 # Value of g  View Notes

Value of g in fps

The acceleration felt by a free-falling object due to the gravitational force of the mass body is called gravitational acceleration and is expressed by g calculated using SI unit m / s2. The value of g depends on the mass of the body and its size, and its value varies from body to body. The value of g in the moon is constant.

Acceleration Due to the Gravity of the Moon

The acceleration due to the gravity of the moon or the magnitude of g on the moon is 1,625 m / s2

Calculate the acceleration due to the gravity of the moon

The acceleration due to the formula of gravity is indicated by

G = GM / R2

Where,

• G is the universal gravitational constant, G = 6.674 x 10-11 m3 kg-1 s-2

• M is the mass of the body measured using kg.

• R is the mass body radius measured by m.

• g is the acceleration due to the gravity determined by m / s2.

The mass of the moon is 7.35 × 1022Kg.

The radius of the moon is 1.74×106m

Substituting the values in the formula we get-

g= 6.67×10−11 × 7.35 × 1022 / (1.74×106)2

Thus, the value of g on the moon is g=1.625m/s2.

The Acceleration Due to Gravity also Follows the Unit of Acceleration

Newton's Law of Gravitation as applied to the Earth is F = G m M / r2, where F is the gravitational force acting on the body of mass m, G is the universal gravitational constant, M is the mass of the Moon, and r is the distance of the body from the center of the Sun. g is the factor in equation F = m g, so g is given as follows:

g = G M / r2

Both G and M are empiric constants, and g has an inverse-square relationship to r, the distance from the center of the earth's mass.

There are two consequences of this:

• Since the Earth is an ellipsoid, the distance from the center of a point on the surface decreases with the latitude, increasing g.

• The rotation of the planet creates an anti-gravity centrifugal effect, which is at the highest at the equator and zero at the poles.

These two effects are conspiring to generate a g rise in latitude. Their magnitudes are easily determined by simple geometry.

• Altitude

The effect of latitude is calculated on the basis of the standard surface of the geoid, which is the spheroid at sea level. Points above sea level are progressively further away from the center of the earth, so g decreases with altitude in a predictable manner.

• The anomaly of Gravity.

In practice, the value of g varies somewhat from the geometrically predicted value to latitude and altitude. Positive variation is caused by the following:

• the mass of the local above-sea-level topography

• above-average density of underlying rocks

The components of the structure of the Earth have a variety of densities. The geometric model of gravity conceives the Earth as a collection of onion-skin layers, each with a uniform density (and this is almost the case). Each individual sheet, because of its uniform density, has its center of mass corresponding to that of the Earth. Nonetheless, if the layer has a small patch of higher density material, the center of the mass is shifted towards the patch, decreasing r, and thereby increasing g.

## The Table Below Shows the Value of g at Various Locations from Earth's Center.

 Location Distance from Earth's Center(m) Value of g(m/s2) Earth's surface 6.38 x 106 m 9.8 1000 km above surface 7.38 x 106 m 7.33 2000 km above surface 8.38 x 106 m 5.68 3000 km above surface 9.38 x 106 m 4.53 4000 km above surface 1.04 x 107 m 3.70 5000 km above surface 1.14 x 107 m 3.08 8000 km above surface 1.44 x 107 m 1.93 9000 km above surface 1.54 x 107 m 1.69 10000 km above surface 1.64 x 107 m 1.49 50000 km above surface 5.64 x 107 m 0.13

As can be seen from both the equation and table above, the value of g varies inversely with the distance from the center of the earth. In fact, the variation in g with distance follows an inverse square law, where g is inversely proportional to the distance from the center of the earth. This inverse square equation means that, if the gap is doubled, the value of g decreases by a factor of 4. As the distance is tripled, the value of g is reduced by a factor of 9. And so on, too. This inverse square relationship is shown in the right-hand graphic.

Q1. Value of g is… ?

Ans . The expression for  acceleration due to gravity is given by g= GM = R2

as we know the radius of the earth at poles is minimum, so acceleration due to gravity will be maximum at poles because g is inversely proportional to R2.

Also, the net acceleration due to gravity is affected (decreases) by centrifugal force due to rotation of the earth.

poles are less affected by rotation also their radius is small so poles have a maximum value of g.

Q2. How do you Calculate the Value of g?

The acceleration due to the formula of gravity is indicated by

G = GM / R2

Where,

• G is the universal gravitational constant, G = 6.674 x 10-11 m3 kg-1 s-2

• M is the mass of the body measured using kg.

• R is the mass body radius measured by m.

• g is the acceleration due to the gravity determined by m / s2.

The mass of the moon is 7.35 × 1022Kg.

The radius of the moon is 1.74×106m

Substituting the values in the formula we get-

g= 6.67×10−11 × 7.35 × 1022 / (1.74×106)2

Thus, the value of g on the moon is g = 1.625m/s2.

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