Escape velocity depends only on the planet
v_esc = √(2GM/r) = √(2gr) contains no property of the escaping object. It is independent of the object's mass (m cancels) and independent of launch direction (energy is a scalar). It depends only on M and r of the planet.
What this actually means
Both independences come straight out of the derivation, so quote the derivation rather than asserting the result. ½mv² = GMm/r, m cancels, therefore mass-independent. Energy is a scalar with no direction, therefore direction-independent.
The direction claim surprises people. Fire straight up, or fire at 45°, and the minimum SPEED is the same, because the energy equation never mentions angle. In practice you would aim to avoid hitting the ground, but the threshold speed is unchanged.
Do state the assumptions if asked: no air resistance and no help from the planet's rotation. Real launches do exploit the Earth's spin by firing eastwards from near the equator, which is why launch sites cluster there.
The classic application: 'explain why the Moon has no atmosphere'. Escape speed at the Moon's surface is only about 2.4 km s⁻¹, comparable with the mean speed of gas molecules, so gas molecules escape over time. Earth's 11 km s⁻¹ is far above molecular speeds, so our atmosphere stays.
It is also worth saying it is a SPEED. Calling it a velocity is standard usage, but if a question asks you to comment on the name, the point is that direction is irrelevant.
Claiming a heavier rocket, or a rocket launched vertically rather than at an angle, needs a different escape speed.
Prove it — watch it be true
- In the escape-velocity launcher, find the threshold launch speed that first gives the escape outcome.
- Change the projectile mass and confirm the threshold speed is exactly the same.
- Switch to a different planet preset and confirm the threshold DOES move, since it depends on M and r.