1.

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth R_(e ), the radius of the planet would be

Answer»

`2R_(E )`
`4R_(e )`
`(1)/(4)R_(e )`
`(1)/(2)R_(e )`

Solution :From equation of acceleration due to gravity.
`g_(e )=(GM_(e ))/(R_(e )^(2))=(G(4//3)piR_(e )^(3))/(R_(e )^(2))rho_(e )`
`g_(e )prop R_(e )rho_(e )`
Acceleration due to gravity of planet `g_(p) prop R_(p)rho_(p)`
`:. R_(e )rho_(p) rArr R_(e )rho_(e )=R_(p)2rho_(e ) rArr R_(p)=(1)/(2)r_(e )`


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