Know the Variation in the Acceleration

Gravitation is a force that claims that all objects on Earth and in space are attracted to one another. The gravitational force exerted on an object is proportional to its mass; the greater the mass of an object, the greater the gravitational force exerted on it by other objects. All visible items, such as a pen, eraser, planets, mobile phone, watch, and refrigerator, are attracted to one other in some way. With the Electromagnetic Force and the Nuclear Force, gravity is one of the non-contact forces.

Acceleration

Acceleration is the rate at which velocity changes with time in terms of both speed and direction. A point or object is said to be accelerated if it moves faster or slower in a straight line. Because the direction of motion on a circle is continually changing, the motion is accelerated even if the speed remains constant. Both effects contribute to acceleration in all other types of motion.

Acceleration is a vector quantity since it has both a magnitude and a direction. Velocity is a vector quantity as well. Acceleration is defined as the change in velocity vector over a time interval divided by the time interval. When velocity is measured in m/sec, acceleration is measured in  m/sec2

Acceleration due to gravity

When a body falls towards the earth, its acceleration changes due to the gravitational force of the earth. This type of acceleration is known as acceleration due to gravity.This is the acceleration that an object experiences as a result of the gravitational force.

Since, acceleration due to gravity has both a magnitude and a direction. As a result, it’s a vector quantity.

Formula for acceleration due to gravity

Mathematically, the acceleration due to gravity is directly proportional to the mass of the body and inversely proportional to the distance from the centre of mass. 

Therefore, the formula for acceleration due to gravity can be given as:

g=GM/r2

Here,

g = gravitational constant= 6.67×10-11 Nm2/kg2

M = mass of the body

r = distance from the centre of mass

The S.I. unit of acceleration due to gravity is m/sec2

Variation in the Acceleration

There are a number of factors that influence the value of g, such as:

• Variation of g with Height: Since the value of g is inversely proportional to the height above the earth’s surface, it reduces as the height rises. 

Consider a mass (m) at a height (h) above the surface of the earth. Now consider the gravitational force acting on the test mass is given by:

F=GMm/(R+h)2  

R = radius of the earth

M = mass of the earth

Now, the acceleration due to gravity at a given height h is given by:

mgh=GMm/(R+h)2

gh= GMm/[R2(1+h/R)] ———– (1)

We know that the value of acceleration due to gravity g is given by:

g= GM/R2————– (2)

Dividing Equation (2) by Equation (1),

gh=g(1+h/R)-2   ——— (3)

At a height above the earth’s surface, this is the acceleration due to gravity. According to the formula above, the value of g reduces as an object’s height increases and eventually becomes 0 at an infinite distance from the earth.                  

1. Variation of g with Depth: The value of g is proportional to the depth below the earth’s surface; hence it rises with increasing depth but falls to zero at the earth’s centre.

Assume a body of mass m lies at a position B (black dot in the above diagram), where B is at a depth of d from the earth’s surface and R – d is its distance from the centre.

Now the value of g at point B is given by:

gd=g(R-d)/R

In the above relation, 

d<R

 Also, [(1-h/R)<1]

Therefore, we can conclude that gd<g

And as we get closer to the centre of the earth, the acceleration  gd  reduces, which is only felt when we are extremely close to the centre of the earth. As a result, the value of g changes as a function of height and depth.

However, the value of g varies even on the earth’s surface. For instance, g is maximum near the poles and least at the equator.

1. Variation of g due to shape of the Earth:  At the equator, the value of g is lower than the value of g at the poles.

The earth is an oblate spheroid, not a fully spherical sphere. The earth’s polar radius (radius at the poles) is 21 kilometres less than its equatorial radius (near the equator). The Earth is not spherical, but rather bulged out as depicted below.

As we know, the acceleration due to gravity is inversely proportional to the square of the earth’s radius. At the equator, the earth’s radius is greater and g is smaller. At the poles, the situation is vice versa.

Assume the earth’s shape is somewhat elliptical. As a result, the distances from the pole and the equator to the centre will be different.

As shown in the figure, the distance between the pole and the centre of earth is Rp  and the distance from the equator to the centre of the earth is  RE. The relation can be given as,

RE>Rp

According to the formula for the value of acceleration due to gravity we know that,

g ∝ 1/R2

Putting the value of REand Rp we can write the relation as:

gp∝ 1/Rp2

gE∝ 1/RE2

gp>gE

 The ratio of gpand gE can be given as:

gp/gE =RE2/Rp2

As a result, the gravitational acceleration at the equator is lower than that at the pole.

1. Variation of g due to Rotation of Earth: The value of g falls as the earth’s rotation speed increases.

The change in g caused by the centrifugal force acting on the earth’s rotation. When the earth rotates, all objects are subjected to a centrifugal force that acts in the opposite direction of gravity.

Consider a sample mass (m) that is located on a latitude that intersects the equator at an angle. When a body spins, every particle within it moves in a circular motion around the axis of rotation, as we’ve seen. The earth spins at a constant angular velocity, while the test mass travels in a circular path of radius r at an angular velocity.

Because this is a non-inertial frame of reference, the sample mass is subjected to centrifugal force mr2. The test mass is drawn towards the planet’s centre by gravity mg. Both of these forces are referred to as co-initial forces because they act from the same point, and coplanar forces because they fall on the same plane.

If two coplanar vectors form two sides of a parallelogram, the resultant of those two vectors will always be along the parallelogram’s diagonal, according to the parallelogram law of vectors. We may compute the magnitude of the apparent value of gravitational force at the latitude using the parallelogram law of vectors:

(mg’)2=(mg)2+(mr)2+2(mg)(mr2)cos(180-)   —–(1)

Here,

Therefore, Equation (1) can be written as:

g’ = g-R2cos2

where g′ is the apparent amount of gravity acceleration at the latitude due to earth rotation, and g is the real magnitude of gravity at the latitude without taking earth rotation into account.

Conclusion

The downward acceleration of a free-falling object is 9.8 m/s2(on Earth). This numerical figure for the acceleration of a free-falling item is such an essential value that it is given a distinct name. It’s known as acceleration due to gravity, and it’s the acceleration of any object moving only owing to gravity. In fact, the acceleration of gravity is such a significant amount that scientists have given it its own symbol, g = 9.8 m/s2 is the most precise numerical value for the acceleration of gravity. This numerical figure (to the second decimal place) has minor fluctuations that are mostly determined by altitude.