A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.
(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is one-fifth of the escape speed from Earth?
____ times RE
(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is one-fourth of the kinetic energy required to escape Earth?
_____ timesRE
(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
P.S.Bobpursley, sorry,I don't really understand the last question you answered....
To answer these questions, we need to consider the concepts of escape speed, kinetic energy, and mechanical energy.
(a) To find the radial distance a projectile reaches if its initial speed is one-fifth of the escape speed, we can start by understanding what escape speed is. Escape speed is the minimum speed an object needs to move away from a planet's gravitational pull and not fall back.
The escape speed from Earth can be calculated using the formula:
Escape Speed = √(2 * gravitational constant * Earth's mass / Earth's radius)
Let's assume that the escape speed is v_esc.
So, the initial speed of the projectile is (1/5) * v_esc.
Now, to find the radial distance the projectile reaches, we can use the concept of potential energy.
The potential energy of an object in the Earth's gravitational field can be calculated using the formula:
Potential Energy = (gravitational constant * Earth's mass * object's mass) / radial distance.
The potential energy will be equal to zero when the projectile reaches its maximum distance.
For this projectile, the potential energy at the maximum distance is given by:
Potential Energy = 0
Using the above formula, we can equate the initial kinetic energy to the potential energy at the maximum distance:
(1/2) * (object's mass) * (initial velocity)^2 = 0
Simplifying the equation, we get:
(initial velocity)^2 = 0
So, the radial distance the projectile reaches is 0 times the Earth's radius, i.e., zero.
Therefore, the answer to part (a) is 0 times RE.
(b) To find the radial distance a projectile reaches if its initial kinetic energy is one-fourth of the kinetic energy required to escape Earth, we need to associate kinetic energy with the initial speed of the projectile.
The kinetic energy of an object can be calculated using the formula:
Kinetic Energy = (1/2) * (object's mass) * (velocity)^2
Let's assume the initial kinetic energy of the projectile is K.
The kinetic energy required to escape Earth can be calculated by substituting the escape speed into the kinetic energy formula.
So, the kinetic energy required to escape Earth is:
(1/2) * (object's mass) * (escape speed)^2
We are given that the initial kinetic energy is one-fourth of the kinetic energy required to escape Earth.
So, K = (1/4) * (1/2) * (object's mass) * (escape speed)^2
Now, we can equate the initial kinetic energy to the potential energy at the maximum distance to find the radial distance the projectile reaches:
(1/2) * (object's mass) * (initial velocity)^2 = (gravitational constant * Earth's mass * object's mass) / radial distance
Simplifying the equation, we get:
(initial velocity)^2 = (2 * gravitational constant * Earth's mass) / radial distance
Substituting the initial velocity of the projectile, we get:
[(1/5) * escape speed]^2 = (2 * gravitational constant * Earth's mass) / radial distance
Simplifying, we get:
(radial distance) = (2 * gravitational constant * Earth's mass) / [(1/5)^2 * escape speed^2]
Therefore, the radial distance the projectile reaches is (2 * gravitational constant * Earth's mass) / [(1/5)^2 * escape speed^2] times the Earth's radius, RE.
(c) To find the least initial mechanical energy required at launch for the projectile to escape Earth, we need to consider the total mechanical energy of the projectile.
The mechanical energy of a projectile in the Earth's gravitational field is the sum of its kinetic energy and potential energy.
For the projectile to escape Earth, its total mechanical energy should be equal to zero (no net mechanical energy).
So, the initial mechanical energy required can be calculated by equating the initial kinetic energy to the potential energy at the maximum distance:
(1/2) * (object's mass) * (initial velocity)^2 = (gravitational constant * Earth's mass * object's mass) / radial distance
Since the initial kinetic energy is (1/2) * (object's mass) * (initial velocity)^2, and the radial distance is the maximum distance the projectile reaches before coming back, we can set them equal to zero:
(1/2) * (object's mass) * (initial velocity)^2 - (gravitational constant * Earth's mass * object's mass) / radial distance = 0
Simplifying the equation, we get:
(initial velocity)^2 = (2 * gravitational constant * Earth's mass) / radial distance
Substituting the initial velocity of the projectile, we get:
[(1/5) * escape speed]^2 = (2 * gravitational constant * Earth's mass) / radial distance
Simplifying, we get:
(radial distance) = (2 * gravitational constant * Earth's mass) / [(1/5)^2 * escape speed^2]
Therefore, the least initial mechanical energy required at launch for the projectile to escape Earth is given by (2 * gravitational constant * Earth's mass) / [(1/5)^2 * escape speed^2] times the Earth's radius, RE.