Projectile fired from earth. Conservation of energy

[auk411]

Homework Statement

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.
What multiple of Earth's radius R_e gives the radial distance a projectile reaches if its (a) initial speed is .5 the escape speed from earth, (b) its initial kinetic energy is .5 of the kinetic energy required to escape Earth? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Homework Equations

v =$$.5\sqrt{2GM/R}$$

The Attempt at a Solution

Total mechanical energy = 1/2 mv^2 - GMm/R =?

I know that I'm going to need to plug in the escape velocity for v^2. However, that is all I know.
(As for b and c, I don't even know what the relevant equations are. )

 

[Bhumble]

I'm not sure why it states mechanical energy in part of the problem but I'm assuming this is kinetic energy.
$$K = \frac{1}{2} mv^2$$
$$v = \sqrt{\frac{2GM}{R}}$$

a) $$v = 0.5\sqrt{\frac{2GM}{R}}$$ Solve for R
b) $$\frac{1}{2}mv^2 = \frac{1}{2}\frac{GMm}{R}$$ Solve for R
c) Start with your given formula for v (without the 0.5 that you added) and get to $$\frac{1}{2}mv^2$$ on the LHS

 

[auk411]

So does anyone know how to answer this cause the person before doesn't know how to answer it.

For one thing, part a is just wrong. That is not true equation.

... Unless I'm just totally confused.

 

[Bhumble]

You're correct that part a) is wrong. I should have had the 0.5 on the RHS of the equation. And the same for part b). I'll edit my original post to reflect this.

 

[rwisz]

I would respond by providing a detailed explanation and solution to the problem. Here is my response:

First, we need to consider the conservation of energy in this scenario. The total mechanical energy of the projectile at any point during its trajectory is equal to the sum of its kinetic and potential energies. We can express this as:

Total mechanical energy = Kinetic energy + Potential energy

Now, let's look at the three parts of the question separately:

(a) If the initial speed of the projectile is 0.5 times the escape speed from Earth, we can calculate the radial distance it will reach by using the equation for escape velocity:

v = √(2GM/R)

Where v is the escape velocity, G is the gravitational constant, M is the mass of Earth, and R is the radius of Earth.

Substituting 0.5v for v in the equation, we get:

0.5v = √(2GM/R)

Squaring both sides, we get:

0.25v^2 = 2GM/R

Rearranging the equation, we get:

R = 8GM/0.25v^2

Now, we know that the radius of Earth is R_e, so we can express the radial distance reached by the projectile as a multiple of Earth's radius:

R/R_e = 8GM/0.25v^2R_e

Simplifying the equation, we get:

R/R_e = 32v^2/0.5v^2R_e

Therefore, the radial distance reached by the projectile is 32 times Earth's radius.

(b) For the second part of the question, we need to calculate the initial kinetic energy of the projectile and compare it to the kinetic energy required for escape. The kinetic energy of the projectile is given by:

Kinetic energy = 1/2 mv^2

Substituting 0.5v for v, we get:

Kinetic energy = 1/2 m(0.5v)^2 = 0.125mv^2

Now, the kinetic energy required for escape is equal to the total mechanical energy at escape, which we can calculate using the formula we derived earlier:

Total mechanical energy = 1/2 mv^2 - GMm/R

Substituting 0.5v for v, we get:

Total mechanical energy = 1/2 m(0.