Universe: Stars And Galaxies
Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 4, Problem 50Q
To determine

The escape speed for Earth and Mars.

The planet for which it would be easier to eject rocks.

Expert Solution & Answer
Check Mark

Answer to Problem 50Q

The escape speed for Earth is 11.2 km/s and for Mars is 5 km/s

It is easy for rocks to escape from planet Mars.

Explanation of Solution

Given:

Gravitational constant, G= 6.67×1011Nm2/kg2

Mass of Earth, M=5.974×1024kg

Radius of the Earth, R=6.67×1011Nm2/kg2

For Mars

Mass of Mars, M=3.397×106kg

Radius of Mars, R=6.418×1023m

Formula Used:

An object needs particular speed to escape from the atmosphere which is called escape speed of the planet.

The escape speed is calculated as Vescape=2GMR

The escape speed is calculated as Vescape=2GMR

For Earth, substitute 6.67×1011Nm2/kg2 for G (gravitation constant), 5.974×1024kg for M (mass) and 6.378×106m for R (radius)

Vescape= 2( 6.67× 10 11 N m 2 /k g 2 )( 5.974× 10 24 kg ) 6.378× 10 6 mVescape= 2( 6.67× 10 11 ( ( N )( kg.m/ s 2 N ) ). m 2 /k g 2 )( 5.974× 10 24 kg ) 6.378× 10 6 mVescape=(11,178m/s)( 1km 1000m)Vescape=11.178km/s

Round off the result to three significant figures the escape speed of the earth is 11.2 Km/s

Substitute 6.67×1011Nm2/kg2 for G, 3.397×106kg for M and 6.418×1023m for R

Vescape= 2( 6.67× 10 11 m 2 /k g 2 )( 6.418× 10 23 kg ) 3.397× 10 6 mVescape= 2( 6.67× 10 11 ( ( N )( kg.m/ s 2 N ) ). m 2 /k g 2 )( 6.418× 10 23 kg ) 3.397× 10 6 mVescape=(5020m/s)( 1km 1000m)Vescape=5.02km/s

Round off the result to three significant figures the escape speed of mars is 5 km/s.

Conclusion:

From the above calculations, it is clear that it is easy for the rocks to escape from the planet Mars because it has a speed of 5 km/s only.

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