Home
Class 11
PHYSICS
Find the depth at which the value of g b...

Find the depth at which the value of g becomes 25% of that at the surface of the earth. (Radius of the earth = 6400 km)

Text Solution

Verified by Experts

g at a depth `g_(d) = g(1-(d)/(R))`
In this problem, `g_(d) = (25)/(100)g = 0.25 g`
substituting `g_(d) = 0.25g` and R = 6400 km
we get d = 4800 km
Promotional Banner

Similar Questions

Explore conceptually related problems

Find out the capacitance of the earth ? (Radius of the earth = 6400 km)

Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the earth. (Radius of the earth = 6400 km)

A what height above the earth's surface the value of g becomes 25% of its value on the earth if radius of the earth is 6400km.

If the value of g at the surface of the earth is 9.8 m//sec^(2) , then the value of g at a place 480 km above the surface of the earth will be (Radius of the earth is 6400 km)

At what depth from the surface of earth the time period of a simple pendulum is 0.5% more than that on the surface of the Earth? (Radius of earth is 6400 km )

How high a man be able to jump on the surface of a planet of radius 320 km, but having density same as that of the earth if he jumps 5 m on the surface of the earth? (Radius of earth = 6400 km)

Calculate the height at which a man's weight becomes ( 4//9) of his weight on the surface of the earth, if the radius of the earth is 6400km.

How much below the surface of the earth does the acceleration due to gravity (i) reduced to 36% (ii) reduces by 36% , of its value on the surface of the earth ? Radius of the earth = 6400 km .

How much below the surface of the earth does the acceleration due to gravity (i) reduced to 36% (ii) reduces by 36% , of its value on the surface of the earth ? Radius of the earth = 6400 km .

The height of the point vertically above the earth's surface, at which acceleration due to gravtiy becomes 1% of its value at the surface is (Radius of the earth =R)