Apply to Amity University, Noida B.Tech Admissions 2024
Variation in 'g' due to height, Variation in 'g' due to Rotation of earth are considered the most difficult concepts.
Acceleration due to gravity (g) are considered the most asked concepts.
75 Questions around this concept.
The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The values of 'g' and 'R' (radius of the earth) are 10 m/s^{2} and 6400 km respectively. The required energy for this work will be 6.4 x 10^{n} joules. Then the value of 'n' is :
The spring balance is graduated at sea level. If the body is weighed by this spring scale at gradually increasing heights from the earth's surface, then the weight indicated by the spring scale
Which of the following statements about gravitational acceleration is true?
Announcement: Register for India's Biggest Education Expo
Latest: Top colleges accepting JEE Main rank between 50k to 2 Lakhs
Don't Miss: JEE cutoffs for top IITs | JEE Main cutoff for top 10 NITs
Recommended: JEE Main college Predictor
Which of the following factors affects the value of 'g'?
The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by
(R=Earth’s radius) :
Apply to Amity University, Noida B.Tech Admissions 2024
How does the value of 'g' change with increase in altitude?
The height at which the acceleration due to gravity becomes ( where = the acceleration due to gravity on the surface of the earth ) in terms of ( the radius of the earth) is
How does the value of 'g' change with increasing depth below the Earth's surface?
How does the value of 'g' change at the equator due to the Earth's rotation?
The Gravitational Force exerted by the earth on a body is known as the gravitational pull of gravity. And this force will produce an acceleration in the motion of a body.
And this is known as the acceleration due to gravity.
This is denoted by g.
Let gravitational force exerted by the earth on the body of mass m resting on the surface of the earth is given by
…(1)
Where M = mass of the earth and R = radius of the earth
And If g is the acceleration due to gravity then this F can be written as
F=(mass)*(acceleration)=mg ...(2)
On Comparing
We get
Now density of earth
Then
So
Its average value is on the surface of earth.
It is a vector quantity and its direction is always towards the center of the earth/Planet.
Dimension-
Its value depends upon the mass, radius, and density of the Earth/Planet.
Means Planet having more value of than earth will have a greater value of acceleration due to gravity (g) than the earth
I.e
It is independent of mass, shape and density of the body situated on the surface of the Earth/planet.
Value of g will be the same for a light as well as heavy body if both are situated on the surface of the Earth/planet.
The value of acceleration due to gravity (g) changes its value due to the following factors
The shape of the earth
Height above the earth's surface
Depth below the earth's surface
Axial rotation of the earth.
Variation of 'g' due to the shape of the earth
Earth has an elliptical shape as shown in fig.
Where Equatorial radius is about 21 km longer than the polar radius.
Or
Where Radius of the equator
Radius of pole
So
In fact
Or we can say Weight increases as the body is taken from equator to pole.
Let's study Variation in 'g' with height
Value of g at the surface of the earth (at distance r=R from earth center)
Value of g at height h from the surface of the earth (at a general distance r=R+h from earth center)
Where
As we go above the surface of the earth, the value of g decreases
So
Where gravity at a height h from the surface of the earth.
The radius of earth
height above the surface
Value of 'g' at ∞
if
No effect of earth gravitational pull at infinite distance.
Value of g when h < < R
Formula
Value of g
So
The absolute decrease in the value of g with height
The fractional decrease in the value of g with height
Percentage decrease in the value of g with height
Let's study Variation in 'g' with depth
Value of g at the surface of the earth (at d=0)
Value of g at depth d from the surface of the earth (at a general distance r=(R-d) from earth centre)=g'
And
Means Value of g' decreases on going below the surface of the earth.
So
Value of 'g' at the centre of the earth
At the centre
So
i.e., Acceleration due to gravity at the centre of the earth becomes zero.
The absolute decrease in the value of g with depth
The fractional decrease in the value of g with depth
The value of g decreases with depth.
Percentage decrease in the value of g with depth
Note- Rate of decrease of gravity outside the earth (h<<R) is double that of inside the earth
Let's study Variation in 'g' due to Rotation of earth
As the earth rotates about its axis
Let its angular velocity is about an axis as shown in the figure.
So if a body placed on its surface then it will move along the circular path.
So the apparent weight of the body will decrease as it will experience a centrifugal force due to rotation.
We can calculate this apparent weight of the body using force balance.
Let calculate its value for a body of mass m situated at a point P (having latitude=)
As shown in the figure.
=The angle between the equatorial plane at that point and line joining that point to the centre of the earth.
For the poles =90 and for equator =0
From applying newton's 2nd law along the line joining point P and center.
Where g' is the value of acceleration due to gravity at point P.
So we get
Apparent weight of body decrease with an increase in angular velocity ()
Apparent weight of the body varies from point to point because each point has different latitude and magnitude of centrifugal force varies with the latitude of the place.
For Pole, =90
So
I.e value of g at the poles is independent of angular velocity of earth.
For equator,=0
I.e Decrease in the value of g is maximum at the equator
Weightlessness due to rotation of the earth-
Weightlessness means g'=0
So
As (For equator)
Where Angular velocity for which a body at the equator will become weightless
The time period of Rotation of earth for which body at the equator will become weightless
Where Radius of earth
And using
We get
And
Relation of gravity at the poles and equator
After considering the effect of rotation, and elliptical shape of the earth
Where gravity at the pole
gravity at equator
"Stay in the loop. Receive exam news, study resources, and expert advice!"