What is the fractional decrease in the value of free-fall acceleration g for a particle when it is lifted from the surface to an elevation `h`? `(h lt lt R)`
If R is the radius of the earth, then the value of acceleration due to gravity at a height h from the surface of the earth will become half its value on the surface of the earth if
If g denotes the value of acceleration due to gravity at a point distance r from the centre of earth of radius R . If r lt R , then
At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on the surface ( R = radius of earth)
At what height over the Earth's pole the free-fall acceleration decreases by one per cent, by half?
If R= radius of the earth and g= acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r lt R) from the centre of the earth is proportional to
Assertion : If rotation of earth about its own axis is suddenly stops then acceleration due to gravity will increase at all places on the earth. Reason : At height h from the surface of earth acceleration due to gravity is g_(h)=g(1- (2h)/R_(e)) .
The acceleration due to gravity is g at a point distant r from the centre of earth of radius R . If r lt R , then
Given, acceleration due to gravity on surface of earth is g. if g' is acceleration due to gravity at a height h above the surface of earth, then
What will be the acceleration due to gravity at height h lt lt R . Where R is radius of earth and g is acceleration to gravity on the surface earth
A solid sphere of mass M and radius R is surrounded by a spherical shell of same mass and radius 2R as shown. A small particle of mass m is relased from rest from a height h (lt lt R) above the shell. There is a hole in the shell. Q. What time will it take to move from A to B?
