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Question

(a) Is the acceleration due to gravity of earth 'g' a constant? Discuss.
(b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4×1022 kg and a radius of 1.74×106m(G=6.7×1011Nm2/kg2). Which satellite do you think it could be?

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Solution

(a) No, the value of acceleration due to gravity (g) is not constant at all the places on the surface of the earth. Since the radius of the earth is minimum at the poles and maximum at the equator, the value of g is maximum at the poles and minimum at the equator. As we go up from the surface of the earth, the distance from the centre of the earth increases and hence the value of g decreases. The value of g also decreases as we go down inside the earth.


b right parenthesis A c c e l e r a t i o n space d u e space t o space g r a v i t y g equals fraction numerator G M over denominator R squared end fraction M a s s comma M equals 7.4 cross times 10 to the power of 22 k g r a d i u s equals 1.74 cross times 10 to the power of 6 m g r a v i t a t i o n a l space c o n s tan t space G equals 6.67 cross times 10 to the power of negative 11 end exponent space bevelled fraction numerator N m squared over denominator K g squared end fraction g equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 7.4 cross times 10 to the power of 22 over denominator open parentheses 1.74 cross times 10 to the power of 6 close parentheses squared end fraction g equals fraction numerator 6.7 cross times 7.4 over denominator 1.74 cross times 1.74 cross times 10 end fraction g equals 1.63 space bevelled m over s squared a s space t h e space v a l u e space o f space g equals 1.673 space bevelled fraction numerator m over denominator s squared comma w h i c h space o n e space s i x t h space t h e space v a l u e space o f space g space o n space e a r t h comma t h e space s a t e l i t e space c o u l d space b e space m o o n end fraction


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