A planet has mass eight times that of the earth and radius twice that of the earth. Taking the earth's escape velocity as 11.2 km/s, what is the escape velocity for this planet?

1. Escape velocity is $v_{esc} = \sqrt{\dfrac{2GM}{R}}$. 2. For the planet, $M' = 8M$ and $R' = 2R$, so $v'_{esc} = \sqrt{\dfrac{2G(8M)}{2R}} = \sqrt{\dfrac{8}{2}\cdot\dfrac{2GM}{R}} = \sqrt{4}\,\sqrt{\dfrac{2GM}{R}}$. 3. Therefore $v'_{esc} = 2\,v_{esc}$. 4. Substitute the earth's value: $v'_{esc} = 2 \times 11.2 = 22.4\text{ km/s}$. 5. The factor is 2, not 4, so 44.8 km/s is wrong (rule out). 6. Hence the planet's escape velocity is about 22.4 km/s. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 10 "Space Missions", p.153_