Question

A certain planet has an escape speed V. If another planet has twice size and twice the mass of the first planet, its escape speed will be

Answer 1

The escape speed of the second planet will be 2 * √(2) times the escape speed of the first planet.

Step-by-step explanation:

Escape speed is the minimum velocity an object needs to achieve to escape the gravitational pull of a planet or celestial body. It is given by the equation:

Vesc = √(2 * g * R)

where Vesc is the escape speed, g is the acceleration due to gravity, and R is the radius of the planet. In this case, if the second planet has twice the size and twice the mass of the first planet, its radius and acceleration due to gravity will also be twice that of the first planet. Therefore, its escape speed will be:

Vesc2 = √(2 * 2g * 2R) = √(8 * g * R) = √(8) * √(g * R) = 2 * √(2) * Vesc

So, the escape speed of the second planet will be 2 * √(2) times the escape speed of the first planet.

Answer 2

Step-by-step explanation:

Given:

M₂ = 2 × M₁

R₂ = 2 × R₁

Escape velocity, V = √2GM/R

Let V₁ and V₂ be the escape velocity of the first and second planet respectively.

V₁ = √2GM₁/R₁

V₂ = √2GM₂/R₂

Equating V₁ and V₂, we have:

(V₁)^2 × R₁ /(2 × M₁) = (V₂)^2 × M₂/(2 × R₂)

(V₁)^2 × R₁ /(2 × M₁) = (V₂)^2 × 2 × R₁ /(2 × 2 × M₁)

(V₁)^2 = (V₂)^2 × (2 × M₁)/R₁ × 2 × R₁/(4 × M₁)

(V₁)^2 = (V₂)^2

V₁ = V₂