What is meant by acceleration due to gravity?Derive an expressio
Discuss the variation of 'g' with angle of latitude.

The acceleration due to gravity changes with the angle of latitude due to shape and rotation of the earth about its own axis.
Variation of g due to shape:
The earth is not a prefect sphere. It is flat at the pole and bulges out at the equator as a result polar radius is smaller than equatorial radius. As the acceleration due to gravity is inversely proportional to the square of radius of the earth, therefore the value of g increases as one moves from equator to pole.
Variation of g due to rotation:
We know that the earth rotates about its own axis as a result of which the objects on the earth experience the centrifugal force. This centrifugal force decreases the acceleration due to gravity and magnitude of change in the value of g due to rotation of the earth depends on the value of the angle of latitude.
Let us consider the earth to be a homogenous sphere of mass Mand radius R. The earth rotates about polar axis and let ω be the angular velocity of rotation of the earth. All the objects at rest on earth also revolve about its polar axis with same angular velocity ω. Let us consider a body of mass m placed at a point P on the earth at angle of latitude λ.

The acceleration due to gravity changes with the angle of latitude du
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#6 {main}</pre> with center at O. In rotating frame the earth, body experiences pseudo force mrω squared along space OP with rightwards arrow on top. Now the apparent weight of the body is resultant of force mg directed towards the center and force mrω squared along stack OP. with rightwards arrow on top
Applying parallelogram of vector addition, we get
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#6 {main}</pre>
or    straight g subscript straight lambda equals square root of straight g squared plus left parenthesis rω squared right parenthesis squared minus 2 left parenthesis grω squared right parenthesis space cosλ end root
             equals straight g square root of 1 plus open parentheses rω squared over straight g close parentheses squared minus 2 rω squared over straight g cosλ end root
As rω squared over straight g less than less than 1 comma  therefore space space space open parentheses rω squared over straight g close parentheses squared can be neglected.
∴           straight g subscript straight lambda equals straight g square root of 1 minus 2 rω squared over straight g cosλ end root
                    equals straight g open parentheses 1 minus rω squared over straight g cosλ close parentheses [using binomial exp]
or          straight g subscript straight lambda equals straight g minus rω squared cosλ
Substituting straight r equals Rcosλ comma we get
   straight g subscript straight lambda equals straight g minus Rω squared cos squared straight lambda



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The weight of a body at depth d is equal to the weight of the body at height h above the surface of the earth. What is the ratio of h to d, for small value of h and d?
                                 OR
Prove that the distance we have to cover into the earth below its surface is twice the distance we have to cover above the surface of the earth to get the same change in the weight of a body?


For small value of h, the acceleration due to gravity at height h is, 


                

∴             

Similarly, at the depth d the weight of body is, 

                   

Since,    

i.e.,  

     

∴                           

                        
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How much faster than its present rate should the earth rotate about its axis, so that the bodies at the equator experience weightlessness? Also find the duration of day. Take g = 9.8 m/s2.

We know that acceleration due to gravity at equator is, 

                    

where,

 is the angular velocity of the earth.

Angular velocity of Earth=  

If the acceleration due to gravity at the equator becomes zero, the body at equator experiences weightlessness.

Let  be the angular velocity of rotation at which the acceleration due to gravity at the equator becomes zero. 

That is,

                 

             

Now,
           

              

Thus, the earth should rotate 17 times faster than the present rate.

Rotation speed increases by a factor of 17.

Therefore, the duration of day decreases by factor 17.

New duration of day, 

              

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What is meant by acceleration due to gravity?
Derive an expression for it in terms of mass of the earth and gravitation constant.


According to Newton’s law of gravitation, every mass exerts gravitational force on every other mass. Thus the earth exerts force on every mass placed surrounding it. This force of attraction exerted by the earth on a body is called force of gravity and the acceleration possessed by a body moving under the force of gravity is called acceleration due to gravity.
The value of acceleration due to gravity does not depend on the mass of a body but it depends on the mass of the earth and distance of the body from the center of the earth.
Let us consider the earth to be a uniform sphere of mass M and radius R. Let a body be placed at a distance r from the surface of the earth. The force of gravity on the body is,
straight F equals straight G Mm over straight r squared
The acceleration produced in the body due to force of gravity i.e. acceleration due to gravity is,

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#6 {main}</pre>

At the surface of the earth, the acceleration due to gravity is,
straight g equals GM over straight R squared

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If acceleration due to gravity at the surface of the earth is g then what is its value at depth d?

Let the earth be homogenous sphere of radius R with density ρ. The acceleration due to gravity at surface of the earth g is,
straight g equals GM over straight R squared equals 4 over 3 πGρR

Let the earth be homogenous sphere of radius R with density ρ. The a
Let at depth d the acceleration due to gravity be gd. At depth d, the gravity is only due to mass of inner sphere of radius R-d.
∴    straight g subscript straight d equals fraction numerator straight G left square bracket mass space of space sphere space of space radius space left parenthesis straight R minus straight d right parenthesis right square bracket over denominator left parenthesis straight R minus straight d right parenthesis squared end fraction
            equals fraction numerator straight G begin display style 4 over 3 end style straight pi left parenthesis straight R minus straight d right parenthesis cubed straight rho over denominator left parenthesis straight R minus straight d right parenthesis squared end fraction
             space space equals 4 over 3 πGρ left parenthesis straight R minus straight d right parenthesis
or        space space space space space space space space space space space straight g subscript straight d equals 4 over 3 πGρR open parentheses fraction numerator straight R minus straight d over denominator straight R end fraction close parentheses
                           = straight g open parentheses 1 minus straight d over straight R close parentheses

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