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Assuming the earth to be a uniform sphere of radius 6,380 km and density 5.5 g cm-3 ,find the value of acceleration due to gravity on its surface. Given that gravitational constant, G=6.66 * 10-11 N m2 kg-2 . Dear meritnation ,can u please solve this sum step by step b coz i don't know how to solve it?

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LET RADIUS OF THE EARTH BE RE ,RE=6380 km = 6380 x 10^3m
THEREFORE VOLUME OF THE EARTH(AS IT IS ASSUMED A PERFECT SPHERE) = 4/3x xRE3= 1.088x10^21m3
WE KNOW THATMASS = DENSITYx VOLUME
THEREFOREMASS OF EARTH = DENSITY OF EARTH xVOLUME OF EARTH
= 5.5 x 103 kgm3 x 1.088x 10^21
= 6 x 1024 kg (approx)

g(acceleration due to gravity) = ( G x MASS OF THE EARTH) / (RADIUS OF THE EARTH)^2
=G X ME / RE
= 6.66 x 10-11 x 6 x 1024 / (6380 x 103 )2
= 9.81 m / s2

HOPE THIS HELPS!
BEST OF LUCK!





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