Question 1:
Drill a hole all the way down to a depth 0.5 metre above the centre of earth. For that we need a long casing to stop sloughing and ground water into the hole. This is an ideal closed end casing which is strong enough to withstand high pressure and temperature inside the earth. Now put a sphere of 1 kg mass of 1 metre diameter at/ on the base of hole. The distance between the centre of the earth and the centre of the sphere is 1 metre.
Now apply the same universal law of gravitation to the two masses. Can I calculate the value of “g” (as Newton drive for his equation on the surface of the earth ) at this point by the following equation.
F = G M m/R^2 g = GM/R^2 , Where R = 1 metre, therefore R = GM, g= 3.98 x 10^14 m/sec^2. So Weight of this mass at o.5 metre above centre of earth =mg = 3.98 x 10^14 N. If not, please explain and why g = 10.7 m/sec^2 in inner core?
Question 2:
Now push the casing further down and past the centre of earth by half a metre. Now Again put a sphere of 1 kg mass of 1 metre diameter at/ on the base of shaft such that the centre of the earth and the centre of the sphere coincides. (Just like white and yolk of boil egg)
Apply the same universal law of gravitation to the two masses. What will be the value of “g” and the gravitational force?
F = G M m/R^2 g = GM/R^2 , Where R = zero metre, therefore g= 0 m/sec^2 or infinity & F=zero or infinity
If not, please explain why?
What is the value “g” at the centre of gravity of the earth?
Question 3:
Now push the casing further down and past the centre of gravity of earth by one km. Now remove the sphere and drop a stone from ground surface or the top of casing. So where will be the resting position of stone, centre of gravity of the earth or at the base of hole?
2007-11-01
13:15:16
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Astronomy & Space