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23

But can someone please explain why this is without using pure algebra? I will try without a single formula. In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass; the more gravitational mass, the more the gravitational force. In Newtonian mechanics, the acceleration of a particle, for a given ...


19

The force you can exert is your mass times your acceleration. The force is only limited by your ability to push yourself off the willow shoot. Imagine that you lie down next to the shoot, holding it in both hands. If you now pulled yourself up rapidly (the way some circus acrobats can pull themselves up a rope while appearing to hang horizontally) then you ...


16

For simplicity, consider a perfectly circular orbit; the gravitational acceleration is always at a right angle to the velocity vector. This means that the speed cannot change despite the fact that there is constant acceleration. Note that for the speed to change, there must be a non-zero component of acceleration parallel (or anti-parallel) to the velocity ...


11

You certainly know that all things fall at the same rate regardless of their mass (neglecting friction). An orbiting body is not different from a falling body in that the only force acting on it is the gravity of the thing it orbits, so there is no reason its mass should influence its orbit.


9

Why does the mass of the orbitting object have no effect on its revolution at all? It does have an effect! However, the effect is immeasurably small if the orbiting object itself has a very small mass compared to the object it is orbiting. The most massive object we humans have put into orbit is the International Space Station, with a mass of 419.5 ...


8

Firstly, the gravitational field inside the Earth, decreases with depth. To a first approximation, you can use the shell theorem for spherically symmetric mass distributions to argue that the gravitational field at some depth is due only to the mass enclosed within a sphere interior to that depth. If we further make the crude assumption that the Earth's ...


6

... you aren't envisioning a giant tunnel in space that the moon would pass through, are you? Like the crude drawing below? We don't have the materials strong enough to attach it to earth with. We don't have enough fuel to launch that into space The turbines would... collide with the surface of the moon to generate electricity? Even if all this worked, ...


5

It sounds like you are imagining that what satellites do is go up through the atmosphere, break though into outer space, and hang there. That is not right. If you simply go straight up to outer space (say 300 km above Earth's surface), gravity will pull you right back down, even if you've left the atmosphere, and you'll crash back into the Earth. Gravity is ...


5

You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since ...


5

Orbit is free-fall around a body, so it is for the same reason a feather falls as fast as a bowling ball (in a vacuum). In free-fall, $F = mg$ And we know that, $F = ma$ So we can substitute, $ma = mg$ And divide by $m$, $a = g$ Thus, no matter what mass is, acceleration equals $g$.


5

You raise an interesting point about the role of experiment and falsifiability in science. Despite a long-standing anomaly in Mercury's perihelion, Newton's theory of gravity itself wasn't heavily questioned, let alone rejected or falsified. Rather, auxiliary assumptions were concocted that saved Newton's theory, such as an erroneous mass of Venus, a planet ...


4

You are no doubt familiar with the apocryphal experiment of Galileo showing that falling bodies fall at a rate independent of their "weight". [ We should really say mass.] An orbiting body is just a special kind of falling body, albeit one that manages to miss the ground due its sideways motion. Hence the term "free fall" as used regarding astronauts or ...


4

You can get this more "intuitively" (idiosyncratically): the flux of this force in closed surface is equal to the quantity of source inside (is a Gauss's Law). This source could be a mass or a charge. The physical picture is: the pressure applied in a closed surface by the field-force is proportional to the quantity of source inside. You can get the ...


4

No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ...


4

If you think about Newton's third law and standing still vs jumping. When you stand still the ground exerts a reaction force on you which is equal and opposite to your weight by Newton's third law. If you jump upwards at the point where you begin to drive upwards you are applying a greater force on the ground than the standing still case, this difference is ...


4

Use a lever. For an application like driving branches in the great outdoors, you would need to come prepared, or locate the site next to something heavy. Anchor one end of the lever under something heavy. Attach something to the branch to press against, and put the branch between you, at one end, and the fulcrum at the other. The lever acts as a force ...


4

If the Earth were a perfect sphere, it would be the same everywhere on Earth's surface. This is known as the shell theorem. It's not too hard to show mathematically, but you can think of it as the fact that all the mass that is close to you balances roughly with the mass that is far away. If you were to tunnel into the Earth, however, you would only ...


3

In a simple universe where the only thing that had gravity on you was the Earth, there is no such limit. As you seem to have seen from Newton's law of gravitation, there is no finite $R$ such that $R^{-2}$ is zero, and that's the only thing that changes. In real life, you can find places where other objects in space have equal but opposite gravitational ...


3

I'm assuming you haven't taken any physics courses, so let's start by explaining the concept of a force. Forces are the central focus of classical mechanics. Basically, a force is a push or pull on an object as a result of its interaction with another object. When applied to an object with mass, a force causes the object's velocity to change in some way. ...


3

To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ...


3

One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ...


3

I will be assuming that the system has some angular momentum about the center of mass initially. If the system has no angular momentum then both the stars would accelerate towards each other and end up colliding. The problem is a two body problem. In such cases, both the stars would revolve around the center of mass. If the distance between two stars is ...


3

The best paper on I've found on tidal tails is Reshetnikov & Sotnikova (2000). Their simple description of tidal tail formation is: To understand the development of tidal tails, one must recall how the water surface of the oceans get stretched radially by differential gravitational attraction exerted on it by our Moon. The differential forces ...


3

I like Kieran Hunt's answer but I'm going to give a different answer, even though I agree with what he said. In a very real sense, our solar system doesn't obey Kepler's laws because there are many bodies. The planets and even more so, the moons in our solar system don't precisely follow Kepler's 3 laws, but they mostly follow it pretty close. Our moon ...


3

In classical physics mass has two definitions: It measures the amount of inertia that you have. In order to accelerate something you have to apply a force to it. The heavier your thing is, the less it will accelerate, $$ a = \frac{F}{m} \, .$$ If you know the force and can measure the acceleration, you have access to the mass. In the physics of gravity, ...


2

I'm guessing you're actually asking how you prove that the motion is chaotic rather than periodic. If so, the answer is with great difficulty! The three body problem was solved by Karl Sundman in 1909, but the n-body problem was only solved in 1991 by Qiudong Wang$^1$. I say solved but atually both proofs really only showed the motion could be described by ...


2

Let's consider a hover craft or a rocket that will accelerate up to a certain elevation and keep hovering at it, as a better vehicle for our purposes than a plane. Let's forget about air momentarily. What will happen is that the moment our air craft leaves contact with the ground, its initial tangential velocity will be the same tangential velocity of ...


2

For two objects to remain in a stable circular orbit, the force acting on them must be equal to the centripetal force corresponding to their rotation. $$F=\frac{mv^2}{r}$$ or in terms of angular velocity $$F=m\omega^2r$$ where $r$ is the radius of orbit in this case. As the gravitational force acting on the two stars is the same. $F$ is equal in both ...


2

In general the acceleration of gravity at the surface of a planet depends on both its radius and its mass (density times volume): $$g=\frac{GM}{R^2}= \frac43\pi G\rho R$$ For a given amount of work done by the athlete ($F\Delta x$), height jumped scales roughly with $g$. A vertical jump looks like this: From a standing start you jump up and measure ...


2

Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding ...



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