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. $5.6$
• B. $11.2$
• C. $22.4$
• D. $16.5$

Answer 1

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 $v_e=\sqrt{\frac{2GM}{r}}$.
where G is the universal gravitational constant ($G=6.67\times {10}^{-11}{m}^3/kg{s}^2$), 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 $5.52g/{cm}^3$.
So we can calculate the mass of the planet :
$Mass=Density\times Volume=d\times \left(4/3\right)\times \pi \times r^3=4,7834\times {10}^{25}$ or 8 times more than Earth.
Substituting these values in the escape velocity formula we get, $v_e=22,378.4m/s$.
Hence, the escape velocity on planet in km/s will be 22.4 km/s.