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

The escape speed on the other planet is V

Explanation:

Escape velocity V = √2GM/R where M = mass of planet and R = radius of planet. G = gravitational constant

Let  V₁ and V₂ be the escape speeds on planet one and two respectively.

if V₁ = √2GM₁/R₁ and V₂ = √2GM₂/R₂ and M₂ = 2M₁ and R₂ = 2R₁ (since it is twice the size of the first planet)

V₂ = √2GM₂/R₂ = V₂ = √(2G × 2M₁/2R₁) = √2GM₁/R₁ = V₁ = V

So the escape speed on the other planet is V

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

it is the second option.

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right