Consider the Earth, and a bowling ball held 186,000 miles (1 light second) above it. When the ball is released, it will start to fall vertically downwards towards the Earth.  Now consider the case if the Earth is moving sideways at 1000 miles/second. The bowling ball is released just as the Earth passes directly underneath.   Does the ball fall   a) vertically again, or   b) does it fall towards where the Earth was 1 second before?  Gravity propagating at the speed of light, would suggest answer b) but as most of the matter in the universe is travelling at very high speed, and planetary orbits are circularish, I think the answer is a)
6070
A thought experiment: Imagine the Sun is suddenly removed. We wouldn't notice a difference for 8 minutes, because that's how long light takes to get from the Sun's surface to Earth.   However, what about the Sun's gravitational effect? If gravity propagates at the speed of light, for 8 minutes the Earth will continue to follow an orbit around nothing. If however, gravity is due to a distortion of spacetime, this distortion will cease to exist as soon as the mass is removed, thus the Earth will leave through the orbit tangent, so we could observe the Sun's disappearance more quickly.  What is the state of the research around such a thought experiment? Can this be inferred from observation?
5603
The idea that as you move through the earth you get a symmetric cancelling of gravitational acceleration which approaches zero acceleration due to gravity at the center of the earth makes a lot of sense I think. I also think that is a pretty easy idea to understand physically.  On the other hand, I am just beginning to study and understand the general relativistic explanation for gravity and I'm wondering how this same idea is explained using general relativity. That is, what is the explanation for why gravitational acceleration decreases (linearly if I'm not mistaken) as you approach the center of the earth using general relativity? Specifically, I would like to understand what is going on with the curvature of spacetime in this scenario? Is it becoming less curved? Some math would be appreciated in answers!
57534
I read in  in this site that the KE a free-falling ball acquires is not originated by the attracting body but that energy was actually stored in the ball when it had been lifted to the height it dropped from.  In this way, it was said, gravity is subject to the conservation of energy principle and cannot change the total energy of an object.  Consider now the maneuver known as a  (also gravity assist) used by space probes such as the . A space probe approaches a planet with velocity $v$, slingshots around, and ends up with velocity $v+2U$, where $U$ is the velocity of the planet.  Consider the energy of the probe. Before, it was $E_i=\frac{1}{2} mv^2$ and after it is $E_f=\frac{1}{2} m(v+2U)^2$. It looks like $E_f$ is much bigger than $E_i$ - but where did the additional energy come from?   Is this not a violation of the conservation of energy principle?
113354
If an object is say thrown down (vertically) at an initial speed that is faster than its terminal velocity, what would happen to that objects speed? Would it slow down?
77058
A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.   Is my solution wrong? How come I can't get the correct answer?  x=horizontal axis , y=vertical axis Area=XY/2, when X=3,  A=6m^2, dx/dt=0.4m/s  Y=2A/X  dy/dx=-2A/x^2.....A=6,x=3      =-12/9  dy/dt=dy/dx * dx/dt      =-12/9 * 0.4      =-0.533333
93613
Imagine, if you will, there exists somewhere out there a supermassive planet. For some reason, this planet is only a shell of its former existence and all that is left is a crust of substantial size to keep it from collapsing in on itself (let's imagine it is stable at 5000' thick). Now, this planet was HUGE! How large, you ask? Large enough that the remaining shell is still able to maintain a gravity well strong enough to keep you anchored to the surface.  Pretty neat, huh?  Well, since this is a sight to be seen by everyone, you stroll along and, not looking where you are going, you tumble into a chasm and straight down into a hole that permeates the thickness of the shell. As you are falling down and down and down, you think:  "At what point do I stop falling down and I start falling up? And would I be able to retain the velocity, since gravity is a conservative force, and fall back through another hole on the surface on the opposite side? Where would the center of gravity even exist? I mean, aaaaahhhhhhh!"  What say you, browsers and scientists? A stretch of the imagination, sure, but is it physically possible? Could there exists a configuration similar to this where gravity could be concentrated to a point but still allow free travel up to the focus?
35176
This is kind of an odd question, but can somebody please tell me that I am crazy with the following question, I did the math, and what I am told to prove is simply wrong:  Question: Show that a ball dropped from height of h feet and bounces in such a way that each bounce is $\frac34$ of the height of the bounce before travels a total distance of 7 h feet.  My Work: $$\sum_{n=0}^{\infty} h \left(\frac34\right)^n = 4h$$  Obviously 4 h does not equal 7 h .  What does the community get?  I know that my calculations are correct, see Wolfram Alpha and it confirms my calculations, that only leaves my formula, or the teacher being incorrect...  Edit: Thanks everyone for pointing out my flaw, it should be something like: $$\sum_{n=0}^{\infty} -h + 2h \left(\frac34\right)^n = 7h$$  Thanks in advance for any help!
111065
In my Computer science class I was given a problem where I have to simulate a bouncing ball using "real physics". I have been trying to find a equation that will simulate the height of the bounce given a gravity and an arbitrary mass. And I will need to calculate the next bounce and it's height. A lot of the equations I've found require a time. But I don't really care about a time, all I want is to get the next height after the previous bounce until it finally hits height of 0 or close to it. I haven't taken a physics class since high school( 5 years ago) and that was basic physics.
75618
You swing a ball on the end of a lightweight string in a horizontal circle at constant speed.  Can the string ever by truly horizontal?  If not, would it slope above the horizontal or below the horizontal?  Why?
99658
I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
28797
I understand that nothing can move faster than light due to time dilation. I want to build upon my understanding of Einstein's theory of Special Relativity, so I came up with this hypothetical problem for myself:  If both particles are moving at 0.9c towards each other (according to an observer on Earth), what speed does each particle have relative to the other?  I'm stumped since I am only familiar with Newtonian classical mechanics to solve this (which would be wrong). I am aware of Lorentz' factor $\gamma =\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ and the time dilation equation $t_m = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}}$, but I'm not sure how to apply it in this instance. I'm just looking for a gentle push in the right direction. Edit: My prior research:  This article, ,  talks about the effect of a baseball thrown at 0.9c would have as it traveled towards a batter. This is not what I'm looking for since it doesn't explain the math behind finding the collision speed between two high velocity entities.  This question on Yahoo, , asks a similar question to mine, but no math is involved to demonstrate the calculations behind the reasoning.  This question on Physics.SE, , does not answer my question either. I have not been able to find a similar Physics.SE question which can provide me to clues to my own hypothetical question.  I have also Googled the following phrases:    i. &quot;collision faster than the speed of light&quot; ii. &quot;collision of high speed particles&quot; iii. &quot;lorentz transformation in a collision&quot; iv. &quot;collision analysis at light speed&quot;
118492
If we are in a free fall which implies we are accelerating at 9.8 m/sec every sec. And let's say that we are falling into a pit that has enormous depth. So isn't this be possible that we may accelerate and surpass the magnitude of speed of light?
16367
This is a question from my Calculus book.  I was showed the correct answer but I really need to learn what the process is to achieve the answer.  Here is the question:  If a ball is thrown into the air with a velocity of $40 \frac{ft}{s}$, its height in feet $t$ seconds later is given by $y=40t - 16t^2$ .  Find the average velocity for the time period beginning when $t = 2$ and lasting   (i) 0.5sec   (ii) 0.1sec   (iii) ...
43666
When I shoot a ball vertically upward, its velocity is decreasing since  there is a downward acceleration of about $9.8\,\mathrm{ms}^{-2}$.   I have read that at the top most point, when $v = 0$, the acceleration is still $9.8\,\mathrm{ms}^{-2}$ in the downward direction where $v=0$. That is, the acceleration is still the same.  But at the highest point, the ball is stationary, so it is not even moving. How can it accelerate?
66864
The picture shows an accelerating spaceship with two clocks inside it. It is so far away from all other bodys that gravity is of no importance.   Will the bottommost clock be slower than the topmost one? Or will both clocks have the same speed?
121441
Earth's orbit is a slight ellipse, so to conserve momentum its speed increases when it is closest to the Sun. If the speed changes there is an acceleration. If there is an acceleration there is a force. Even if the change is small and gradual, wouldn't we experience a force because the Earth is so massive?
9487
I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km.  The true height is about 20km.  My question is: why is there this discrepancy?  The calculation I did was to imagine placing a ball down on the spinning Earth.  Wherever I place it, it shouldn't move.  The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.  Gravity wants to pull the ball towards the poles, away from the bulge.  It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole.  This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.  If the ball doesn't move, this potential must be constant, so  $$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$  we might as well let the constant be zero and write  $$h = \frac{(\omega R \sin\theta)^2}{2g}$$  For the Earth,   $$R = 6.4*10^6m$$  $$\omega = \frac{2\pi}{24\ \textrm{hours}}$$  $$g = 9.8\ m/s^2$$  This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.    , the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected.  (Wikipedia cites diameter; I estimated radius.)  Where is the extra bulge coming from?  My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so.  It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).  (Wikipedia also includes , without derivation, that agrees with mine.)