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Calculate the minimum speed required by a rocket to pull out of the gravitational force of Mars. Given that the earth has a mass `9` times and radius twice of the planet Mars. Escape speed on the surface of earth is `11.2 km s^(-1)`.

Text Solution

Verified by Experts

`g_(m) = g_(e) ((M_(s))/(M_(e))) ((R_(e))/(R_(m)))^(2) = g_(e) xx (1)/(4) xx 4 = (4)/(9) g_(e)`
On earth, `upsilon_(e) = sqrt(2G_(e) R_(e))`,
On Mars, `upsilon_(m) = sqrt(2g_(m) R_(m))`
`:. (upsilon_(m))/(upsilon_(e)) = upsilon_(e) xx (sqrt(2))/(3) = (11.2 xx 1.414)/(3)`
`= 5.28 km s^(-1)`
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Knowledge Check

  • A planet has mass equal to mass of the earth but radius one fourth of radius of the earth . Then escape velocity at the surface of this planet will be

    A
    `11.2` km/s
    B
    `22.4 ` km/s
    C
    `5.6` km/s
    D
    `44.8` km/s