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The escape velocity of a body from the e...

The escape velocity of a body from the earth's surface is `11.2km//s` . The mass of the moon is `(1//81)^(th)` of that of earth. The radius of the moon is `(1//3.7)^(th)` that of earth . Find the escape velocity from moon's surface .

Text Solution

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Given : Escape velocity on earth's surface `(v_(esc)) = 11.2km//s`
ratio of moon and earth's mass `(M_m//M_e)=1//81`
Ratio of moon and earth's radius `(R_m//R_e) = 1//3.7`
To find : Escape velocity `(v_e)_m`
Formula : `(i) v_(esc) = sqrt((2GM_m)/(R_m))`
Calculation : From formula (i) and (ii)
`((v_(esc))_m)/(v_(esc)) = sqrt((M_mxxR_e)/(M_exxR_m))`
`=sqrt(1/81xx3.7)`
=0.214
`therefore (v_(esc))_m = v_(esc) xx 0.214`
`= 11.2 xx 0.214`
`=11.2xx0.214`
=2.39 m/s
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Knowledge Check

  • The escape velocity of a body from the surface of earth is

    A
    11.2 km/s
    B
    11.4 km/s
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  • The escape velocity of a body from the surface of the earth is equal to

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    3 times critical velocity of a body close to surface of the earth
    B
    `sqrt2` times critical velocity of a body orbiting very close to surface of the earth.
    C
    critical velocity of a body orbiting very close to surface of the earth.
    D
    `1/2` times critical velocity of a body orbiting very close to surface of the earth

  • The escape velocity from the surface of the Earth is

    A
    11.2km/s
    B
    11.2m/s
    C
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    D
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