If a planet of mass `6.4xx10^(23)` kg can be compressed into a sphere such that the escape velocity from its surface is `8xx10^(4)` m/s, then what should be the radius of the sphere ? ( Gravitational constant, `G=6.6xx10^(-11) "Nm"^(-2)kg^(-2)`)

Options

A. 40.4 km
B. 13.2 km
C. 20.4 km
D. 6.8 km

Answer

B

Explanation

As, we know, the escape velocity of a planet is given as
`v_(e)=sqrt((2GM)/(R))`
or `v_(e)^(2)=(2GM)/(R)`
Radius of sphere `(2GM)/(v_(e)^(2))`
where, `M=6.4xx10^(23)kg, v_(e)=8xx10^(4)m//s` and `G=6.6xx10^(-11)N-m^(2)kg^(-2)`
So, `R=(2xx6.6xx10^(-11)xx6.4xx10^(23))/(64xx10^(8))=13.2xx10^(3)`
`rArr R=13.2` km
Hence, option (b) is correct.