A planet of radius R $$=$$ (1/10) (radius$$ of $$Earth) has the same mass density as earth. Scientists dig a well of depth $$R/5$$ on it and lower a wire of the same length and of linear mass density $$10$$−$$3kg/m$$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is:(Take$$ the radius of Earth $$=6$$×$$106m$$ and the acceleration due to gravity of Earth is $$10m/s2$$ $$)

Question

A planet of radius R $$=$$ (1/10) (radius$$ of $$Earth) has the same mass density as earth. Scientists dig a well of depth $$R/5$$ on it and lower a wire of the same length and of linear mass density $$10$$−$$3kg/m$$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is:(Take$$ the radius of Earth $$=6$$×$$106m$$ and the acceleration due to gravity of Earth is  $$10m/s2$$ $$)

Infinity Learn · Accepted Answer

As $$R=RE10---(1)$$  ρ$$=M4$$π$$/3R3=ME4$$π$$/3RE3=density$$ ⇒$$M=ME103$$  $$g=GMR2$$ $$g=G(ME/103)(RE/10)2$$ $$g=GME10RE2$$ ⇒$$g=gE10---(2)$$  $$gdepth=g1$$−dR $$gdepth=gR$$−dR $$gdepth=gxR---(3)$$ mass per unit $$length=$$λ$$=dmdx$$ ⇒$$dm=$$λ$$dx---(4)(where$$ x is the distance from the centre of the $$planet)$$ force $$F=mg$$ force of attraction element of wire, $$dF=(dm)gdepth$$ substitute, $$F=$$∫$$4R/5R$$λdxgxR $$F=$$λgR∫$$4R/5Rxdx$$ $$F=9$$λ$$gR50$$ substitute equation (1),(2)  $$F=9$$λ$$50gE10RE10=9$$λ$$gERE5000$$ $$F=910$$−$$3kg/m310m/s26$$×$$106m5000$$ $$F=108N$$

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