We are sending huge amount of debris into space from earth, and also very heavy satellites and rockets, then the mass of earth must be decreasing over time. If the mass will decrease, then gravitational attraction between earth and sun must decrease, and its angular momentum must decrease, which would result in greater orbit of earth and earth being cooler over the time, but instead the temperature of earth is increasing, why is it so?
48246
We walk or ride on our vehicles to our destinations daily. Does our movement have any effect on the rotation of the earth according to Newton's law? What will be the effect if we move all the peoples along with their vehicles at their maximum velocity in one line in one direction along the equator? How much effect will it make?
121442
The Earth turns with a very high velocity, around its own axis and around the Sun.  So why can't we feel that it's turning, but we can still feel earthquake.
23004
I'm sorry if the answer is obvious for you guys, but why don't we all (including buildings, road, people, the ground) collapse to the center of the earth because of gravity? Is it because we have velocity, just like the earth not falling to the sun (or electrons not falling to the protons)? But that analogy doesn't sound right because those involve wide empty space.
71739
Assume the Earth to be a uniform sphere of mass M and radius R. Also let the Earth be stationary initially. Assume further that there is all other stellar bodies are very far away so as to have no influence in this problem.  A ball of mass $m$ is to be thrown from the surface of the Earth so that it ultimately ends up moving in a circular orbit at a height $h$ from the surface of the Earth. How should the ball be thrown, as in, what should be its initial velocity and, in particular, what should be the angle $\theta$ of the initial velocity with the radial vector from the Earth's center to its initial position on the surface of the Earth?  My attempt at the solution:  Since the Earth-ball system is isolated, its energy would be conserved. Let the initial speed of the ball be $v_0$ and the final speed be $v_f$. We'll have the following equation:  $$ \frac{1}{2}m v_0 ^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 - \frac{GMm}{R+h}$$   Next, we know that when the ball will finally be in circular orbit, the gravitational force will be providing the necessary centripetal acceleration for it to maintain circular motion, which gives us the following equation:  $$ \frac{mv_f^2}{R+h} = \frac{GMm}{(R+h)^2} $$  Besides, since gravity is a central force, the angular momentum of the ball about the center of the Earth will remain conserved, which yields the following equation:  $$ mv_0R \sin(\theta) = mv_f(R+h) $$  That completes setting up three equations for three variables $- v_0, v_f, \theta $. Solving them, we would obtain the following values for each:  $$ v_o = \sqrt\frac{GM(R+2h)}{R(R+h)} $$   $$ v_f = \sqrt\frac{GM}{R+h} $$  $$ \theta = \sin^{-1} \Biggl( \frac{1+\frac{h}{R}}{\sqrt{1+\frac{2h}{R}}} \Biggr) $$  However, it is observed that for $\theta$, the argument of $\sin^{-1}$ is always greater than or equal to one, being one only if $h$ is zero. The proof may be carried out by initially assuming that the argument of $\sin^{-1}$ is less than or equal to one for $h$ greater than or equal to $0$, and solving the inequality will yield that $h$ must be less than or equal to zero. So the only acceptable solution is for $h$ to be zero. This just winds up being the case of throwing the ball with sufficient speed tangentially on the surface of Earth so that it keeps orbiting on the Earth's surface itself. Thus, there is no possible way to just throw a ball so that it ends up in a circular orbit of certain desired radius.  The problem is that I don't see any physical intuition for why this should be true; either that or I am doing something wrong. Any guidance or help would be greatly appreciated!
5385
The force of gravity is constantly being applied to an orbiting object.  And therefore the object is constantly accelerating.  Why doesn't gravity eventually "win" over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas?  I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it?
15190
In an older question from June 2011, , apparently revived by the system, a user is asking about the origin of the rotation of celestial bodies.  The  (with 17 votes) states the following:     When things collapse under their own gravity in space (i.e. clouds of gas and dust), any small amount of asymmetry in the collapse will be enough start it spinning.   And then it goes on saying:     Even if it spins by a tiny amount, as it collapses, angular momentum     conservation will mean it spins more and more quickly.   Is all this really correct?  I could try to ask in a comment to the answer. I actually did. But the author is no longer registered.  Hence I am asking it as a question.  But that is not the end of the story. As I was about to ask my question, I looked for other relevant questions, and there are some. I did not make an exaustive survey, but I did not find one that had satisfactory answers:    (december 2011) has just one accepted answer (0 votes) stating essentially that the solar system has angular momentum because it started from a cloud that had angular momentum. No comment.  (july 2011) is questionning about everything from galaxies to electrons. The answers are of a much higher quality, but while they seem to justify the possiblility of spin and angular momentum (and its conservation), there are not clear on how it appears, at least for celestial bodies.  (june 2013): That was my own very naive first question on this site. It is related but does not directly address spin.   One answer I found is that momentum appear from bodies in relative motion past each other. Nice, but not really an answer: where did they get their relative speed to begin with. Unfortunately the answer is: from the angular momentum of larger structures they belong to (well, it can be more comples). So this is just chicken and egg explanation. But how did it get started?  Note that my question is not a duplicate, since it is primarily whether the above statements are correct, which is not considered in these other questions.  Furthermore, these questions are fairly old, and given that none had received answers that seem to answer the question of the origine of existence of celestial angular momentum (not in a way I can understand anyway), it may be appropriate to attempt restarting the process, with the above discussion to motivate that an answer should explain the physical mechanism that started the existence of spin of celestial bodies.
9361
Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.  Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)  This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?    Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right.
108164
Suppose that a satellite circles around a planet that exerts $2000N$ of gravitational force on the satellite.    I understand the fact that since the circular motion and the centripetal force are always perpendicular to each other, the work done by gravity is 0.  However, the satellite is being moved from one point to another, so Intuitively I am thinking "there must be some work done to move the satellite from point A to point B".  But besides the gravitational force, there is no force acting on the satellite.  Can some one explain me the gap with my logic and intuition?
20029
This diagram from  shows the gravitational potential energy of the sun-earth two body system, and demonstrates clearly the semi-stability of the $L_1$, $L_2$, and $L_3$ lagrangian points. The blue arrows indicate lower potential energy, red higher - so any movement in the plane perpendicular to the masses would require energy and without it, an object there would settle.    However, the $L_4$ and $L_5$ are claimed to be stable, even though the direction arrows indicate it's at the top of a gravity well, and would fall into lower potential energy in any direction. What is it that makes these points stable if not gravity? What am I missing?
10851
The question I am working on is, &quot;Consider the following. (a) Find the angular speed of Earth's rotation about its axis. rad/s (b) How does this rotation affect the shape of Earth?&quot;  I am fully capable of solving part (a); however, I am not sure how to describe the effect earth's rotation on its shape. I tried to search my textbook for the answer, but could not find anything. Is there an actual effect on the shape?
26404
What's the weight of the earth?  That is, what's the weight of the earth in the gravitational field of the sun? I imagine it's either   $0$, because there's no net fall toward the sun, so the downward component of acceleration is $0$ and $F=ma=0$, or $Gm_1m_2/r^2\approx7\cdot10^{-11}\cdot6\cdot10^{24}\cdot2\cdot10^{30}/(1.5\cdot10^{11})^2\approx4\cdot10^{22}$ Newtons,   but which, and why?
21824
Pardon the misleading title.  It is to my understanding that moving/heavy clocks run slow. The Earth itself is under gravitational influence from many sources, and is moving. Is there a way to know how much 'faster' a clock would run if those influences were removed? I don't need a precise number, an order of magnitude is fine, just to get the idea.  Any insight is appreciated.
85223
I can't get something strait... Lets use the usual example: A man is flying away from earth on a very fast spaceship. So, His time is now not moving in the same "speed" like earth time but slower. But why do we say that the man is moving faster than earth? why is it not that earth is moving faster than the man? I mean, it's not like there is a direction that if you move that way then you are faster and the other way means you moves slower.  So, how can we assume that's time is moving slower for the man and not on earth?  Maybe the man is flying the "other" way and now he is slower than earth...  If the man is not accelerating than it could appear to him that earth is moving fast and he is stationary... isn't it?  So, I can sum it as: In relation to what do we say that something is faster than some other thing?  Thanks!
111636
My understanding of Quantum physics and String Theory is very basic and I don't yet have a grasp on the maths, but in my research I have come up with a question.  Does a decrease in temperature also signify/create a decrease in mass?  I am quite willing to believe the concepts I think I am beginning to understand are totally misconceptions on my part.  But, as I see it mass is a product/mark of disruption to the Higgs field as a particle moves.  If these particles are ultimately composed of vibrating Strings, their mass is given by the rate at which they vibrate and how it interacts with the Higgs field.  I don't have any idea if, as the temperature (as we know it) approaches 0k, the string/quark/sparticle would change its state in any way or if the changing of the state of the String would change what type of particle it represents.  It seems I'm more full of questions than answers, but that's why I love these topics!  I know what I'm learning is new if the best and brightest are still working on the 'basics' to just make their systems work :).
80258
The earth is not a sphere, because it bulges at the equator.   I tried fiddling with centripetal force equations and gravity, but I couldn't derive why this bulge occurs.    Is there   (a) a mathematical explanation using forces (not energies) and   (b) a simple intuitive explanation to explain to others why the bulge occurs?
100151
Hypothetically, if you were to travel say, 6 months into the past with no spatial displacement, you would end up floating in space because the Earth would be on the opposite side of the sun.  I know when it comes to sci-fi, they can't account for everything and I don't expect them to. So I'm not nit-picking or anything, I'm just wondering if this is something that has been addressed to any detail, as it seems like it would have to be a major component of any time travel.
110675
Suppose the Earth is not rotating. As usual, the Moon follows its normal path around the Earth. Let's assume it's a circular motion and that there are no other gravitational influences.  A test particle on the surface of the Earth directed to the Moon on the line that connects the middle points of both will feel the Earth's gravity directed to its center and the Moon's gravity directed to the Moon.   A test particle on the opposite side of the Earth will also feel the Earth's gravity directed to its center (equal but opposite to the gravity experienced by the above-mentioned test particle) and the (smaller because of the greater distance to the Moon) gravity caused by the Moon.  So the surface water on Earth is attracted on both sides where the test particles reside with equal strength to the Earth. But the gravitational influence of the Moon on the water is bigger on the side directed to the Moon than it is on the opposite side. The difference is the greatest for the two test particles.  You would expect that because the water is pulled to the Moon by a little bigger gravitational force on the side of the Earth facing the Moon than on the opposite side (not facing the Moon), a bulge of water, directed to the Moon, will emerge that rotates around the Earth in sync with the Moon's motion.   But we must of course not forget that in this case there are also centrifugal forces that pull on the water due to the rotation of the Earth and the Moon around their CM (this is of course not the rotation of the Earth around its axis which I put zero). The CM between the mass and the Moon lies at 4600(km) from the center of the Earth in the direction of the Moon. The centrifugal force is bigger on the part of the Earth the furthest from the CM, where the gravitational effect of the Moon is the smallest (when the Earth is rotating also the centrifugal forces due to the rotation of the Earth itself come into play, which makes the situation more complicated).   So the question reduces to: What is the ratio between the centrifugal force plus the gravitational force caused by the Moon on the furthest part on Earth to the Moon and the centrifugal force plus the gravitational force caused by the Moon on the closest part of the Earth.   On the far side (from the Moon) of the Earth the Moon's gravitation is smaller but the centrifugal force bigger, while on the close side the Moon's gravitation is bigger and the centrifugal force smaller. Anybody who can do this quite straightforward calculation (for finding the ratio)? One thing is sure: two opposite bulges will develop which each go around the Earth in the same time as the Moon makes one complete cycle around the Earth. The ratio gives us information about the height of the bulbs.   EDIT I made an obvious error (to make my error clear I let the question as it is). The Earth isn't rotating (as can be read in my question), so no centrifugal forces are present. It's true that, in this case, the CM (4200(km) from the center of the Earth) rotates around the center of the Earth in sync with the rotation of the Moon, but the Earth doesn't rotate around the CM. So there are only tidal forces at work which causes the two bulges rotating around the Earth in one Moon cycle.