Question

When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is very far away from the planet is 32.3 m/s. Use this information to determine the escape speed of the planet.

Answer 1

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$K.E=\dfrac{1}{2}\times m\times(32.3)^2$

We need to calculate the escape speed of the planet

Using formula of kinetic energy

$\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}$

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2$

$\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2$

$v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}$

$v=41.29\ m/s$

Hence, The escape speed of the planet is 41.29 m/s.