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The escape velocity for a body projected...

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

A

`11 // sqrt2 km//sec `

B

`11 sqrt2 km//sec `

C

`2km//sec `

D

`11 km //sec `

Text Solution

Verified by Experts

The correct Answer is:
D
`v_e = sqrt((2GM)/(R ))`
Since the escape velocity is independent of direction of projection, therefore escape velocity of the body, projected at an angle of `45^@` with the vertical will be same, i.e., 11 km/sec.
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Knowledge Check

  • 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

  • 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`

  • 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`

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