The escape velocity for the moon.

Question

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Answer 1

Question

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

(b)

To determine

The velocity of the particle in the earth’s environment.

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$

Ratio is given as below:

$K.EfK.Ei=12mpvf212mpvi2=vf2vi2$

$K.EfK.Ei=vf2vi2=G( M m R m + M E R E ) 2G M m R m K.EfK.Ei=( M m R m + M E R E )×Rm2Mm$

$K.EfK.Ei=(1+MERERmMm)×12$

Calculation:

Putting the values of radius and mass of earth and moon, we get,

$K.EfK.Ei=(1+( 5.972× 10 24 6371 × 1737 3.342× 10 22 ))×12K.EfK.Ei=24.86$

Conclusion:

The ratio of final to initial kinetic energy of the particle is $24.86$ .

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Answer 2

Question

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

(b)

To determine

The velocity of the particle in the earth’s environment.

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$

Ratio is given as below:

$K.EfK.Ei=12mpvf212mpvi2=vf2vi2$

$K.EfK.Ei=vf2vi2=G( M m R m + M E R E ) 2G M m R m K.EfK.Ei=( M m R m + M E R E )×Rm2Mm$

$K.EfK.Ei=(1+MERERmMm)×12$

Calculation:

Putting the values of radius and mass of earth and moon, we get,

$K.EfK.Ei=(1+( 5.972× 10 24 6371 × 1737 3.342× 10 22 ))×12K.EfK.Ei=24.86$

Conclusion:

The ratio of final to initial kinetic energy of the particle is $24.86$ .

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Answer 3

Question

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

(b)

To determine

The velocity of the particle in the earth’s environment.

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$

Ratio is given as below:

$K.EfK.Ei=12mpvf212mpvi2=vf2vi2$

$K.EfK.Ei=vf2vi2=G( M m R m + M E R E ) 2G M m R m K.EfK.Ei=( M m R m + M E R E )×Rm2Mm$

$K.EfK.Ei=(1+MERERmMm)×12$

Calculation:

Putting the values of radius and mass of earth and moon, we get,

$K.EfK.Ei=(1+( 5.972× 10 24 6371 × 1737 3.342× 10 22 ))×12K.EfK.Ei=24.86$

Conclusion:

The ratio of final to initial kinetic energy of the particle is $24.86$ .

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Answer 4

Question

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

(b)

To determine

The velocity of the particle in the earth’s environment.

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$

Ratio is given as below:

$K.EfK.Ei=12mpvf212mpvi2=vf2vi2$

$K.EfK.Ei=vf2vi2=G( M m R m + M E R E ) 2G M m R m K.EfK.Ei=( M m R m + M E R E )×Rm2Mm$

$K.EfK.Ei=(1+MERERmMm)×12$

Calculation:

Putting the values of radius and mass of earth and moon, we get,

$K.EfK.Ei=(1+( 5.972× 10 24 6371 × 1737 3.342× 10 22 ))×12K.EfK.Ei=24.86$

Conclusion:

The ratio of final to initial kinetic energy of the particle is $24.86$ .

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Answer 5

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

(b)

To determine

The velocity of the particle in the earth’s environment.

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$

Ratio is given as below:

$K.EfK.Ei=12mpvf212mpvi2=vf2vi2$

$K.EfK.Ei=vf2vi2=G( M m R m + M E R E ) 2G M m R m K.EfK.Ei=( M m R m + M E R E )×Rm2Mm$

$K.EfK.Ei=(1+MERERmMm)×12$

Calculation:

Putting the values of radius and mass of earth and moon, we get,

$K.EfK.Ei=(1+( 5.972× 10 24 6371 × 1737 3.342× 10 22 ))×12K.EfK.Ei=24.86$

Conclusion:

The ratio of final to initial kinetic energy of the particle is $24.86$ .

Answer 6

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

Answer 7

Chapter 7, Problem 63P

(a)

To determine

The escape velocity for the moon.

Answer 8

(a)

Expert Solution

Answer to Problem 63P

The escape velocity of Moon is 2.38 km/s.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Formula Used:

The escape velocity of a particle to escape the gravitational field of the Moon is given as

$v=2GMR$

Where, G is the gravitational constant of Moon is $6.67×10−11 N⋅m2⋅kg−2$ , 1737.1 km is the radius of Moon, mass of Moon is $7.342×1022kg$

Calculation:

Substitute the values:

$v= 2GMR= 2×6.67× 10 −11 ×7.342× 10 22 1737.1× 10 3 =2.38 km/s$

Conclusion:

The escape velocity of Moon is 2.38 km/s.

Answer 13

(b)

To determine

The velocity of the particle in the earth’s environment.

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

Answer 14

(b)

To determine

The velocity of the particle in the earth’s environment.

Answer 15

(b)

Expert Solution

Answer to Problem 63P

The velocity withwhich particle will enter the Earth’s surface is $1.85×106m/sec$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Data:

The energy conservation is given as

$Initial Total Energy==Final Total EnergyP.Ei+K.Ei=P.Ef+K.Ef (1)$

Particle at moon starts with a $K.Ei$ with an escape velocity $vm= 2G M m R m$ ,where G is the gravitational constant whose value is $6.67×10−11 Nm2kg-2$ , $Mm$ is the mass of the moon and $Rm$ is the radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. Therefore, $P.Ei=0$ . As the particles reaches the Earth’s surface, the particle and earth potential energy is final and given as $P.Ef=−GMEmpRE$ , $ME$ and $RE$ is the mass and radius of the Earth respectively and $mp$ is the mass of the particle.

Formula Used:

Using equation 1, write

$K.Ei=P.Ef+K.Ef12mpvi2=−GMEmpRE+12mpvf2$

Putting the initial velocity equal to the escape velocity from moon, we can write:

$12mp( 2G M m R m )2=−GMEmpRE+12mpvf2$

$12mp2GMmRm=−GMEmpRE+12mpvf2$

Solving for final velocity $vf$

$vf=G( M m R m + M E R E )$

Calculation:

$vf=6.67×10−11( 7.342× 10 22 1737.1× 10 3 + 5.972× 10 22 6371× 10 3 )$

$vf=1.85×103m/s$

Conclusion:

The velocity withwhich particle will enter the Earth’s surface is $1.85×103m/s$

Answer 20

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$

Ratio is given as below:

$K.EfK.Ei=12mpvf212mpvi2=vf2vi2$

$K.EfK.Ei=vf2vi2=G( M m R m + M E R E ) 2G M m R m K.EfK.Ei=( M m R m + M E R E )×Rm2Mm$

$K.EfK.Ei=(1+MERERmMm)×12$

Calculation:

Putting the values of radius and mass of earth and moon, we get,

$K.EfK.Ei=(1+( 5.972× 10 24 6371 × 1737 3.342× 10 22 ))×12K.EfK.Ei=24.86$

Conclusion:

The ratio of final to initial kinetic energy of the particle is $24.86$ .

Answer 21

(c)

To determine

The ratio of final to initial kinetic energy of the particle.

Answer 22

(c)

Expert Solution

Answer to Problem 63P

The ratio of final to initial kinetic energy of the particle is 24.86.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given data:

Particle at moon starts with a $K.Ei$ with a escape velocity $vm= 2G M m R m$ ,where G is thegravitational constant whose value is $6.67×10−11 Nm2kg-2$ . $Mm$ is the mass of the moon and $Rm$ radius of the moon, $vm$ is the escape velocity from the moon which is the initial velocity of the particle.

Formula Used:

When particle is at the surface of moon, then the potential energy of particle and Earth is neglected as it is far from the Earth. The velocity with which particle will enter the Earth’s surface is $vf=G( M m R m + M E R E )$