If the earth were a solid sphere of iron of radius. 6.37×10 power of 3 km and of density 7560 kg/m cube what would be the value of g at its surface ( G=6.67×10 to the power of -11 SIU)
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Answer:
Correct option is B)
radius =6.37×10
6
m
Volume =
3
4
πr
3
V=
3
4
×π×(6.37×10
6
)
3
Mass =5.975×10
24
kg
density (ρ)=
V
mass(m)
ρ
=
4×π×(6.37×10
6
)
3
5.975×10
24
=5.52×10
3
kgm
−3
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