The acceleration due to gravity at the poles is `10ms^(-2)` and equitorial radius is `6400 km` for the earth. Then the angular velocity of rotaiton of the earth about its axis so that the weight of a body at the equator reduces to `75%` is

What should be the angular velocity of rotation of Earth about its own axis so that the weight of body at the equator reduces to 3/5 of its present value? (Take R as the radius of the Earth)

If omega is the angular velocity of rotation of the earth about its axis and R is radius of the earth, to decrease the weight of a body near the equator by 40% , then the new angular speed should be

The earth rotates about its own axis, then the value of acceleration due to gravity is

In order to make the effective acceleration due to gravity at the equator to zero, the numerical value of the angular velocity of rotation of the earth should be [Radius of the earth=6400 km, g=10m/s^(2) ]

The mass of the Earth is 6 xx 10^(24) kg and its radius is 6400 km. Find the acceleration due to gravity on the surface of the Earth.

What is the linear velocity of a person at equator of the earth due to its spinning motion? (Radius of the earth =6400km )

Assertion : The difference in the value of acceleration due to gravity at pole and equator is proportional to square of angular velocity of earth. Reason : the value of acceleration due to gravity is minimum at the equator and maximum at the pole.

The acceleration due to gravity at a depth d below the surface of the earth is 20% of its value at the surface. What is the value of d if the radius of the earth=6400km ?

Acceleration due to gravity at earth's surface is 10ms^(-2) . The value of acceleration due to gravity at the surface of a planet of mass (1/5)^(th) and radius 1/2 of the earth is