Question

The escape velocity from a spherical planet is about vo km/s. The escape velocity from a planet
having twice the radius and half the mean density
​

Answer 1

I hope that you got your required answer.
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Refer ncert textbook for formula
Ve=[8piG(rho)(R)^2]^1/2

Answer 2

Given :

• Escape velocity from first planet = v₀
• The second planet has twice the radius and half the mean density of the first planet.

To find :

• Escape velocity from the second planet.

Solution :

• Let the radius and mean density of the first planet be R₁ and d₁ respectively and that of the second planet be R₂ and d₂ respectively.
• According to the question, R₂ = 2R₁ and d₂ = 0.5d₁
• The escape velocity is the velocity of projection with which the body will go out of the gravitational field of the earth and never return.
• The escape velocity is given by :

         $Ve = \sqrt{\frac{2GM}{R} }$     where G is the gravitational constant, M is the mass and R is the radius of the planet .

• As the planet is spherical, $M = \frac{4}{3} \pi R^{3} . d$     where d is the density.

• Therefore, $Ve = \sqrt{\frac{2G}{R}\frac{4}{3}\pi R^{3} d }$
• From the above equation, we can see that Ve is directly proportional to R and $\sqrt{d}$ .
• $\frac{V1}{V2} = \frac{R1}{R2} \sqrt{\frac{d1}{d2} }$

• $V0 = \frac{1}{2} \sqrt{\frac{1}{0.5} } V2$

• V₂ = $\sqrt{2}$ v₀

Answer :

• The escape velocity from the planet having twice the radius and half the mean density will be $\sqrt{2}$ v₀ .