If everyone on earth except one person were to gather up in one location, say the North Pole, and jump at the exact same time and stay airborne for about 1.5 sec. And that one person who did not jump would stand on a weigth scale, would he/she see any difference to their weight assuming the scale can messure weight to a $10^6th$($10^6 $ because I figured that a  number with a higher accuracy would change due other changes in the earth or the solar system, such as the gravitational pull of other celestial masses, no?) of a kilogram?  My assumption here is that the force acting on an object from the earth differs if a big mass would suddenly rise from the surface of the remaining mass(the  earth), does that make sense? $$F = \frac{Gm_{earth}m_{standing\space person}}{r^2}$$ $F$ would become smaller when $m_{earth}$ becomes decreases due to the jump.
120609
Escape velocity depends upon the mass of earth. my doubt is whether human mass adds into it or not? If it add then Earth's escape velocity would be increasing!
90871
I've recently been contemplating things like artificial gravity in a rotating space ship (for example, the ) and learning about the Coralis effect and other interesting fictitious forces that appear in rotating reference frames.  It occurred to me that, if you were living on a space station spun for artificial gravity, it might be possible to become weightless in such an environment by travelling fast enough against the spin of the ship. My friend, who I was discussing this with, thinks otherwise, that you would actually be under more g's as you tried to gain the speed to lift off from the inner edge of the ship.  In terms of concrete numbers, say the ship is 4 kilometers in radius (as an O'Neill Cylinder would). At this size, the rate of rotation to generate 1g in centrifugal force (the artificial gravity) is under 0.5 rpm. If a vehicle travels on the inside of the cylinder against the spin fast enough, eventually the vehicle would cease moving when looked at from a static reference frame. I would think at this point, the vehicle could simply push off from the edge of the cylinder and float towards the center of the station. Is this correct?  Furthermore, would this method of "getting into air", as it were, be easier than it would be to counteract real gravity on Earth (with things like planes that generate lift with their wings)?
36801
Can someone help me giving a qualitative answer to this problem in General Relativity:  Imagine you are on earth with two perfectly synchronized clock's. If you hold on in your hand but you throw the other one in the air and catch it after a certain time interval, would they still give the same time?  I think that the one in the air would run faster but I cannot explain why.  Thank you!
81576
$M$ = mass of the Sun  $m$ = mass of the Earth  $r$ = distance between the Earth and the Sun  The sun is converting mass into energy by nuclear fusion.   $$F = \frac{GMm}{r^2} = \frac{mv^2}{r} \rightarrow r = \frac{GM}{v^2}$$  $$\Delta E = \Delta M c^2 = (M_{t} - M_{t+\Delta t}) c^2 \rightarrow \Delta M = \Delta E / c^2$$  $$\rightarrow \frac{\Delta r}{\Delta t} = \frac{G}{v^2 c^2}.\frac{\Delta E}{\Delta t}$$  Sun radiates $3.9 × 10^{26}~\mathrm W = \Delta E/\Delta t$  Velocity of the earth $v = 29.8 \mathrm{km/s}$  There is nothing that is stopping the earth from moving with the same velocity so for centripetal force to balance gravitational force $r$ must change.   Is $r$ increasing? ($\Delta r/ \Delta t = 3.26070717 × 10^{-10} \mathrm{m/s} $)
7117
When I'm rotating, I feel centrifugal force. But if I were the only one in the universe and rotating, wouldn't I just kinda be still (since I'm not rotating with respect to anything) or would there just be a force pulling me out? What if there were another person a billion miles away, this should not change anything, right?
3850
When a person lifts a book while standing on the earth's surface. Why is his force considered external to the book-earth system? Isn't his force internal like the case of the forces that cause the exploson of a bomb. I think the forces the person exerts on both the book and the earth are internal since they are equal but opposite therefore resulting in the position of the centre of mass of the system being maintained and if i'm right is the mechanical energy of the system constant?  Am i wrong? Please correct me.
612
The common understanding is that, setting air resistance aside, all objects dropped to Earth fall at the same rate. This is often demonstrated through the thought experiment of cutting a large object in half. The halves clearly can't then fall more slowly just by being sliced in two.  However, I believe the answer is that when two objects fall together, attached or not, they do "fall" faster than an object of less mass alone does. This is because not only does the Earth accelerate the objects toward itself but the objects also accelerate the Earth toward themselves. Considering the formula: $$ F_{\text{g}} = \frac{G m_1 m_2}{d^2} $$  Given $F = ma$ thus $a = F/m$, we note that the mass of the small object doesn't seem to matter as when calculating acceleration the force is divided by the $m$ term, its mass. However, this overlooks that the force is actually applied to both objects, not just to the smaller one. The acceleration on the second, larger object is found by dividing $F$, in turn, by the larger object's mass. The two objects' acceleration vectors are exactly opposite, so closing acceleration is the sum of the two:  $$ a_{\text{closing}} = \frac{F}{m_1} + \frac{F}{m_2} $$  Since the Earth is extremely massive compared to everyday objects, the acceleration imparted on the object by the Earth will radically dominate the equation. As the Earth is $\sim 5.972 \times {10}^{24} \, \mathrm{kg} ,$ a falling object of $5.972 \, \mathrm{kg}$ (just over 13 pounds) would accelerate the Earth about $\frac{1}{{10}^{24}}$ as much, which is one part in a trillion trillion.  Of course in everyday situations, we can for all practical purposes treat objects as falling at the same rate because of this negligible difference—which our instruments probably couldn't even detect. But I'm hoping not for a discussion of practicality or what's measurable or observable, but what we think is actually happening.  Am I right or wrong?  What really clinched this for me was considering dropping a small Moon-massed object close to the Earth and a small Earth-massed object close to the Moon. This made me realize that falling isn't one object moving toward some fixed frame of reference, but that the Earth is just another object, and thus "falling" consists of multiple objects mutually attracting in space.
18076
So I came across this picture on facebook   My logic says it should show 200N, because you are pulling it with 100N on each side. On the other hand (as one of the facebook comments also stated, also saying is he is wrong he wasted 20k$ on physics degree), if you replace the right side with a wall you get this situation:    In this case I guess everyone agrees that the dynamometer would show 100N force. But if you isolate the dynamometer and "cut" the ropes on either picture, you would get the same thing. Because the system needs to be balanced, there would be 100N on each side.    And I study engineering too but I am confused by this. Why do we get the same distribution of forces, but dynamometer shows a different force?
112221
As small as it may be, does every 'thing' have a gravitational pull?  That is, something with mass or energy. No matter how obsolete or negligible it may be, is it there? If so, how is it calculated? What does 'it' affect?
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?
17142
I had a recent conversation with my girlfriend, who is a physics grad student. She was kind enough to listen to me rant about an idea concerning escape velocity. Unfortunately, I am still thinking about this question, but don't want to bother her with it.  My thought is that we shouldn't actually need to reach 11km/sec to leave the earth's gravity. Instead of this, can we simply apply enough force to overcome earth's gravity without reaching that speed? How we would do this is not a part of my question.  Orbital altitude is $99$ miles. Travelling $60$ mph straight up should get you there in $1.65$ hours, granted you have enough force to push. I know some of the simpler math like $F=ma$, but that didn't help me understand.  I browsed some articles about space elevators and couldn't find my answer.  Then I thought about a regular elevator and the fact that it has enough force to move it away from the earth successfully. Even if the elevator moved at $1$ nanometer/sec, it would still move away from the Earth, overcoming the minute pull of gravity at that altitude.  If you missed it, my question was:  can we simply apply enough force to overcome Earth's gravity without reaching that speed?  Edit: In another example, let's say we could drive a car from here to low Earth orbit. As long as we had enough torque, we should be able to overcome gravity, much like a pickup truck pulling a bunch of lumber. Is that correct?
111374
There are 2 balls in a vacuum, next to each other but not touch. They are on the edge of a surface they will both leave the table at exactly the same time. One gets pushed harder than the other. The earth is completely flat. Apparently both balls will hit the floor at exactly the same time and even further be at the same vertical height from the floor at any point in time. How is this possible?  If you push the one ball much much much harder than the first so that it is going at a considerable speed 100m/s compared to 1m/s say. The balls can't possible hit the floor at the same time, right?  Now the earth is round. You hit the one ball so hard it goes into orbit the other ball just falls of the table straight down like before. As the one ball has gone into orbit it will never reach the ground. However the slow ball will reach the ground much faster. How can a rule be different? If it doesn't work at the extreme then it shouldnt work at all.
114407
A thought experiment has been bothering me. Assume an enclosed box free-falling in a constant gravity field. The box contains a person and a bowling ball at the top end. The person throws the bowling ball down (in the direction the system is falling/the direction gravity is pulling). Obviously, being a closed system, forces are all internal, so there's no net effect in that sense. But, the center of mass has shifted downward some amount more than what is due to gravity.  The question, then, is this: What does someone outside the box observe? Part of me says they don't see anything, the box appears to fall under the influence of gravity with no visible change. The other part says there's a period corresponding to the flight of the bowling ball where the box falls faster than can be accounted for by gravity.   A follow-up question that I believe is closely related is the following: Imagine jumping off a high-dive board in a cannonball position. Shortly before hitting the water, you extend your legs downward. Do you hit the water faster, slower, or with the same velocity as someone who kept their legs extended throughout the jump?
68330
I've always been under the impression that jumping in an elevator wouldn't help at all, especially after reading a snippet of physics where Einstein said that free fall was identical to zero gravity.   I just watched an episode of QI (a British game show) where they said that jumping would only take off about five pounds of force... How would jumping help at all? If the elevator is in free fall, then you're not really jumping--you're just pushing, right?
3173
Clearly, the Earth is accelerating (as it has centripetal acceleration), and thus it is a non-inertial frame of reference. Then, why don't we use pseudo forces all the time? My teachers told me that the effect produced by the centripetal acceleration of the Earth on our calculations is negligible and hence we can safely neglect pseudo force. But how? I want rigorous proof with examples that show exactly how much we are compensating in exchange for neglecting that the Earth is a non-inertial frame of reference.
10731
I have asked a question on I  pushing the wall,where the force of me is not zero but as per the formula F=m.a,the force is zero and then I got the answer , it is because here the net force is acting on the body.But I a doubt that then a body of mass 5kg and moving with an acceleration 2m/s^2 will experience a force = 10N,Is in this case also the net force acting? Or  the net force is acting only when the body doesn't move? If this is correct then my question why?
4134
Let us say a block of mass is placed on the surface of earth. Then while drawing the forces on that body, we say:   Force $F = mg$ acting towards the center of Earth. Normal reaction $N$ offered by the surface of Earth.   If no other forces are acting, we say $F=N$. But what about the centrifugal force $m\omega^2R$ . Why don't we ever bring that into picture? What am I missing?