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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

    A
    v
    B
    `v cos theta`
    C
    `v sin theta`
    D
    `v tan theta`

  • The escape velocity of a body projected vertically upward from the earth's surface is 11.2 " kms"^(-1) . If the body is projected in a direction making 30^(@) angle to the vertical, its escape velocity in this case will be

    A
    `11.2 " kms"^(-1)`
    B
    `(11.2)/(2) " kms"^(-1)`
    C
    `11.2 xx (sqrt(3))/(2) "kms"^(-1)`
    D
    `(11.2)/(3)"kms"^(-1)`

  • The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of 45^(@) with the vertical, the escape velocity will be

    A
    `11/sqrt(2)` km/s
    B
    `11sqrt(2)` km/s
    C
    22 km/s
    D
    11 km/s
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