Obtain a formula for the variation of 'g' below the surface of earth. Hence show that 'g' vanishes at the centre of earth.

If V is the gravitational potential on the surface of the earth, then what is its value at the centre of the earth ?

A person weighs 98 N at the surface of earth. Its mass at the centre of earth will be

Define gravity and explain its variation, above and below the surface of earth.

The value of 'g' on earth surface depends :-

The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is

Draw g' versus d and g' versus h graph. Here, 'd' is depth below the surface of earth and h is the height from the surface of earth.

The Value of g on the surface of earth is smallest at the equator because

A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:

A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:

(a) Earth can be thought of as a sphere of radius 64 00 km . Any object (or a person ) is performing circular motion around the axis on earth due to earth's rotation (period 1 day ). What is acceleration of object on the surface of the earth (at equator ) towards its centre ? What is its latitude theta ? How does these accelerations compare with g=9.8 m//s^2 ? (b) Earth also moves in circular orbit around sun every year with an orbital radius of 1.5 xx 10 ^(11) m . What is the acceleration of earth ( or any object on the surface of the earth ) towards the centre of the sun ? How does this acceleration compare with g=9.8 ms^2 ?