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Question

The escape velocity on the earth is 11.2km/s. A planet has twice the radius of earth and same mean density as earth Then the escape velocity on planet in km/s will be :

A
5.6
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B
11.2
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C
22.4
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D
16.5
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Solution

The correct option is C 22.4
Answer is C.
For a spherically symmetric body, the escape velocity at a given distance is calculated by the formula ve=2GMr.
where G is the universal gravitational constant (G=6.67×1011m3/kgs2), M the mass of the planet, star or other body, and r the distance from the center of gravity.
In this case, r is double from the one of earth. So, radius of the planet is 2*6,371 kilometers = 12,742 kilometers.
The density of Earth is 5.52g/cm3.
So we can calculate the mass of the planet :
Mass=Density×Volume=d×(4/3)×π×r3=4,7834×1025 or 8 times more than Earth.
Substituting these values in the escape velocity formula we get, ve=22,378.4m/s.
Hence, the escape velocity on planet in km/s will be 22.4 km/s.

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