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The escape speed of a body on the earth'...

The escape speed of a body on the earth's surface is `11.2kms^(-1)`. A body is projected with thrice of this speed. The speed of the body when it escape the gravitational pull of earth is

A

`11.2 kms^(-1)`

B

`22.4sqrt(2) kms^(-1)`

C

`(22.4)/(sqrt(2)) kms^(-1)`

D

`22.4sqrt(3) km s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
(b): Let v be the speed of the body when it escapes the gravitational pull of the earth and u be speed of projection of the body form the earth's surface.
According to law of conservation of mechaninacal energy,
`(1)/(2)mu^(2)-(GM_(E)m)/(R_(E))=(1)/(2)mv^(2)-0`
where m and `M_(E)` be masses of the body and earth respectively and `R_(E)` is the radius of the earth.
`or v^(2)=u^(2)-(2GM_(E))/(R_(E))`
`v^(2)=u^(2)-v_(e)^(2)`
` v=sqrt(u^(2)-v_(e)^(2))=sqrt((3v_(e))^(2)-v_(e)^(2))" " ( :. v_(e)=sqrt((2GM_(E))/(R_(E))))`
`=sqrt(8)v_(e)=11.2sqrt(8)km s^(-1)=22.4sqrt(2) km s^(-1)`
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