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The escape velocity of a body on the sur...

The escape velocity of a body on the surface of the earth is `11.2" km"//s` If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

A

`22.4" km"//s`

B

`44.8" km"//s`

C

`5.6" km"//s`

D

`11.2" km"//s`

Text Solution

Verified by Experts

The correct Answer is:
A
Escape velocity of a body `(v_(e ))=11.2" km"//s`, new mass of the earth M`=2M_(e )` and radius of the earth (R )`=0.5 R_(e )`.
`:. "Escape velocity " (v_(e ))=sqrt((2GM_(e ))/(R_(e ))) prop sqrt((M_(e ))/(R_(e )))`
Therefore `(v_(e ))/(v._(e ))=sqrt((M_(e ))/(R_(e ))xx(0.5 R_(e ))/(2M_(e )))=sqrt((1)/(2))=(1)/(2)`
`v._(e )=2v_(e )=22.4" km"//s`
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