Our weight is more at the pole than at the equator (because it is closer to the center). Does that mean if we go down into the earth (some kind of bore), will our weight increase further? I remember reading somewhere that this is not the case. Is that true?
22310
Would the effect of gravity on me change if I were to dig a very deep hole and stand in it? If so, how would it change? Am I more likely to be pulled downwards, or pulled towards the edges of the hole? If there would be no change, why not?
35568
If you could travel to the center of the Earth (or any planet), would you be weightless there?
70500
I know there are similar questions on StackExchange but I think it is different and detailed.  The earth is spinning 465 meters/second so why don't we jump and land on a different location?  I googled about this question and I got some answers:   When we jump, we (and the atmosphere itself) also spin along with the earth so we don't land at a different location. (But why we are also spinning along with earth ?) The earth is so big and we are very small relative to earth so the tiny jump won't make any difference.   But according to answer 1, We are also spinning with the earth. But the question is, "why are we also spinning along with earth" ??  Regarding the answer 2, it won't make any difference but logically it doesn't make sense or I just don't understand it.  Here is the followup question,   "When we jump why don't we thrown out of earth due to centrifugal force of the spinning earth ??"  Please provide some detailed answer to these questions. With logic if possible.   This phenomenon is not possible I know, but even if it exists, then traveling will be so easy, we just have to hang on space (with help of helicopter etc.. ) and we can travel the whole world in 24 hours :D
38413
I have demonstrated that  only is measured based on the gravitational pull of where you live.  For example, the gravity on the surface of Mars is three times smaller than the gravity on the surface of Earth.  So my question is simple: If we take into account that weight is only measured based on gravitational pull on that , if there was no gravity how would we define mass?  That is, would mass weigh anything if no gravitational pull acted on it? Then what would be mass?
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.
15882
As a physics newbie I'm trying to get a grip on basic orbital mechanics. I think I'm beginning to get grasp on how bodies interact with each other. When a body approaches another body it accelerates due to gravity. It can reach a point where its velocity is high enough to keep falling but also keep missing the object it is falling towards. What keeps it from accelerating (because of gravity) and eventually reaching escape velocity? I feel like I'm either looking at things the wrong way or I have the entire thing wrong.
68853
I am a student of class 9. When I was going through magnetism and read that an earth is a magnet I got some doubts. My question is: is earth really a magnet? Doesn anyone have any proof that earth is a magnet? Is there a magnetic core at the center of the earth? Has anyone reached the core of the earth?
11591
...And you jumped in.  What would happen when you got to the middle of the Earth? Would you gradually slow down, until you got to the middle and once you were in middle would every direction feel like it was up?
54254
As I was trying to work out the expression for the apparent "g"-in case of the rotation of Earth-I was considering the centripetal force (m$w^2$r) and the weight (mg) all the time, thinking these were the forces that were acting on the body.     But later I found out that it should have had used centrifugal force instead of centripetal force. But why? We should be considering the forces acting on the body right? And it's the centripetal force that is acting on the body (with direction towards the center of the small circle).
55417
According to ,  will require $1.711 \cdot 10^{32}~\text{J}$ to shatter the Earth into a gravitationally unbound primordial dust cloud.  However, this energy is over 30% lower than the gravitational binding energy derived from the average density model .  Which seems irrational:  if the Earth were in a lower energy state at a uniform density distribution, how did its density gradient evolve in the first place – wouldn’t it be evolving to a homogenous density distribution in that case?  Curtis Sexton and  seem to agree with my reasoning on this point:   the ’s beam weapon output would have to exceed $2.4 \cdot 10^{32}~\text{J}$ to impart escape speed to all of the matter comprising an Earth-like planet.  They can’t both be right, so who wins?  My money’s on the Death Star, but I haven’t found the mistake in Dave’s calculations for Marvin yet.  Does anyone here have the analytical chops to see where he went wrong, or alternatively, where I’ve spaced out?  Here's Dave's mathematical estimate for Marvin:  First he offers an equation for compiling the layers of various densities from the  into a sum of     where $r$ is the radial distance from the Earth’s center and     the various constants are noted below:  $i$. Layer; Height $h_i$ (m); $a_i$ ($kg \cdot m^{-5}$); $b_i$ ($kg \cdot m^{-4}$); $c_i$ ($kg \cdot m^{-3}$)  1  Inner core; 1.2215×10$^{6}$; -2.1773×10$^{-10}$     1.9110×10$^{-8}$ 1.3088×10$^{4}$  2  Outer core; 3.4800×10$^{6}$; -2.4123×10$^{-10}$;    1.3976×10$^{-4}$;  1.2346×10$^{4}$  3  D'' layer; 3.6300×10$^{6}$; 0.00; -5.0007×10$^{-4}$; 7.3067×10$^{3}$  4  Lower Mantle; 5.7010×10$^{6}$; -3.0922×10$^{-11}$; -2.4441×10$^{-4}$; 6.7823×10$^{3}$  5  Inner transition zone 1; 5.7710×10$^{6}$; 0.00; -2.3286×10$^{-4}$; 5.3197×10$^{3}$  6  Inner transition zone 2; 5.9710×10$^{6}$; 0.00; -1.2603×10$^{-3}$; 1.1249×10$^{4}$  7  Outer transition zone; 6.1510×10$^{6}$; 0.00; -5.9706×10$^{-4}$; 7.1083×10$^{3}$  8  Low velocity zone &amp; lid; 6.3466×10$^{6}$; 0.00; 1.0869×10$^{-4}$; 2.6910×10$^{3}$  9  Inner crust; 6.3560×10$^{6}$; 0.00; 0.00; 2.9000×10$^{3}$  10  Outer crust; 6.3680×10$^{6}$; 0.00; 0.00; 2.6000×10$^{3}$  11  Ocean; 6.3710×10$^{6}$; 0.00; 0.00; 1.0200×10$^{3}$  Using the real density distributions for each shell layer:    And calculating for each point $h$ at any given radial distance from the outside in, with the inner and upper bounds $h_1$ and $h_2$ respectively, gives:    So for each $i^{th}$ layer:   And for each point at a radial distance of $r$ we have:   where $ΔU_0$ = 0 and the indexed constants for each piece of the function are described in the table above.  This yields:  -1.711×10$^{32}~\text{J}$  Many thanks to anyone who can explain how this came out wrong!
121885
For a project into balloon simulation I'd like to know how the force on a balloon changes with altitude:  I know that the Buoyant force on a balloon is:  $F = (\rho_{air} - \rho_{helium})gV$  Using changing air density from the international standard atmosphere and changes in gravitational acceleration I can calculate the buoyant force and add this to the (negative) force of gravity mg.  However I know that it isn't this simple, the balloon changes volume with altitude as the air pressure outside the balloon changes.  I'd like to know how the balloon inflates - this would be a great help!  I think it may be something to do with ideal gas laws -  I can calculate air temperature and pressure from the international standard atmosphere but do I use pressure as the air pressure? - or air pressure minus balloon pressure?
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?
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.)
22284
I was watching Google map. Saw a line which is joinging north to south pole. It is going between middle of China and Mexico. But, the line is not straight. Tried to search in net but not able to get anything. Can anyone check and let me know  or
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?
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Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.
115376
If we had an airtight container that was large enough, say extended out to the Karman line, would there still be a pressure gradient? If we assume that the pressure at the bottom was $1\ \mathrm{atm}$, would it have the same exact pressure gradient at 6 miles up on the outside? This is assuming that the gravity is still $9.8\ \mathrm{m/s^2}$.