Find the escape speed for a rocket leaving the moon. the acceleration of gravity on the moon is 0.166 times that on earth and the moon's radius is 0.273 re. answer in units of km/s.

Radius of Earth = 6,378 Km

Radius of the Moon = (6,378 Km)(0.273) = 1,741.19 Km or 1,741,190 m

G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the moon = 7.348 x 10²² KG

Formula:Escape Velocity Ve = √2 GM/R

Ve = √2 (6.67 x 10⁻¹¹ N.m²/Kg²)(7.348 x 10²² Kg)/1,741.19 /1,741,190 m

Ve = √5,629,616.53 m²/s²

Ve = 2,372.68 m/s or

Ve = 2.37 Km/s

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asuwafohjames asuwafohjames · Answer 2 · 2022-03-01T12:12:27+01:00

The escape velocity of the rocket from the moon is 2.38 km/s.

What is escape velocity?

This is the minimum velocity required by an astronomical body to escape the force of gravity.

To calculate the escape velocity of the rocket, we use the formula below.

Formula:

Vs = [tex]\sqrt{2gR}[/tex]................... Equation 1

Where:

Vs = Escape velocity of the rocket leaving the moon
g = acceleration due to gravity of the moon
R = Radius of the moon

From the question,

Given:

g = 9.8(0.166) = 1.6268 m/s²
R = 0.273(6371) = 1739.283 km = 1739283 m

Substitute the given values into equation 1

Vs = [tex]\sqrt{2*1.6268*1739283}[/tex]
Vs = [tex]\sqrt{5658931.1688}[/tex]
Vs = 2378.85 m/s
Vs = 2.37885 km/s
Vs ≈ 2.38 km/s.

Hence, the escape velocity of the rocket from the moon is 2.38 km/s.
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