6

Newton's theory does imply that the earth will fall into the sun - if it never had any tangential velocity with regards to the sun. As it obviously does have such velocity, it follows an elliptical path, as described by Newton's laws. As to why the earth does not follow a spiral path into the sun, that would indeed happen on condition that a) there is ...


5

Yes, the two-body problem for an inverse-square force is considered completely solvable. The equation you wrote is the radial infall equation. It has an analytic solution for $t(r)$. See Wikipedia. If you actually meant $r$ to be a vector, there is also an analytic solution. A bound orbit is an ellipse, with a circle as a special case of zero eccentricity. ...


5

This actually isn't too hard. The Maxwell-Boltzmann distribution provides the average velocity of gases. The most probable speed (which is only good for order of magnitude estimates) is: $v_p=\sqrt{\frac{2kT}{m}}$ You can put in the numbers: $T$ is the temperature of the Earth's surface, about $300K$, and $m$ is the mass of whatever gas you're interested ...


3

First a nitpick: the mars-sized planet called Thea which is hypothesized to have collided with Earth long ago did not do so because it used to orbit earth and this orbit decayed. Instead it is assumed that the turbulent interactions of all the planets in the early solar system just put it onto a collision course by chance. Your question has several layers ...


3

Both approaches are wrong because both assume a circular orbit. In the first one you make the assumption that the force is $mv^2/R$ which is only correct for uniform circular motion. In the second you assume that the speed is $2\pi R/T$ which is also only correct for uniform circular motion. You can start with the vis viva equation. The most useful form ...


3

You're missing a negative sign. The gravitational potential energy $U$ of two masses $m$ and $M$ separated by a distance $r$ is given by $$U=-G\frac{mM}{r}$$ As the objects get closer together, the gravitational potential energy becomes more negative, which is another way to say that it decreases. As the potential energy decreases, the kinetic energy ...


3

It's because what you feel is distortions to your body's shape. When in free fall, when air resistance is low, every part of your body is accelerating very nearly the same amount. No difference in force between any parts of your body means your body doesn't have to exert any forces to keep its shape, so you don't feel anything. If the gravitational field ...


2

By setting $E=0$ you are saying that the object's speed is equal to escape velocity $v_e(r)=\sqrt{\frac{2GM}{r}}$. By fixing angular momemtum $l_0$ you are saying that the angular speed of the object is $v_{\theta}(r)=\frac{l_0}{mr}$. But if $r \lt \frac{l_0^2}{2GMm^2}$ then $\frac{l_0^2}{m^2r^2} > \frac{2GM}{r}$ so $v_\theta(r) > v_e(r)$. All this is ...


2

The fields are different but I would say you have to look to their interaction. The two objects A nd B might have a different mass and shape but if you look at the smallest relevant parts (the elementary particles with a mass > 0) then you see easily that for each of those "paricles" from A which "pulls" a "particle" from B the inverse is true too. Just ...


2

When in free fall, neglecting air drag, the only force acting on your body is gravity, which is a non contact force, and you feel weightless. In other words the feeling of “weightlessness” is that of not experiencing the sensation of any contact forces. Stand on the ground and you feel the upward contact force of the ground on your feet that opposes and ...


2

As I can understand from the question you are asking if there exists such an upward velocity for which the object trains afloat at the same height? The answer then is no. The reason being that from any frame (as for here the earth) if an object posses some velocity then motion takes place (the very definition of velocity is dependent on motion!) ,and an ...


2

Here is a diagram showing the orbit of a satellite around the North Pole. The gravitational force $\vec F$ can be split into two components $\vec {F'}$ and $\vec {F''}$ such that $\vec {F}=\vec {F'} + \vec {F''}$. The component in the plane of the orbit $\vec {F'}$ provides the centripetal acceleration around the North Pole $r\omega^2$. However ...


2

The sum of the gravitational forces on a satellite from all the atoms in the Earth is a net force toward the center of the Earth. (This is a consequence of spherical symmetry. It can also be shown by using integral calculus.) Thus any satellite’s orbit must accelerate toward the center and not toward, say, a pole.


2

Your proposed path is one in which the gravitational interactions from the Earth, the Moon, and the Sun are all important. Kepler's Third Law assumes a two-body system where the satellite is one of the bodies, and you have a four-body system, so don't expect it to work very well, or at all.


2

They both have the same inverse square law, but the difference is electric charges can be positive and negative, so the force can be attractive or repulsive. Mass can only be positive, and two masses attract whereas two positive charges repel. Electric fields can also be circular if there are changing magnetic fields around. That does not happen with ...


2

In principle they do, although I do not know how measurable is the effect at the range of speeds a car can go. In order to depress the soil the wheel makes a force, let us call it the normal. This normal force is independent of speed.Once the soil feel the force, it will move down a distance $d$ at some rate, for instance $d=1/2N/mt^2$ is the simplest ...


2

Kepler's second law states that the curve $\mathbf{r}(t)$ followed by a planet sweeps out equal areas in equal times. That is $$ \frac{\text{d}A}{\text{d}t} = \text{constant.} $$ The infinitesimal area element swept out in a time $\text{d}t$ by a planet travelling on such a path is $$ \text{d}A = \frac{1}{2}|\mathbf{r}\times\text{d}\mathbf{r}| $$ so that $$...


1

Look at this example: The position vector to the mass is: $$\vec{R}=\left[ \begin {array}{c} x\\ y \left( x \right) \end {array} \right] $$ with $y(x)=a(\alpha)\,x$ $$\vec{R}=\left[ \begin {array}{c} x\\ a(\alpha)\,x \end {array} \right] \tag 1$$ To obtain the equation of motion I will use the NEWTON method. the generalized coordinate is $x$ and the ...


1

The thing you're missing is that the force of gravity isn't the only force involved. The potential $U(y)=mgy$ is a function of $y$, not directly a function of $x$. In order to get a potential as a function of $x$, we need a function that gives us $y$ as a function of $x$, so we can plug in this function and define $U(x)$ as being equal to $U(y(x))=mgy(x)$. ...


1

Non-hierarchical multiple star systems (where all members have similar masses and similar orbital periods) are possible in theory, and occasionally observed in reality. Their relative rarity compared to hierarchical systems suggests that non-hierarchical systems may be unstable (or, at least, have very small domains of stability) and so mostly short-lived. ...


1

It is absolutely false that the orbit of a satellite should always be a curve in a plane. And it is also false that the direction of the force should always point towards the center of mass of the attracting body. As far as the first statement, it should be noticed that in the case of central forces the existence of a plane where the orbit is confined to ...


1

It's not immediately obvious that your question is posed in a meaningful way, because you assume that quantities like force and energy have meaning in a completely electrostatic classical universe, or in a classical universe where the only interactions are purely static gravitational ones. In such a universe you can't build a spring scale or a calorimeter, ...


1

That's the tangential velocity for a circular orbit of radius $r$ subject only to gravitational force. If you have an elliptical orbit, then the gravitational force is changing both the radial and tangential components of the velocity. If the path is hyperbolic or parabolic, the same is true. Also, if you have drag, it's going to affect the tangential ...


1

There is absolutely no difference between charge and mass. Just as Earth exerts an attractive force on each object, proportional to that object's mass (and Earth mass, too), irrespective of the number or masses of the objects around, so the force exerted by a proton on each electron is independent on the presence of more electrons. However, electrons exert a ...


1

The potential at a point is derived with calculus,you cannot multiply $\frac{GM}{r^2}$ with the difference in positions because the gravitational field varies with distance from centre of mass of the body.Change in potential energy is defined as negative of the work done by conservative force so $$-F_{conservative} \Delta x = \Delta U$$$$F_{conservative}=- ...


1

A conclusion that I am getting with that assumption is that every object would turn into a black hole( I concluded this because the escape velocity that I'm getting is infinite). Let me explain how: In the case when gravity follows inverse square law then escape velocity is given by $$v_{escape}=\sqrt {2G \frac {M}{R}}$$ but here it is different. (What's ...


1

If we ignore the cases of elliptical orbits, the earth gives the satellite a gravitational attraction towards its centre, which is perpendicular to the velocity of the satellite. In your question, such a motion requires a force towards the poles of our earth, which is impossible to satisfy because a component force of the gravity would pull the satellite ...


1

You have a sign error. The potential energy for gravitation is lower the closer two objects are to each other, not higher. The potential energy of two objects separated by $r$ is $$U = U_0 -\frac{GM_1 M_2}{r}$$ where $U_0$ is the potential energy at infinite separation. $U_0$ is conventionally taken to be zero, but it doesn't really matter what value it ...


1

Think of one particle of a layer. It is moving in simple harmonic motion around the center, since $$a\propto x$$ Now think of the whole layer. The whole layer is moving together. If you can analyze how one particle of this layer moves than you know how the whole layer moves. If you understand how each layer moves then you understand how the whole sphere ...


1

In Newtonian Mechanics the Acceleration of an object is equal to the rate of change vector in the objects velocity, or to put it another way is the second derivative of an objects position with respect to time. Also $$\vec{a}=\frac{\vec{F}}{m}$$ with $\vec{a}$ being the acceleration, $\vec{F}$ being the force exerted on the object, and $m$ being the mass of ...


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