New answers tagged

1

In figure A the satellite is orbiting in a circular orbit. By the satellite increasing speed it would turn the orbit into a more egg shaped orbit as shown in B. This would rather increase the time for orbit revolution rather than decreasing it When the satellite is approaching B.1 it's velocity will begin to slow down as earths gravity begins to pull it. ...


4

My simple answer to this is that high and low tides are a bit arbitrary; they're the turning points of a continuous function over time. At the places you've mentioned, the (usually most significant) bi-daily tidal harmonic is "swamped out" by the (usually less significant) daily harmonic. It's still there – it's almost certainly non zero, but it's just not ...


0

I'm not quite sure how he gets a value of $36.36$ for that constant, because it depends a bit on exactly how he does his approximations and what values he assumes for different constants. But basically it's this: For a spherically symmetric mass distribution like the one considered here, the gravitational acceleration is simply, from Newton's law for the ...


0

As far as I understand your question, you can consider the Moon as continuously falling down to the Earth (or continuously following the moving Earth, if you want) due to the gravity. Nothing compensates gravity. However the problem is that the Moon aims not precisely into the Earth, but a bit away - that was original "error" from the Moon's first day. The ...


0

Lets distinguish 'rotation' into 'spin' (rotating around itself), and 'orbit' (orbiting around another object). The orbital motion is due to gravity. The 'centrifugal force' (instead of 'kinetic energy' of rotation, per se) is balanced by the 'gravitational force' [1]. The spin of an object doesn't need an external force (gravity from another object) to ...


21

The tides are a result of the response of the Earth's oceans to the tidal forces exerted on the water by the Moon and the Sun. The responses are vastly complicated by the Earth's rotation about its axis, by the physical geography of the Earth, and by the nature of the orbits of these bodies. Of key interest with regard to this question are the inclination of ...


9

OK, here's a theory: It's all (when I say all I mean mostly) got to do with the fact that the moon's plane of revolution is inclined about 20 degrees or so relative to the earth's equator. That causes differences in the frequency of tides in various latitudes. In order to see how that happens, here's a simplified example: Draw a cross, so that you have 4 ...


0

I'm no expert on this, and the detail of tides is complex, but basically the situation is a =s follows. The moon (and sun to a lesser extent) provide a gravitational forcing with a period of just over 12 hours. Then the resultant ocean movement interacts with the local sea floor geography, which modifies the response. In most places the result is the ...


2

Hints: Note that the derivative of the sign function $$ {\rm sgn}^{\prime}(z)~=~2\delta(z) \tag{A}$$ is twice the Dirac delta distribution. This fact seems to be at the heart of OP's question. Repeated differentiations of the Mestel disk potential $$\Phi~:=~ v_0^2 \ln(r+|z|), \qquad r~:=~\sqrt{R^2+z^2}, \tag{B}$$ leads to $$\frac{\partial \Phi}{\partial ...


2

Just to more fully answer your question, here is an example of what differences in distances can mean mean as far as they affect gravitational forces. The planet Jupiter is extremely massive, and one side of one of it's moons, Io, feels a slightly larger gravitational pull than the opposite side. This difference in distance results in a gravitational force ...


4

That formula holds for a simple pendulum of length $L$ in a gravitational field $g$ and released from rest with an amplitude $\theta_0$. Since this system is conservative its mechanical energy is constant and equals the gravitational potential energy when it is released. Setting the zero of potential energy at the fixed point of the pendulum, the mechanical ...


2

Yes. All objects are gravitationally attracted to one another. Even people. Therefore, the earth will draw you to itself no matter how far out you are.


1

Gravity is an attractive force, and classical Newtonian theory is adequate to answer in the affirmative: starting from two objects in space at rest, the attractive potential is 1/r , r their distance, and they will fall on each other. The reason planets and satellites have a stable orbit is due to the solutions of the kinematic equations when there exists ...


3

Tl;DR: It is mainly do to the skier's positioning on the ski and how being thrown off-balance by the avalanche affects it. Concerning the first part of the first question: 1) If all free falling objects accelerate at the same rate (this was on a fairly steep mountain section), why did we get "trapped" into the avalanche, when our acceleration already ...


2

But my problem is that I don't understand how a force can be a vector, in my head I see it as a direction vector and some power number Right. If the direction is a "unit" vector, then you can compare the magnitudes of different forces to compare the strengths. But you can multiply the magnitude and the direction to get a new vector that contains ...


1

Force has both a magnitude and direction, which are the properties of a vector. The magnitude can be given by $\vec{F} = G\frac{M_1M_2}{\vec{r^2}}$ where $G$ is the gravitational constant and $M_1,M_2$ are the masses. the distance from each other is represented by the vector $\vec{r}$ which will be the displacement from the origin.


0

Not sure about the etiquette of this but I think I can now answer my own question. Please post if there is a better answer. The problem is that I misunderstood the meaning of the statement "the geodesic is the path the optimises the proper time". What this means is that given two endpoints, say $x_0^\alpha$ and $x_1^\alpha$ a geodesic is a path ...


1

@DavidZ comment in the suggestions is correct: there is no unique, general solution to this problem. Depending on the context, dozens of different heuristic approaches are used in simulations. This is part of the general 'N-body' problem. The most literature is in the context of large N-body, cosmological simulations where the problem is called often ...


3

The Sun pulls on the Earth as well. So both Earth and Moon are "falling towards the Sun" all the time, just as they are moving in almost the same orbit. Earth causes the orbit of the Moon to "wobble" a little bit. If you were simply given the coordinates of the Moon as it moves around the Sun, you would notice there is a deviation from the expected ellipse ...


2

I assume you are asking about a light sphere falling from infinity towards a heavy one. Well, the potential energy at $r$ is $$P = - \frac{GMm}{r}$$ And kinetic energy is $$K = \frac{mv^2}{2}$$ Total energy is zero, so $P=-K$ or $$v^2 = \frac{2GM}{r}$$


0

My guess is that it would be impractical. Your reasoning is basically correct: a longer right arm will grant you a larger $\Delta h$, increasing the speed of the projectile. But there would be some issues: The structure would have to be higher, otherwise the weight would smash into the ground. Hence, it will be heavier and more difficult to carry. The ...


3

If I understand your notation right, then by $h_L$ you mean the height of the left weight? In that case your formula doesn't make sense: To account for energy conservation, you would have to take the heights before and afterwards of the right weight. Then you get \begin{align} m_R*g*h_{before} = m_R*g*h_{after} + \frac{1}{2} m_L * v^2 \end{align} You ...


1

The equations look like the displacement of an object subject to zero acceleration (first equation) or the acceleration of gravity (second equation). If we assume that the first equation represents horizontal motion and the second, vertical motion (suggested by the 9.80 m/s$^2$), we can try to come up with a scenario. By inspection of equation (1) we see ...


3

These are a couple of classical equations of motion derived from Newton's laws, dealing with the motion of a body with initial velocity of 50 m/s at an angle of $\theta$ degrees (or radians) with respect to the horizontal/ground/$x$-axis, assuming that the coordinate system is flat. The object is subject to a downwards acceleration, which we infer is due to ...


2

Since you have an equilibrium situation the net force on any portion of the water is zero. In region $B$ the horizontal forces acting on the water are either provided by the walls of the container in the left-hand diagram of by the water in regions $A$ and $C$ in the right-hand diagram. So as far as the water in region $B$ is concerned it makes no ...


0

Only if the situation is symmetric and the masses are aligned like an equilateral triangle all masses will meet each other at the same time: If the initial distances are not the same the point masses will miss each other and start orbiting around each other. For example the initial configuration of an isosceles triangle: Here the two smaller masses ...


0

In a vacuum the rod will keep its orientation because of the equivalence principle, but with air resistance the heavier part will tilt toward the ground because its mass and therefore force is higher. The force depends not only on the mass of the planet but also on the mass of the falling object, and since air resistance is also a force that acts on the ...


0

Because a Newtonian liquid's molecular characteristics, they tend to possess a definite volume without any specific shape. The important aspect is the height of the liquid not the shape of the container. If the height of a liquid with a specific gravity of 1 is 2.31 feet high (rounded off), The weight of that column is one pound over an area of 1^2 inch. ...


2

I think you are confusing the escape velocity with terminal velocity. While for the local velocity the limit is only the speed of light, terminal velocity is achieved much sooner because of the air resistance. The equation can be found here and depends on the shape, size and density of the asteroid. For particles travelling near the speed of light you have ...


2

For the sake of simplicity (at the expense of real-life accuracy), let's assume that an asteroid is already travelling at some speed $v_0$ directly toward the Earth, and it never deviates away from that direct line path (despite the revolution of the Earth around the Sun). Let's also assume the distance from Earth is very large compared to the radius of ...


-5

No, you won't speed up and the fastest speed the earth can accelerate you to is at escape velocity.


0

To begin with, $v dv= a ds$ is the differential equation that relates velocity with variable acceleration. $a=(G*\mathrm{massOfEarth})/s^2$ Substitute it in the first equation and integrate for $v$, 0 to $v$ and $s$, distance between the moon and earth to 0 Hope this is right


1

Instead of forces, you work with conservation of potential and kinetic energy. In this case, the kinetic energy that gives this force is $-\frac{Gm_1 m_2}{r}$. The difference in potential energy is transformed into kinetic energy. With forces, you'd have to go the long way around (integrate the acceleration over time to get the velocity and figure out the ...


2

This question is easilly answered by considering the gravitational potential of earth, and invoking conservation of energy. The potential is $V(r) = -G\frac{m_1 m_2}{r}$ and the kinetic energy of the moon will be given by $$ K = V(r_{moon}) - V(r_{impact}) $$ where $r_{moon}$ is the current distance from the center of the earth to the center of the moon, ...


2

The basic equations, assuming no air drag($^*$) are as follows. At $t=0$ we drop the object from height $H$, we assume its initial speed is zero ($v_0=0$). Only gravity ($mg$) is acting on it. If the object free falls to height $h_1$, energy conservation then shows us that its speed has now become: $$v_1=\sqrt{2g(H-h_1)}$$ At height $h_1$ a braking ...


1

Many ways of looking at this: If there were any differences in pressure in horizontal direction, the pressure difference would mean a force that would induce motion of the fluid until the pressure difference would drop to 0. So in stationary state, pressure must be constant horizontally, and vertically, the difference in pressure between different heights ...


0

I cannot comment (yet) - that is why I have to ask via an answer... Please let me know if I understand your question right: You want to know why the pressure on the lower, inner surface of the bucket is the same, considering that there is more water in the bucket with the inclined side surfaces... If this is your question think about that: There is more ...


0

First, calculus isn't just really small steps: I can show you limit processes that disagree with any really small step based solution, like making a staircase with smaller and smaller treads. The total "tread plus rise" size remains 2k, while the limit is a line with length sqrt(2)k. However, almost all of the parts of calculus that work in predicting ...


3

UPDATED ANSWER : The centre of gravity will always be the same as the centre of mass in a uniform gravitational field (constant in magnitude and direction). This applies for bodies with non-uniform density as well as those with uniform density. The Earth's gravitational field can be considered uniform if the dimensions of the object are much smaller than ...


4

Since you still seem puzzled I'll try a different tactic here: You're showing the tangential velocity (A) and the radial acceleration (B) and adding them to get the green arrow. What you're missing is that this occurs in a gravity field. As the initial path climbs away from the object it's orbiting it loses velocity. This shows up as a third arrow ...


3

After reading the answers from other people, I see that you are still confused, so I thougth i coult take my chances on clearing up your confusion. It seems that your confusion is why does the orbiting body not accelerate to infinite velocities (or crashes to body A) eventually, and the origin of your confusion is in your assumption that the resulting ...


5

Your vector sum drawing is incorrect. Considering the simple case of a perfectly circular orbit, there is a tangential velocity vector and a perpendicular (inward) acceleration vector. Strictly speaking, you can't add them because they are different quantities (acceleration vs velocity). At some point in the orbit, the object has an initial velocity V in ...


14

Because the direction of the velocity changes. The velocity will start to point less and less 'towards' point A and when the distance between A and B is the smallest, the velocity will make a right angle with the radius, which means acceleration vector also makes a right angle with the velocity. At this point the radial component of the speed is zero and the ...


12

This question points out the importance of symplecticity in physics. In an orbital simulation, suppose one simply advances state via $$\begin{align} \boldsymbol x(t+\!\Delta t) &= \boldsymbol x(t) + \boldsymbol v(t)\, \Delta t \tag 1 \\ \boldsymbol v(t+\Delta t) &= \boldsymbol v(t) + \boldsymbol a(t)\, \Delta t \end{align}$$ where $\Delta t$ is a ...


20

You have to consider the limit of infinitesimally short time, in which the (vertical on the paper) component of velocity is infinitely short, and thus also the angle changes for an infinitesimal amount. In this limit, the correction to the length is quadratic in the time step and vanishes exactly in the physical limit of continuous time. Pythagoras: ...


5

Any body travelling with increasing velocity increases its kinetic energy $KE$. Since in your system: $$KE+PE=\text{constant}$$ where $PE$ is potential energy. Therefore increase in $KE$ results into decrease in distance between the objects (so as increase $PE$). Note: $KE$ is always positive. $PE$ can be positive or negative. $PE$ is negative for bound ...


7

It's true that if you know the masses of, e.g. two orbiting stars $M_1$ and $M_2$, their orbital period $T$, and the distance $d$ between them, then you know $G$. And we can measure $T$ pretty well and $d$ fairly well. But how do you think we figure out the masses of the stars? We can't just count the amount of stuff in them; we have to infer the mass from ...


1

Let's look at F = ma. This is not true for each force individually, for sure - each particle experiences many forces but only one acceleration. It's really Σ(all forces)F = ma. Even if we say just the force of gravity, we're looking at Σ(other objects)Fg + Σ(other forces)F = ma. So let's think about applying MOND. We have a particle that's ...


5

If you're thinking about stable orbiting systems the big difference between gravity and the magnetic force is that magnetic monopoles do not exist. The simplest source of a magnetic field is the magnetic dipole. By contrast gravitational monopoles exist but gravitational dipoles do not. The Sun and the Earth are both (approximately) gravitational monopoles, ...


3

I think the two big factors would be that the Earth would 'want to' become tidally-locked to the Moon and the Sun. The Moon would win here, which is easy to see because tides are caused more strongly by the Moon than by the Sun. So in due course the Earth would end up tidally-locked to the Moon with a rotation period which would be the same as a lunar ...



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