gravity that would do work on your drone, and what you need is for the drone to do work to counter balance that. In other words, power is always work over ...


1

The problem is that finite wings inherently induce drag (called lift induced drag). Even if you could assume the wing infinite you still have parasitic drag. So you will always need to supply energy to keep the plane flying at a constant speed (and as a consequence to keep generating lift). Without drag it could, in principle, keep flying without power. It's ...


0

In principle, you are right that it is possible to keep an object in the air without expending any energy. Think about a blimp, you can fill it with helium, and then just let it float. It doesn't have to do anything to float, it just floats because it is lighter than air at ground level (the air gets less dense as you go up). You can even make heavier than ...


2

In my experience, in brushes the strongest force acting on the water is the capillary force due to the surface tension in the liquid and the proximity of the hairs in the brush. Surface tension will cause the water to try to "wet" as much of the brush hair as possible - regardless of orientation. If there is excess water, such water will be pulled down by ...


0

You should let it hang downwards, otherwise the water will drip down onto the ferrule, which can eventually corrode it. There's pretty much no reason to leave them tip-up: in particular any "local pocket" of humid air is unlikely to be rising much rather than expanding outwards in all directions. If you are worried about the local pocket of humid air during ...


4

I'm going to answer the question of your title, and also address the curious statement that "tidal gravity=real gravity". Let's begin with your statement: Tidal gravity, by my understanding, is the difference in gravity between two points. You're very much on the right track here. When people talk of "tidal effects" and "tidal gravity" when not ...


-4

If the two objects are electrons then the gravitational force is always negligible.


5

In classical gravity, the answer is "never". In general relativity, the answer is "never". Now what about a quantum theory of gravity? We don't know how it'll work, but it should reduce to general relativity in the classical limit (i.e. the limit of weak fields and large distances, which is exactly what your question is about). So the answer is still ...


3

No distance is far enough. Among other things, if you are extremely far away, then there is room for lots and lots of things to be far away from you and even if they individually have little effect we can find the net effect of all of them. So we know the effect of each one is not zero. So we can prove the effect of A on B is not zero even when they are ...


1

The question is ill-posed. At the classical level, the force (gravitational or otherwise) between objects never becomes zero. It goes to zero as the distance goes to infinity, but it never really becomes zero. At the quantum level, we don't have a theory of gravity, but already the concept of "distance" doesn't make precise sense anymore, since quantum ...


0

It just depends upon the masses of the two bodies: not on the ground because static friction would also be acting on it.


-1

If there is nothing that stops both masses from accelerating towards each other (like a barrier or other larger gravitational influences, for instance), then yes: they will very slowly accelerate towards each other, albeit depending on actual values of mass and distance at a very, very slow rate. You might want to look the Cavendish' Experiment, sometimes ...


3

If they're at rest to start with, and if there is nothing else in the Universe, then Yes. The gravitational force increases their velocity to a finite value (which still increases) after a finite time which is enough for them to collide at some moment. The time may be very very long if the objects are very light and/or very distant, however. (You may avoid ...


0

TL;DR You would eventually reach the higher velocity, but the hill ends before you get close. No Gear Changes First lets dismiss the notion that the power and torque requirements to ascend a hill at constant speed are different. To ascend a constant grade hill at a constant speed requires the same constant force $F$ regardless of velocity $V$. Let's ...


2

General relativity in just one dimension will always be flat, as all 1D manifolds are diffeomorphic to flat space : \begin{equation} ds^2 = -f(t) dt^2 \end{equation} As you can perform the variable change \begin{equation} \frac{dt'}{dt} = \frac{1}{\sqrt{f(t)}} \rightarrow t' = \int \sqrt{f(t)} dt \end{equation} Giving you \begin{equation} ds^2 = - ...


2

According to Google the mass of the earth is around $5.972 \times 10^{24}$ kg (presumably that figure includes its 7 billion human inhabitants plus everything else were meant to be sharing with). If we take the average weight for a human as given by WolframAlpha as 62 kg this gives humanity a combined weight of $4.34 \times 10^{11}$ kg - so our share in the ...


0

Is it not enough to move the earth from its position (even a bit)? For all practical purposes, nope. Think of it in the following manner: The (Earth + All the Humans) can be viewed as one system, and in this case, the gravitational force exchanged between any one person (e.g. you), and the Earth, is an internal force of this system. And internal forces ...


2

If someone (like superman) could stop the moon from its orbital motion then yes it would fall towards the Earth. Only then the direction of motion would be parallel to gravity. Same with the I.S.S or the satellites orbiting Earth. They could also spiral in and crash because the atmosphere is taking away their kinetic energy. Just like the comment says ...


1

The effect of earth's gravity is not disregarded. It is what keeps the earth in one piece and approximately spherical. This is the dominant effect in the the earth's vicinity. Without it the sun and the moon would most likely stretch and tear it apart, as would earth's own rotation. Tides are a tiny perturbation on top of this effect caused primarily by the ...


1

Let me summarize the different accelerations here: Acceleration of Earth's gravity. This is very important, it points "down" in your coordinates no matter where you are on the Earth, and it keeps the oceans on the planet, so that they roughly describe a round distribution around the Earth. Acceleration of the Moon's gravity on the center of the Earth. This ...


0

But isn't there a difference in Earth's gravity between antipodal points of the earth (opposite directions)? I don't follow your statement above, but my sincere apologies if I have misunderstood your question and I will be happy to delete this answer if it does not work out for you. I think you may be mixing up the Earth's tidal force on the Moon, ...


4

The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...


0

Is it correctly understood that energy is continuously put into the system, in order to maintain the orbit? And that gravity is thus an infinite source of energy? Consider the general case of one particle orbiting another in an ellipse (this is general because we can a always reduce the two body problem to an effective one body problem, and the general ...


0

I think you need to read up a bit on vectors and scalars, how we define things like acceleration. Vectors are physical quantities that have a magnitude and a direction. Velocity is a vector (speed is its magnitude). Acceleration is also a vector. Acceleration is defined as the rate of change of velocity, basically it measures how velocity changes with ...


1

Due to the force of gravity, which goes as the inverse of the square, planets trace out an ellipse in space as they orbit around the sun, which is located at a single focus. The other focus is unphysical. Actually, given two massive bodies, their "difference" vector will trace out an ellipse with the center of mass at the focus. Because the sun is so much ...


0

There is no physical object at the location of the second focus. Newton showed that an elliptical path was the consequence of an inverse square radial force from a fixed point. While you can identify the point that is the second focus, nothing associated with that point is required to create the elliptical motion. Deriving Kepler's Laws from ...


4

Because in the second scenario you described, the exhaust gases from the rocket fly off at high speed into space. This is where the extra energy goes. A better alternative to get circular motion of the satellite with the earth removed, would be a second satellite, connected to the first satellite with a long cable. If, for example, both satelittes have the ...


1

The aircraft gives you a protected area in which you can experience free fall without any drag or other significant forces having a large effect on your body. This is essentially identical to what an astronaut experiences on the International Space Station, only with the drawback that the semi-minor axis of your orbit is so small that it has two intersection ...



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