CBSE Class 9 Answered

Why the value of g at poles is 9.9m/s²?

Asked by manaskumarbiswasbrb | 04 Jan, 2023, 08:13: PM

Expert Answer

Acceleration due to gravity g on sea level of earth surface varies with latitude as shown in fig.(1) .

At equator (latitude 0^o ) g = 9.78 m/s² and at north pole ( latitude 90^o ) g = 9.83 m/s²

Lower value of g at equator and higher value of g at pole is due to two following reasons.

(1) Earth is not a perfect sphere

From Newton's gravitiation theory ,

begin mathsize 14px style g space equals space fraction numerator G space M over denominator R squared end fraction end style

where G is universal gravitational constant, M is mass of earth and R is radial distance from

centre of earth to the surface of earth.

Hence acceleration due to gravity is inversely proportional to square.

Earth is approximately ellipsoidal as shown in fig.(2) . Equator radius R_E is greater than

polar radius R_P by 21 km . Hence we have less value of g at equator and greater value of g at pole.

(2) Earth's rotation about its own axis

Every object on earth's surface is subjected to circular motion due to earths rotation about its own axis.

Radius of this circular motion varies from equator to north pole. At equator, this radius equals radius of earth.

At pole , this radius is zero . At a latitude θ , this radius equals R cosθ as shown in fig.(3) , where R is radius of earth.

If a is centrifugal acceleration at latitude θ, then a = ω² R cosθ

Normal component a_rof centrifugal acceleration is ( a cosθ ) = ω² R cos²θ

Hence, net acceleration a_netexperienced by objects at latitude θ and on earth surface is

a_net = g_o - ω² R cos²θ

where g_o is acceleration due to gravity if there is no earth's rotation about its own axis.

at equator , θ = 0 , hence a_net = g_o - ω² R

at pole , θ = 90^o , hence a_net = g_o

Above explanation shows the reason for lower value of g at equator and greater value of g at pole.

Answered by Thiyagarajan K | 05 Jan, 2023, 11:15: AM

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