# please show the video of how the the value of g (acceleration due to gravity) depends on the latitude and the longitude of the earth? also show the dependence of the value of g upon the spin of the earth ???

### Asked by n jain | 16th Mar, 2011, 12:00: AM

### Dear student,
the value of *g* at any point on the earth's surface is
*g* = (6.67 x 10^{-11}) x *M*/*r*^{2}
where *M* is the mass of the Earth, and *r* is the distance from the centre of the Earth. The value 6.67 x 10^{-11} is known as the *Universal Gravitational Constant* - as its name suggests it is the same everywhere, and was discovered by Newton.value of *g* depends only on the mass of the Earth and the distance from the centre of the Earth. This means that at a given point on the Earth's surface all objects fall towards the centre of the Earth with the same acceleration, and that they therefore fall at the same rate. So if a heavy object and a light object are dropped at the same moment, they will land at the same moment, unless air currents affect things.
The Earth rotates about its axis from west to east. If the Earth is a sphere at rest, the value of'g' will be same at all places on the Earth and acts towards the centre of the Earth. But, due to the rotation of the Earth, the net acceleration acting on a body at various places will be different and as a part of the gravitational force acting on a body is utilised to will be different and as a part of the gravitational force acting on the body is utilised to compensate the centripetal, force developed in it. So, the apparent acceleration due to gravity is different at different latitudes. Let the acceleration due to gravity be g0 at a place where the latitude is 0. If the radius of the Earth is 'R' and its angular velocity is w. We can prove that the value of g is maximi m at poles and minimum at the equator.

Longitude is *the angular distance between a point on any meridian and the prime meridian at Greenwich.*

The expression for acceleration due to gravity is

g_{e} = Re2GMe

Acceleration due to gravity is inversely proportional to the square of the distance between the centre of the Earth and the object. The acceleration due to gravity produced in an object on the surface of the Earth is dependent on the radius of the Earth. Earth is not a perfect sphere (slightly bulging out at the equator) its radius decreases as we move from the equator to the poles. At the equator and at sea level its value is about 9.78 m/s^{2} and at the poles it is 9.83 m/s^{2}. Its mean value is taken as 9.8 m/s^{2} for all calculations.

Similarly the value of g decreases as we go up to the top of a mountain or higher up in the air.

Distance of the Object from the Centre of the Earth

The value of g inside the Earth is directly proportional to the distance from the centre of the Earth. Hence, g decreases as we go down into the Earth till it becomes zero at the centre of the Earth.

Hope this helps.
Thankyou
team
Topperelarning.com

*g*at any point on the earth's surface is

*g*= (6.67 x 10

^{-11}) x

*M*/

*r*

^{2}

*M*is the mass of the Earth, and

*r*is the distance from the centre of the Earth. The value 6.67 x 10

^{-11}is known as the

*Universal Gravitational Constant*- as its name suggests it is the same everywhere, and was discovered by Newton.value of

*g*depends only on the mass of the Earth and the distance from the centre of the Earth. This means that at a given point on the Earth's surface all objects fall towards the centre of the Earth with the same acceleration, and that they therefore fall at the same rate. So if a heavy object and a light object are dropped at the same moment, they will land at the same moment, unless air currents affect things.

The Earth rotates about its axis from west to east. If the Earth is a sphere at rest, the value of'g' will be same at all places on the Earth and acts towards the centre of the Earth. But, due to the rotation of the Earth, the net acceleration acting on a body at various places will be different and as a part of the gravitational force acting on a body is utilised to will be different and as a part of the gravitational force acting on the body is utilised to compensate the centripetal, force developed in it. So, the apparent acceleration due to gravity is different at different latitudes. Let the acceleration due to gravity be g0 at a place where the latitude is 0. If the radius of the Earth is 'R' and its angular velocity is w. We can prove that the value of g is maximi m at poles and minimum at the equator.

Longitude is *the angular distance between a point on any meridian and the prime meridian at Greenwich.*

The expression for acceleration due to gravity is

g_{e} = Re2GMe

Acceleration due to gravity is inversely proportional to the square of the distance between the centre of the Earth and the object. The acceleration due to gravity produced in an object on the surface of the Earth is dependent on the radius of the Earth. Earth is not a perfect sphere (slightly bulging out at the equator) its radius decreases as we move from the equator to the poles. At the equator and at sea level its value is about 9.78 m/s^{2} and at the poles it is 9.83 m/s^{2}. Its mean value is taken as 9.8 m/s^{2} for all calculations.

Similarly the value of g decreases as we go up to the top of a mountain or higher up in the air.

Distance of the Object from the Centre of the Earth

The value of g inside the Earth is directly proportional to the distance from the centre of the Earth. Hence, g decreases as we go down into the Earth till it becomes zero at the centre of the Earth.

### Answered by | 16th Mar, 2011, 02:32: PM

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