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The escape velocity for a body projected...

The escape velocity for a body projected vertically upwards from the surface of the earth is `11.2 km s^(-1)`. If the body is projected in a direction making an angle `45^@` with the vertical, the escape velocity will be

A

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

B

`11.2xxsqrt(2)kms^(-1)`

C

`11.2xx2kms^-1`

D

`11.2kms^-1`

Text Solution

Verified by Experts

The correct Answer is:
D
Change in energy `=1/(2(GMm)//R)=1//2Mv_(e)^(2)`
So, it is independent of angle as gravitational field is conservative in nature.
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Knowledge Check

  • The escape velocity of a body projected vertically upwards from the surface of the earth is v.If the body is projected in a direction making an angle theta with the vertical, the vetical, the escape velocity would be

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    B
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