+22 votes
in Velocity physics by
edited by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
+16 votes
by

The escape velocity is the minimum velocity an object needs to escape the gravitational pull of a celestial body. It is calculated based on the mass and radius of the body.

The formula for escape velocity is given by:

v = √(2GM/r),

where: v is the escape velocity, G is the gravitational constant (approximately 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)), M is the mass of the celestial body, and r is the radius of the celestial body.

If we assume that the mass of the Earth remains the same and only the radius changes, we can compare the values.

Let's denote the original radius of the Earth as r₀ and the new radius as r₁ (which is half of r₀).

Using the equation above, we can compare the escape velocities:

v₀ = √(2GM/r₀), (original escape velocity) v₁ = √(2GM/r₁). (new escape velocity)

Since we're assuming the mass (M) remains the same, we can calculate the ratio of the new escape velocity to the original escape velocity:

v₁/v₀ = √(r₀/r₁).

Substituting r₁ = r₀/2, we get:

v₁/v₀ = √(r₀/(r₀/2)) = √(2).

Therefore, if the radius of the Earth becomes half, the escape velocity will increase by a factor of √2 (approximately 1.414).

Welcome to Physicsgurus Q&A, where you can ask questions and receive answers from other members of the community.
...