# Tag Info

4

The easiest way to calculate escape velocity, is neglicting Earths rotation and assuming the object takes of in a radial direction. Then, indeed, you start from $$E = K_1 + U_1 = K_2 + U_2$$ where $K_1=\frac{mv^2}{2}$ and $U_1=- \frac{GMm}{r}$. Since the range of gravitional forces is infinity, you say (theoretically, not practically) that an object has ...

0

Even without gravity, momentum conservation will still hold. If you elastically scatter an unknown mass $m$ with an initial (known) velocity $v$ against a known mass at rest, for instance we can take the SI standard of $1$ Kg, then from the resulting measured velocities you should be able to find $m$. Thus Yes, there will still be a mass.

0

This scene was clearly shot with a stationary bus oriented (nearly) nose down and a green screen. A backpack and book are clearly dropped on the rider and accelerate at the normal rates. The rider hits seats and can't hold them as he falls. There is no way that air resistance on Earth could provide enough force to change the movement of the bus (and ...

3

No, no you guys (Except Floris and those who up-voted him) have missed an important observation... Look Carefully at the video again. At first the bus just tilts as the bridge bends. When the bus starts tilting (due to friction with the bridge it has not yet started falling) it has not yet obtained considerable vertical velocity. However as the man loses ...

0

You need to measure the torsion constant by timing the oscillation before adding the two big spheres. If the big spheres are present the system is no longer a simple harmonic oscillator because the restoring force is now a complicated function of the rotation angle.

-5

I think everyone is wrong. It's dark matter that causes heaver metals to be pushed to the center and lighter metals/matter to "float" to the surface and beyond. Everything after the ozone is dark matter, an ocean of it! There are several factors to this... (1) The universe is expanding, and with it, dark matter. This causes an equally and accelerating ...

-1

Oh, absolutely. With a Superman-like body you can also 1) Lift an entire mountain in two hands without the rock disintegrating under the local pressure, and without the mountain falling apart due to preexisting weaknesses in the rock, 2) Travel faster than light 3) Travel faster than light and travel backwards in time 4) Hear sounds so faint that they ...

-1

I believe you could overcome gravity (in the atmosphere) in the same way as birds do - you just need enough muscles to do that.

1

First of all let's study an imaginary system where both the bus and the person are not subject to drag forces due to the air: If the person is not bounded to anything he will be subject to free falling and thus to a uniform acceleration $g$. Also the bus will be free falling and thus they fall together with the same velocity. If we take the drag forces into ...

38

The bus experiences considerable drag, and will therefore fall more slowly than a person inside the bus. The scenario is possible in principle - but after carefully viewing the clip and doing some calculations, I believe that the details are inaccurate. Assume the bus has a mass of 5000 kg (pretty light for a bus), and is 3 m wide by 3 m tall - so the ...

21

If the bus was in a vacuum (both inside and outside), then the passenger would float. However, the effects of air resistance on the two objects (passenger and bus) are probably not negligible in such an instance. The bus will be moving relative to the outside air, and so will be accelerating towards the ground at a rate less than $g$. If we then released ...

6

At first, the bus and the person would accelerate at the same rate due to gravity. However, the situation is more complicated due to air resistance. The bus experiences air resistance as it falls. The person inside the bus experiences less air resistance because the air inside the bus moves with the bus. This means that the person does not experience as much ...

1

In a reference frame where the center of mass of the Earth-Sun system is at rest, we will have $$m_\text{Sun} \vec{v}_\text{Sun} + m_\text{Earth} \vec{v}_\text{Earth} = 0 \quad \Rightarrow \quad \vec{v}_\text{Sun} = - \frac{m_\text{Earth}}{m_\text{Sun}} \vec{v}_\text{Earth}.$$ In particular, if this is true at some initial time, then it's true at all ...

2

I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...

3

I am not sure what is the path $C$ you are integrating over? In your definition you evaluate $U(C)$ which in the present case of force is independent on the explicit path you choose but still depends on initial and final point, i.e. $U(p_1,p_2)$. In your final result it seems you are actually 'walking' three times the path $p_1=(-\infty,y,z)$ to ...

2

The black machine is a weight lifting machine. It is self contained with no power source. If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you ...

0

I think the previous answers misunderstood the question. It's rather obvious that a balloon cannot have buoyancy in vacuum, and therefore cannot float into vacuum at steady speed. However, one needs to also consider the possibility that a balloon would achieve enough speed in lower atmosphere to bring it to 2nd orbital velocity. Imagine a huge light ...

1

break earth into 2 parts in which each part is nearly equal to half of volume of the earth Yes and no. No, as in you can't neatly split the planet in half. Most of the earth is liquid and will re-form once the cutting device has passed through it. Much like cutting pudding. Yes, if you want 2 half-earth balls when you are done and don't care about the ...

0

The purpose of all this is to calculate and derive potential equation of the gravitation field. Let us assume that we have symmetrical sphere object that "generates" gravitational field. We want to know if the field depends on the objects "homogeneity" (correct me if this is not quite the right term). So from Gauss theorem (it has different names, but they ...

4

One can answer this question by calculating the energy needed to shift half the Earth's mass so that it is infinitely far from the other half. Let's calculate the gravitational potential energy released as we create a planet: assuming a constant density $\rho$, when the planet is growing and of radius $r$ and thus of mass $M(r)=\frac{4}{3}\pi\,r^3\,\rho$, ...

0

No, because the vast majority of the planet has a molten interior and where it is not in the liquid phase it is held in solid phase by the internal pressure. You could maybe disperse it into space with a big enough bomb, but not actually break it into two parts.

1

For the sake of simplicity assume the binaries' motion to be circular. Then you can use the formalism of the circular restricted 3-body problem (CR3BP) to model the motion of the test particle $m_3$. Your Lagrangian will be time-independent and the conserved quantity (Jacobi constant) can be evaluated at infinity to give an equation for conditions of escape, ...

0

As we go above or below the surface of earth the value of g decreases since g is inversely proportional to height

-1

Your derivation is for equality of force at a certain distance from the center. This equality will not hold at other distances as the particle is released. The correct derivation is the one involving energy and is first one involving a factor of 2.

3

What you have derived is the equation for the orbital velocity for a circular orbit. This is the velocity at which the acceleration required to keep moving in a circle exactly matches the acceleration due to gravity. To derive the escape velocity you need to know that the potential energy for an object of mass $m$ in the gravitational field of a planet with ...

0

What you are basically asking is how to convert initial conditions, position and velocity, to orbital elements. In this case both position and velocity are 2D vectors, with a reference frame positioned at the center of the celestial body, with gravitational parameter $\mu$, and the orbital elements: "longitude of the ascending node" and "inclination" do not ...

3

The $F$ you've found above is not the "real" force in GR, but rather an "effective force" derived from an "effective potential". Basically, we can use the fact that there are multiple constants of the motion to reduce the full three-dimensional problem down to an equivalent one-dimensional problem. This procedure works in Newtonian gravity1 as well, as ...

0

You have to think of what happens when object $B$ is "converted to energy". Since there is no such thing as raw energy (dark energy excluded) $B$ has to be transformed into other particles and or kinetic energy. As many of the other comment have mentioned the kinetic energy also has a gravitational pull, so initially after the "conversion" nothing has really ...

0

By "converting to energy," I'm assuming you mean converting part of the mass of the system to photons. These photons will quickly zoom off to large distances, making them irrelevant gravitationally. The original gravitational potential energy was caused by the original masses A and B warping spacetime. This warping is not instantaneously changed when mass B ...

0

I would need to look into this more, but I believe gravity couples to energy rather than the Lorentz invariant mass $m$. If this is true, then it doesn't matter if you view the mass of particle B as mass or energy.

1

Here's how I demonstrated this concept to my son when he was younger. Take a plastic bottle and put some pebbles or little toys in it. Then toss it in the air and catch it. If you look in mid-flight, you can see the little toys just floating around inside the bottle. But they're still at 1G. And when you look at this example, it's totally obvious what's ...

-2

NOOOOO!!! This is wrong. Granted the pendulum formula (T = 2 * pi * sqrt(L / g)) does not take into account mass of the bob, much less the pendulum, mass can and does affect the pendulum period. The Pendulum Formula is accurate and i give it credit, but its variables are broadly defined. T represents time or period, and g represents gravitational ...

-3

Here you have to be clear how you are looking at the energy. Of course all energy can take the form mass times c2. In this particular case of object A and B we have a total energy equal to mc2 where m is mass of A plus Mc2 where M is mass of B. With this you subtract the gravitational potential energy since it is attractive. You then get the system total ...

2

It is better to think of the equation $E =mc^2$ as a true equality rather than a conversion. Mass is energy. If one has that mindset, then it is intuitive that energy has a gravitational field. A hot cup of tea weighs more than a cold one.

2

There is not any mathematical but beside that Newton combined his laws of motion with Kepler's laws and deduced the law of gravitation. Path of planets around the sun are elliptical so for simplicity we can assume the orbit to be circular. Let us consider $a$ planet of mass $m$ moving with constant speed $v$ in a circular orbit. $$T=\frac{2\pi r}{v}$$ and ...

1

We can't expect to be able to take the principles of mathematics and come to conclusions about physics problems. At some point observation of phenomena is required. We then take these observations and use them to conclude whatever we can to find is true about the world. Historically, two different people played the two roles of data collection and then ...

-3

Gravity is an effect. There is theoretical and interpreted experimental results that point to the concept that gravity is a result of neutrino refraction through matter. if this is true then the universe is comprised of a mat of neutrinos and neutrino strings in constant intertwining motion. a good analogy of gravity is to have two hoses or sprays facing ...

-1

The sun is a ball of gas, so "relative to its surface" and "walk upright" are a bit of a problem. You might fire a rocket engine hard enough to balance the force of gravity, and stand still. The acceleration of gravity gets weaker at large distance from the sun, but never is 0. $a = GM/r^2$. So you could say it is gone when it is too small for you to ...

0

Your problem is that you are trying to do this in Cartesian coordinates (x,y). This makes the math much harder than it needs to be. It is more natural, when dealing with circular motion, to use polar coordinates - we express a position relative to the origin by its distance ($r$) and the angle relative to some reference axis ($\theta$). The relationship ...

1

If you're talking about the weight you'd measure if you stood on a scale, to a very small extent, yes. If there are other objects nearby me, chances are some will have some extra pull upwards, and some extra pull downwards. The direction and magnitude however will be small, if you're talking about everyday objects/people in a room. Say one person, 60kg is ...

3

I suggest that it doesn't make much sense to say that the planets orbit the barycenter of the solar system. Beware that you are going very much against the grain of the best models of the solar system in writing that. All three of the leading ephemeris models (JPL's Development Ephemeris, the Russian Institute for Applied Astronomy's Ephemerides of the ...

0

According to my knowledge, Gauss Law can be applied to any function where the quantity(the force) is inversely proportional to distance squared.Now, is the force required to be conservative?

0

Indeed, we have that: $$\nabla \cdot \vec{g}=-4\pi G\rho$$ Where $\rho$ is the mass density. Integrating both sides, we have: $$\iiint_{\Sigma}(\nabla \cdot \vec{g})\:\mathrm{d}V=-4\pi G\iiint_{\Sigma}\rho(\vec{r})\:\mathrm{d}^{3}\vec{r}$$ But by Gauss' Law (the divergence theorem), we have: \oint_{\partial \Sigma}\vec{g}\cdot\mathrm{d}\vec{A}=-4\pi ...

8

You seem to be groping towards the fact that the gravitational three-body problem is, in general, not solvable. We can get away with saying "the sun is at one focus of an elliptical orbit" in the solar system because the sun is so much larger than anything else around. The sun is 1000 times more massive than Jupiter, so the sun-Jupiter barycenter is about ...

0

It will oscillate around the center of Earth. To find the equation you will need to solve the generic oscillation problem: the gravity force / acceleration depends on the distance to Earth's center while everything which is further from the center will not impact on the acceleration. $g(x)=G\cdot M(x)/x^2$; $M(x)=\mathrm{average~earth~density}\cdot ... 0 Newton formulated his Law of Gravitation from observation and experimentation. He termed this method as 'induction'. 2 To answer the question in your title, he used his newly found fluxions (calculus) to prove that Kepler's laws of planetary motion imply a radial, inverse square law. Feynman's Lost Lecture is a mixture both of Feynman's attempts to give the simplest possible explanation of how one goes about this derivation and his insights into the history of how Newton ... 1 Newton devised a very good law of gravity (until Einstein came along) where the force between the two bodies is scaled by a very small number usually written as a capital G. It's a general law that applies to any two bodies. But if you plug in the mass of the earth, the mass of a test ball, and the distance between the center of earth and the test ball, then ... 0 Ellipses, parabolas and hyperbolas are both: Defined as the conic sections you speak of: work out the intersection between the cone$\vec{R}.(\cos\phi\,\hat{X} + \sin\phi\,\hat{Z}) = \sin\theta\,|\vec{R}|$and the plane$z=const$where$\theta$is your polar angle and$\phi\$ the angle between the cone's axis of symmetry and the slicing plane and you'll ...

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