The value of g (acceleration due to gravity) at earth's surface is 10ms−2. It's value at the centre of the earth which is assumed to be a sphere of radius R metre and uniform mass density is,
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Net force acting on an object placed at the center of the earth is zero since all the forces act on the object spherical symmetrically and hence cancel out.
Since there is no force at the center of earth, there would be no acceleration.
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If g is the acceleration due to gravity on the surface of the earth, it's value at a height equal to triple the radius of earth is (assuming the earth to be a perfect sphere)
The depth ′d′ at which the value of acceleration due to gravity becomes 1n times the value at the earth's surface is (R= radius of earth)
(a) Assuming the earth to be a sphere of uniform density, calculate the value of acceleration due to gravity at a point (i) 1600km above the earth, (ii) 1600km below the earth, (b) Also find the rate of variation of acceleration due to gravity above and below the earth's surface. Radius of earth =6400km, g=9.8m/s2.
The value of acceleration due to gravity, at earth surface is g. Its value at the centre of the earth, which we assume as a.sphere of radius R and of uniform mass density, will be:
The height from earth's surface at which acceleration due to gravity becomes g4 is (where g is acceleration due to gravity on the surface of earth and R is radius of earth).