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Question: The radius of the Moon is 1750 km, and its mass is 7×1022kg. What would be the escape speed from an isolated Moon? Why was a small rocket adequate to lift the lunar astronauts back up from the surface of the Moon?

Short Answer

Expert verified

Answer:

The escape speed is 2310 m/s.

On the Moon's surface, there is no need to reach escape velocity. At a higher altitude, they can achieve a slower escape speed.

Step by step solution

01

Given Data

The radius of the Moon isRmoon=1750km=1750×103m.

The mass is 7×1022kg7×1022kg.

02

Concept of the escape speed and velocity 

The escape speed is defined as the speed at which the object's kinetic energy plus gravitational potential energy equals zero

m12mvescape2+-GMmoonmRmoon=0

The escape velocity is

vescape=2GMmoonRmoon

.

03

Determine the escape speed

It's worth noting that the escape speed is defined as the speed at which the object's kinetic energy plus gravitational potential energy equals zero. Then turn this into an equation

vescape=2GMmoonRmoon

Now determine the velocity,

vescape=2GMmoonRmoonvescape=26.76×10-11N·m2/kg27×1022kg1750×103m=2310m/s

Reason a small rocket can lift lunar astronauts back from moon

For a small rocket, this might be a significant speed to achieve upon takeoff. However, they are able to leave the Moon because they achieve a lower escape speed at a higher altitude.

On the Moon's surface, there is no need to reach escape velocity. The third line of the equation shows that the lower the escape speed, the further away the rocket is from the Moon.

Therefore, it is easier to reach by the rocket.

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