Question
Question: The acceleration due to gravity at the poles and the equator is \({g_p}\) and \({g_e}\) respective...
The acceleration due to gravity at the poles and the equator is gp and ge respectively. If the earth is a sphere of radius RE and rotating about its axis with angular speed ω, then gp−ge is then given by
(A) REω2
(B) RE2ω2
(C) R2Eω2
(D) REω2
Solution
Hint
In order to solve the problem, we have to take into the account of acceleration due to the gravity at the pole and the equator at a certain latitude. We also have to take in the account of the rotation of the earth. The acceleration due to gravity at a place of latitude λ due to the earth’s rotation is;
g′=g−Rω2cos2λ
Where g denotes the acceleration due to gravity, R denotes the radius of the earth, ω denotes the angular velocity,
λ denotes the latitude.
Complete step by step solution
Given data:
Acceleration due to gravity at pole is gp,
Acceleration due to gravity at the equator is ge .
The acceleration due to gravity at a place of latitude λ due to the earth’s rotation is;
g′=g−Rω2cos2λ
At the equator
The acceleration due to gravity;
g′=g−REω2cos2λ
Since at equator the latitude λ=0∘ , cos0∘=1;
By substituting we get;
g′=g−REω2
Since g′=ge ;
ge=g−REω2
At the pole:
The acceleration due to gravity;
g′=g−RPω2cos2λ
Since at the pole the latitude λ=90∘, cos90∘=0∘;
By substituting we get;
g′=g
Since g′=gp we get;
gp=g
Since we need gp−ge
gp−ge=g−g−REω2z
By cancelling the like terms, we get;
gp−ge=REω2
Therefore, the answer is gp−ge=REω2 .
Hence, the option (D) gp−ge=REω2 is the correct answer.
Note
The earth shape is not circular, it is elongated in shape. Since the shape of the earth is not round the radius of the earth differs at the poles and at the equator. Due to this the gravitational force gets affected. In light of this reason the mass placed in the equator and pole gets affected.