# Launching object into space (simple)

1. Oct 7, 2009

### Nal101

1. The problem statement, all variables and given/known data
A package of mass m sits at the equator of an airless asteroid of mass M, radius R, and spin angular speed. We want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed v. We have a powerful spring whose stiffness is ks. How much must we compress the spring? (Use m for m, k_s for ks, v for v, w for , R for R, M for M, and R for R as necessary.)

2. Relevant equations
-refer to 3.

3. The attempt at a solution
I looked through the section in my book, but couldn't find correct equations. I assumed that you could solve for potential energy required to launch the package out of orbit then set it equal to 1/2 * k_s * s^2 and solve for s. The real problem here, I do not know how to calculate the required force/energy to launch the package so it will be at a specific speed where gravity will no longer be affecting it.

My first idea (doesn't seem correct when looking at it though) combine the following equations..
v=wd
w=sqrt(k_s / m)
U=1/2*mv^2
d=2$$\pi$$r
to make..
U = 1/2*m*w^2*(2$$\pi[tex]r)^2 The problem I had with it was that I did not see any correlation to the forces of gravity and centripetal, nor its initial/final velocities. 2. Oct 7, 2009 ### RoyalCat You must consider two things: By "never come back" I assume they mean that the orbit of the package is unbound (What condition must you impose on the total energy of the package for this to occur?) Though it may seem that you're launching the package straight up, it does in fact have a tangential velocity. Consider this when you're calculating the total energy of the package. Also, [tex]v_i=\omega R$$
What you said: $$v_i=\omega 2\pi R$$ is wrong.
$$\omega=\sqrt{\frac{k}{m}}$$ is also very wrong.
You're used to seeing $$\omega$$ in the context of a harmonic oscillator, where it means angular frequency. Here $$\omega$$ is the angular velocity of asteroid about its axis. Consider the 2D image where you see the equatorial cross-section of the asteroid. For the sake of clarification, imagine a disk of radius $$R$$ rotating at an angular velocity $$\omega$$. What tangential velocity will a mass on its rim have?

$$U=\tfrac{1}{2}mv^2$$ is also wrong. The kinetic energy of the package is not part of its potential energy.

For this problem, you must analyze the three kinds of energy you have prior to the launch and compare them with some situation at infinity where you wish to have a desired end result.

Assumptions:
1. The package is on the equator of the asteroid.
2. The package does not slip/slide as it is launched. It retains its original tangential velocity.

Last edited: Oct 7, 2009
3. Oct 7, 2009

### Nal101

I had assumed that the package would be shot out in the direction of its current velocity.

Knowing the tangential velocity, how do I calculate the (kinetic?) energy?
What are the other 2 energies, potential and something bringing in gravity? or is gravity included in the potential?

4. Oct 7, 2009

### RoyalCat

They didn't state the direction in which the spring launches the mass because that is irrelevant. So long as this direction isn't straight into the ground, you're good to go.

The three energies you need to consider are the package's original kinetic energy, the potential elastic energy and the potential gravitational energy.

What is the condition on this total energy for the package to have an unbound orbit?

5. Oct 7, 2009

### Nal101

Would the condition be that the Kinetic energy is larger than potential? So the energy of the system is greater than 0?

Also..
Potential gravitational is.. gravitational constant * mass / radius
Potential elastic would be.. .5 * k * s^2
And original kinetic is.. .5 * m * v^2
whereas v_i = $$\omega$$ R
and v_f is set to the specific indicated speed at which the package needs to travel (do not know equation to incorporate)
Correct?

Last edited: Oct 7, 2009