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Gravitational force is really weak compared to the other fundamental forces, so it's very difficult to measure the gravitational constant. This is how Cavendish did it without knowing the Earth's mass: He put two lead balls on either end of a long bar. He hung the bar at its center from a long twisted wire with known torque. Then, he placed two really ...

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Brionius dealt with the value of G. Cavendish's experiments also confirmed the product of masses term. The inverse-square portion of Newton's Theory of Universal Gravitation was immediately accepted, since it straightforwardly produces Keppler's Laws. Further thought shows that, in order to produce stable orbits, the exponent must be exactly 2 - no more, no ...

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This is explained in the Wikipedia article on the Cavendish Experiment. Cavendish measured the gravitational attraction between two known masses using a torsion balance apparatus, which allowed him to calculate $G$ (to a surprising degree of accuracy given that it was done in the 1790s). That measurement did not involve the mass of the Earth. Once $G$ was ...

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You dont know the mass of earth, but you know the force earth apply to you. This is F=mg, where g is 9.8 m/seg^2 Buut this is equal to the gravitation force, F=GmM/r^2. G is the gravitation constant, M the mass of earth, m your mass and r the radio of earth. Cavendish was who messure the earth mass. In that time, Cavendish knew the radio of earth and G (he ...

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Short version: The Earth is round because it was molten once. It would definitely matter if our planet were a cube, because gravity near the points would be off from "vertical." Longer version: Smaller objects in our solar system are indeed more oddly shaped - they often look like a bunch of small blobs stuck together. They slowly build up in size as bits ...

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No, the Earth's gravity varies over its surface because (a) the Earth is not spherical, (b) it does not have a uniform (or spherically symmetric) distribution of density and (c) of course the "surface" is at different heights. You can find a local gravitational strength (and direction - because the local gravitational vector does not necessarily point to ...

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No, acceleration due to gravity varies, depending solely on the altitude on which Gravitational force is working, the approximate* value can be determined using equations: g(eff, h) = g*[(r/r h)^2] H being the altitude, r being earth's mean radius, g effective is the acceleration due to gravity at an altitude h above sea level. Thanks :) Approximate because ...

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If the Earth were a perfect sphere, it would be the same everywhere on Earth's surface. This is known as the shell theorem. It's not too hard to show mathematically, but you can think of it as the fact that all the mass that is close to you balances roughly with the mass that is far away. If you were to tunnel into the Earth, however, you would only ...

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First, two remarks: The melting of the ice in the Arctic ocean would have zero effect. What matters is the ice over Greenland and Antarctica. There is a very long-term secular slowing of the Earth's rotation rate due to the recession of the Moon from the Earth. This answer ignores that effect. (Alternatively, this answer has this secular effect subtracted ...

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The latter idea you talk about is the Giant Impact Hypothesis. It turns out that you can make a moon in a few easy steps (given the correct conditions): Have a bunch of protoplanets whiz about on semi-chaotic trajectories. Smack two of them together at a 45° angle. Let the bits of the protoplanets that don't merge together be ejected from the resulting ...

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To a first-order effect, there would be no change. But one consequence of melting is that the water moves to other places. Water that moves from the poles to other areas on the surface of the earth would serve to (slightly) increase the moment of inertia of the planet. This is because the mass of the water would be farther from the rotational axis. The ...

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Firstly, the gravitational field inside the Earth, decreases with depth. To a first approximation, you can use the shell theorem for spherically symmetric mass distributions to argue that the gravitational field at some depth is due only to the mass enclosed within a sphere interior to that depth. If we further make the crude assumption that the Earth's ...

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Q1. In the case of a uniform spherical distribution you cannot sense anything further away from the center than you are. This is directly derived from Gauss' law for gravity. At the center you do not feel any net gravitational force from earth at all. Think about it this way: earth is pulling you up from all directions exactly the same way, so all the ...

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