Question #ece0e

Question

575 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-02T21:29:14

#.0017%#

We can consider that body as a mass of density same as earth (i.e. #3000 kgm^-3#) and some extra mass of density #2000 kgm^-3#.

Now, on the surface of earth this extra mass will have an effect as if there is a point mass at the centre of this body. Its entire mass is:

#M =rhor^3 = 2000xx2000^3kg = 1.6xx10^13 kg#

We want acceleration due to gravity of this mass at a distance #r = 2500m = 2.5xx10^3m#

and we know:

#G = 6.67 × 10^-11 m^3 kg^-1 s^-2#

hence, acceleration due to gravity of this mass:

#deltag = (GM)/r^2 = ( 6.67 × 10^-11 xx1.6xx10^13)/(6.25xx10^6) ms^-2 ~~1.7xx10^-4 #

percentage change in g = #(deltag)/g= (1.7xx10^-4)/9.8 xx100% ~~.0017%#