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37

John Rennie already gave the practical answer considering the atmosphere, noting that without doing anything objects near the ISS will deorbit quickly from drag. But that's letting reality get in the way of a good physics problem. I'll show that while a human can't send a ball crashing into the surface in one orbit, they can come close. The ISS is listed as ...


37

You don't need to throw the ball! At the altitude of the ISS the atmosphere is thick enough that it loses 50-100m of altitude every day due to the drag. At that rate over your ten year timescale the ISS would lose 180 to 360km. When you take into account the increased drag at lower altitudes ten years is enough to bring the ISS crashing to a fiery end. So ...


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: ...


17

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 ...


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 ...


8

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 ...


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 ...


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 ...


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 ...


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, ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


3

See moment of inertia is analogous to mass. Moment of inertia can be thought of as a physical "property" of the object similar to that of mass. And as we know that mass does not depend on any force or gravitational field or any other external effect, so does moment of inertia. Hope this answers your question.


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 ...


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

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

Yes the human body has a gravitational field, and yes it's large enough to be measured experimentally (see the Cavendish experiment).


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 ...


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 ...


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 ...


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 $$v^2 = \frac{2GM}{r}$$


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 ...



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