(a)
The escape velocity for the moon.
(a)
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
Where, G is the gravitational constant of Moon is
Calculation:
Substitute the values:
Conclusion:
The escape velocity of Moon is 2.38 km/s.
(b)
The velocity of the particle in the earth’s environment.
(b)
Answer to Problem 63P
The velocity withwhich particle will enter the Earth’s surface is
Explanation of Solution
Given Data:
The energy conservation is given as
Particle at moon starts with a
Formula Used:
Using equation 1, write
Putting the initial velocity equal to the escape velocity from moon, we can write:
Solving for final velocity
Calculation:
Conclusion:
The velocity withwhich particle will enter the Earth’s surface is
(c)
The ratio of final to initial kinetic energy of the particle.
(c)
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
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
Ratio is given as below:
Calculation:
Putting the values of radius and mass of earth and moon, we get,
Conclusion:
The ratio of final to initial kinetic energy of the particle is
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Chapter 7 Solutions
Physics Fundamentals
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