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Joe Smith asks on Gravity, Moon, Ocean Tides and Atmospheric tides

My primary question: If it’s true that oceanic tides can be caused when the moon’s gravity pulls the molecules of ocean water up and away from earth by a certain distance, and if it’s also true that earth’s atmospheric tides can, likewise, also be caused by the moon’s gravity pulling the molecules in the atmosphere up and away from earth by a certain distance, then what stops the atmospheric molecules, once they have accelerated even just a tiny distance in the direction away from earth and towards the moon, from continuing on their journey to the moon?  (Let’s, for example, say that the atmospheric molecules in question are—to simplify matters—the ones at the very top of earth’s atmosphere, so that no other atmospheric molecules are between these particular atmospheric molecules and the moon, as the moon pulls on them.)
The details: In other words, my question is, if the moon’s gravity is strong enough to cause a particular atmospheric molecule to start accelerating towards the moon, then, since this molecule, in the very beginning of this acceleration, will now be slightly further away from earth’s gravity and slightly closer to the moon’s gravity than it was when it was in its original position, what would cause this molecule’s acceleration towards the moon to stop?  As we know, the very instant the molecule’s distance from earth increases by even the tiniest amount, earth’s gravitational pull on it instantly begins to decrease exponentially, while at the same time, i.e., simultaneously, the now shorter distance between this molecule and the moon, would cause the moon’s gravitational pull on it to become even greater (exponentially greater) than it originally was when it first started to pull on the molecule, and these two facts (the now weaker and ever weakening earth gravity, combined with the now stronger and ever strengthening moon gravity) it would seem, would cause the molecule to keep heading in the direction of the moon.  And this “snowball effect” would be exponential, according to the inverse square law of gravity.  What device or mechanism, in the theory of gravity, would stop the moon from stealing this atmospheric molecule from earth?
(A side question I have here, which I hope doesn’t detract attention from my real question, is the following: if the atmospheric molecule accelerating towards the moon is affected by the moon’s gravity according to the inverse square of the distance, and this same molecule, which is also accelerating away from the earth, is affected by the earth’s gravity according to the inverse square of the distance, doesn’t this mean that the above mentioned “snowball effect” on the atmospheric molecule would actually occur not according to the inverse square of the distance, but rather, according to the sum of the inverse square of the molecule’s distance from the earth, and the inverse square of its distance from the moon?  It seems like “summing” is the correct operation to perform on the forces of both the earth’s and the moon’s gravity upon the atmospheric molecule here, rather than multiplying these two forces.  Is this correct?)1393043699
The above, is my primary question.  It is perplexing, because it seems to me that if the moon could cause a single atmospheric molecule to accelerate towards it for even the briefest instant, then the theory of gravity (as far as I am familiar with it) has no way of explaining why this single molecule wouldn’t be stolen by the moon.  (In other words, the “inverse square law” of gravity should lead to a runaway molecule, shouldn’t it?) And if the moon could steal one single atmospheric molecule, then I can see no reason why it wouldn’t eventually steal the whole of earth’s atmosphere.  I understand that the very general answer here is that, ultimately, earth’s gravity is stronger than the moon’s, so the moon can’t steal anything from earth; earth’s pull is stronger than the moon’s.  But, as I hope I’ve demonstrated above, when one doesn’t stop at the general picture of atmospheric tides, but instead delves into the details, problems seem to arise—at least from the perspective of my admittedly not all-encompassing knowledge of the theory of Newtonian gravity.
A secondary, related question is as follows: If, when the moon is overhead, the center of gravity of my body (and, presumably, every single atom in my body) is ever so slightly accelerated away from earth and towards the moon, what mechanism in the theory of gravity reverses this acceleration so that I don’t simply fly upwards to the moon?  After all, it seems to me that the same principle is at work here as in the atmospheric tides.
Another point: I realize that in my primary question, the single atmospheric molecule, and, in my secondary question, the center of gravity of my body (and all the atoms of my body), are not being spontaneously pulled perpendicularly away from the earth by the moon.  In order for this to occur, the moon would have to spontaneously just “materialize” directly overhead.  So, I realize, that instead what is happening in both my primary and secondary questions is that the atmospheric molecule (X) and the center of gravity of my body(Y) (and all the atoms of my body [Z]) are accelerating away from earth and closer to the moon, not perpendicularly to the earth (except when and if the moon is exactly overhead), but rather, X Y and Z are accelerating away from earth and towards the moon in a kind of arc over a period of time roughly equal to the time that the moon is “in the sky”.  Nevertheless, this “arc of acceleration”, rather than acceleration from a spontaneous perpendicular gravitational pull from the moon (as in the case where the moon were to suddenly and spontaneously appear directly overhead), doesn’t seem to me to change or negate the gist of my question.  Furthermore, regarding my entire question here about the workings of gravity with respect to atmospheric tides, even if one were to argue that my “arc of acceleration” scenario (scenario A) was significantly different than my “moon spontaneously appearing overhead” scenario (scenario B) so that scenario A were somehow easily explainable due to some factor I’ve failed to account for, then my response to such an explanation for scenario A would simply be: ok, then what about scenario B?  The reason this would be my response is because, in scenario B, where the moon suddenly materializes directly overhead, my current understanding of Newtonian gravity can offer no reason why, when the moon suddenly appears, that molecules at the top of earth’s atmosphere wouldn’t begin to accelerate towards it; and once this acceleration begins, it seems to me that the above mentioned “snowball effect” of gravity’s inverse square law should take over.
My appeal to those smarter than myself: If anyone sees some faulty assumptions I’ve made above, or some facts I’ve neglected to account for, which would explain not only how it’s possible that atmospheric tides due to the moon’s gravity do in fact exist, but also how the implied acceleration changes and distance changes (with respect to the earth and the moon) of the constituent molecules of these tides (scenario A) would not ultimately lead to the moon stealing the earth’s atmosphere, I would really appreciate it; likewise, even if my question about scenario A is easily explainable due to something I’ve missed, I’d equally appreciate an explanation for scenario B, if scenario A’s explanation doesn’t already cover it.  I’m literally losing sleep over this question, so thanks, in advance.


Joe Smith

Please refer to






Motion under gravity

When two bodies of two different weights(ex..an elephant n a boy) falling from certain height.. Both reaches the ground at the same time because gravity is independent of mass…but when we leave a leaf n a coin from a certain height coin reaches the ground first why?

Asked Harshith

It is the air resistance which plays the VILLAIN’s role here. The leave is affected more by the air resistance than the coin. In other words the magnitude of force exerted by gravity on ball is much greater than buoyant force and air resistance in the case of the coin whereas in the case of a leaf, the air resistance and buoyant force is comparable to its weight.

If these are dropped in a vacuum chamber, both will fall together and will take the same interval of time to reach the ground when dropped from the same height.

Please refer to the following videos



Projectile Motion – Numerical Problems (Assignment for KV Pattom Class XI students)

The students are requested to start solving these numerical problems. More questions will be added till 8 pm on 13/09/2014. There will be more than 20 questions in all. All students are expected to solve at least 20 numericals among these. The deadline form submission is 15/09/2014.

  1. A football player kick a ball at an angle of 37° to the horizontal with an initial speed of 15 m/s. Assuming that the ball travels in a vertical plane, calculate (i) the time at which the ball reaches the highest point. (ii) the maximum height reached by the ball (iii) the horizontal range of the projectile and (iv) the time for which the ball is in air.
  2. A body is projected with a velocity of 20 m/s in a direction making an angle 60° with the horizontal. CCalculate (i) position after 0.5 seconds and  (ii) velocity after 0.5 seconds
  3. The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 m/s, what is the direction of the initial velocity? (g=9.8 m/s)
  4. A bullet fired from a gun with a velocity of 140 m/s strikes the ground at the same level as the gun at a distance of 1 km. Find the angle of inclination with the horizontal at which the bullet is fired. (g=9.8 m/s)
  5. A bullet is fired at an angle of 15° with the horizontal and hits the ground 6 km away. Is it possible to hit a target 10 km away by adjusting the angle of projection assuming the initial speed to be the same?
  6. A cricketer can throw a ball to a maximum horizontal distance of 160 m. Calculate the maximum vertical height to which he can throw the ball. (g=10  m/s)
  7. A football is kicked with 20 m/s at a projection angle of 45°. A receiver on the goal line 25 metres away in the direction of the kick runs the same instant to meet the ball. What must be his speed, if he is to catch the ball before it hits the ground?
  8. A bullet fired at an angle of 60° with the vertical hits the ground at a distance of 2 km. Calculate the distance at which the bullet will hit the ground when fired at an angle of 45°, assuming the speed to be the same.
  9. A person observes a bird on tree 39.6 m high and at a distance of 59.2m. With what velocity the person should shoot an arrow at an angle of 45° so that it may hit the bird?
  10. A ball is thrown from the top of a tower with an initial velocity of 10 m/s at an angle of 30° with the horizontal. If it hits the ground at a distance of 17.3 m from the base of the tower, calculate the height of the tower. (Given g =10  m/s)
  11. Prove that the time of flight T and the horizontal range R of a projectile are connected by the equation gT2 =2R tanθ, where θ is the angle of projection.
  12. Show that the range of a projectile for two angles α and β is same if α+β=90°
  13. A body is projected with velocity of 40 m/s. After 2 s it crosses a vertical pole of height 20.4 m. Calculate the angle of projection and horizontal range.
  14. A plane is flying horizontally at a height of 1000 m with a velocity of 100 m/s when a bomb is released from it. Find (i) the time taken by it to reach the ground. (ii) the velocity with which the bomb hits the target and (iii) the distance of the target.
  15. From the top of a building 19.6 m high a ball is projected horizontally. After how long does it strike the ground? If the line joining the point of projection to the point where it hits the ground makes an angle of 45° with the horizontal, what is the initial velocity of the ball.
  16. A body is thrown horizontally from the top of a tower and strikes the ground after 2 seconds at an angle of 45° with the horizontal. Find the height of the tower and the speed with which the body was thrown. Take g =9.8 m/s)
  17. A ball is projected horizontally from a tower with a velocity of 4 m/s. Find the velocity of the ball after 0.7 s. (Given g =10  m/s)
  18. In between two hills of heights 100 m and 92 m respectively , there is a valley of breadth 16m. If a vehicle jumps from the first hill to the second, what must be the minimum horizontal velocity so that it may not fall into the valley? (Given g =10  m/s)
  19. A mailbag is to be dropped into a post office from an aeroplane flying horizontally with a velocity of 270 km/h at a height of 176.4 m above the ground. How far must the aeroplane be from the post office at the time of dropping the bag so that the bag directly falls into the post office?
  20. An aeroplane is flying in a horizontal direction with a velocity of 600 km/h and at a height of 1960 m. When it is above a point A on ground an object is dropped from it. The object strike the ground at the point B. Find the distance AB.
  21. Two tall buildings are situated 200 m apart. With what speed must a ball be thrown horizontally from the window 540 m above the ground in one building so that it will enter a window 50 m above the ground in the other?


We all know there is no such thing as zero gravity even at the very end of universe there is micro gravity according to formula for gravity…!
Means we are away from certain object so the gravity exist in vacuum so now my question is:-
Q- We all know that there is zero Gravity in center of earth core so if we place an object in center of earth the would Gravity formula apply on that object or that object is in complete Zero gravity…????


Asked Pratik Bhoir






If we simply keep a body at the centre, it will remain there as it is attracted equally in all directions.


A man in an elevator free falls through a plan...
A man in an elevator free falls through a planet to simulate how a gravity train could work. At both ends of the planet the train comes to a complete stop, but free falls through the middle (or wherever the tunnel is constructed) of the planet. (Photo credit: Wikipedia)




Attraction, Gravitation and some doubts

“In universal law of gravitaion what is  the word “attraction” actually refers to?
I am already on the earth surface and attached to it.Then what is the meaning that earth still attacting me accounts for.
Does that mean earth want me to attarct me towards its center point (ie.., into the inner core of the earth’s interior).
Please answer I am puzzled and worried about my analysis.

Explain with reasons…”


Asked Gopi Krishna




Earth's Core


Yes, Earth continues to attract you until you reach its centre.


Every body attracts all other objects towards its centre, not to its surface!


Gravity and Galileo

How Galileo gave acceleration due to gravity before Newton discovered gravity . exactly what happen
Asked Nikhil

Colorized engraving after Enoch Seeman's 1726 ...

Newton stated the Universal Law of gravitation and explained why objects fall to earth when dropped.

But objects were falling even before Newton explained it.

The concept of speed and velocity was known before the Law of Gravitation was formulated.

Weightlessness in a satellite

Guion Bluford Experiences Weightlessness on th...
Guion Bluford Experiences Weightlessness on the KC-135 – GPN-2002-000148 (Photo credit: Wikipedia)


Why does an astronaut feel weightless in a space capsule orbiting the earth?

I’ve gotten a key point saying that the space capsule orbiting the earth has a centripetal acceleration, but I cannot figure out why is it related to the question.


Asked Ganondorf Jallida


Answer: The centripetal acceleration is provided by the force of gravity. Since the force of gravity is completely used up in providing the centripetal acceleration to the space capsule, the astronaut feel weightlessness.



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Half Yearly Exam Count Down

Half Yearly Exam Count Down KVS Ernakulam RegionOctober 17th, 2017
The Half Yearly Examination for Kendriya Vidyalayas of Kerala starts from 17 October 2017
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