News1 min ago
about gravitational force
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why at the centre of the earth ,earths gravitational force is zero?
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Newton tells us this about gravity: F=Gm1m2/r^2, where r=radius... Therefore, at the center of the earth r= 0 and so the gravitational force is 0....
Put another way "...the net gravitational force on an object inside a uniform spherical shell is always zero. At a given radius from Earth's center, the portion of Earth outside of that radius can be considered a shell; the net force of gravity on an object at a given radius from Earth's center is due only to the mass of Earth within that radius. As a result, as the object approaches the center of the planet and the radius decreases, the acceleration due to gravity would also decrease. Although the rate of change of velocity decreases, the object would continue to increase in velocity until it passes through the center of Earth." (Source: Astrophysicist Neil deGrasse Tyson at Nova Now).
Put another way "...the net gravitational force on an object inside a uniform spherical shell is always zero. At a given radius from Earth's center, the portion of Earth outside of that radius can be considered a shell; the net force of gravity on an object at a given radius from Earth's center is due only to the mass of Earth within that radius. As a result, as the object approaches the center of the planet and the radius decreases, the acceleration due to gravity would also decrease. Although the rate of change of velocity decreases, the object would continue to increase in velocity until it passes through the center of Earth." (Source: Astrophysicist Neil deGrasse Tyson at Nova Now).
Clanad surely knows that Newton's formula gives the force between two point masses. It is not sufficient to say that r=0, therefore the force is zero. All that means is that the force between two point masses that are touching is zero. His next paragraph is correct, though harder to understand. The last sentence concerns the theoretical passage of an object dropped through a theortetical hole drilled through the centre of the earth. Try this 'thought experiment': Imagine you are in a cavity at the centre of the Earth. Half of the Earth is on your left hand side, pulling you to the left, half is on you right hand side pulling you to the right. The forces are equal and opposite, so the net force is zero. This is not the same a being in the middle of a tug-of-war with a rope attached to each wrist! The same argument applies to the Earth above your head and below your feet - the net force is zero.
In trying to grasp Bert's point, one has to understand (I think) that Newton "assumed" the mass of the sphere to be concentrated at it's center, so to speak. Therefore, "...for a spherically symmetric mass, the net gravity force on an object from that mass would be only that due to the mass inside its radius, and that would act as if it were a point mass located at the center. When this is analyzed in detail, you find that the gravity at any radius r less than R>Earth will be linearly proportional to the distance from the center..." (Source: My dog eared copy of Hesiod's Anvil). This agrees with Sir Iassac that gravitational force at r0=0... or so it seems to me on solving the equation.
In Bert's second example, the gravitational forces acting on a body at the hypothetically attainable center of the Earth would be equal in all directions, not just from your right or left... and would result in G=0 because and only because r=0...
Interstingly, the period (time) needed for the body to fall through the center of the Earth and emerge at the other side would be the same period as a body orbiting the Earth near the surface (disregarding air resistances)... about 42 minutes, as I recall...
In Bert's second example, the gravitational forces acting on a body at the hypothetically attainable center of the Earth would be equal in all directions, not just from your right or left... and would result in G=0 because and only because r=0...
Interstingly, the period (time) needed for the body to fall through the center of the Earth and emerge at the other side would be the same period as a body orbiting the Earth near the surface (disregarding air resistances)... about 42 minutes, as I recall...
I was just trying to put the answer in terms that -llRocKKinGll- might understand [not wishing to be patronising, but having no idea of -llRocKKinGll-'s knowledge of physics or maths] with no maths whatsoever. I gave the argument for left and right, and for up and down, and would hope that anyone capable of asking the question would be able to extrapolate it to all possible directions. Newton's Law of Universal Gravitation between point masses can be shown, mathematically, to prove that there is no net force on a point mass within a closed spherical shell, and to prove that the force between two massive objects of any size is equivalent to the force between two point masses of the same magnitude and separated by the distance between the centres of mass of the two objects - but it does need maths. The force of gravity as you go down will indeed be proportional to the radius [assuming the Earth has a uniform density, which it doesn't]. this is because the mass "above" you has no gravitational effect. The mass "below" you is proportional to the cube of the radius, while the force exerted by it is inversely proportional to the square of the radius. [radius squared] cancels out, leaving a force proportional to the radius. Therefore, after all the maths, the force = 0 when r = 0. I was also pointing out that the last sentence of Clanad's first answer didn't really have much to do with the rest of his answer, nor the question.
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