# Broken Spaceship Accelerating towards circular sheet of dust

1. Oct 2, 2016

### Dusty912

1. The problem statement, all variables and given/known data
A broken spaceship is located 10 km above the center of a large circular thin sheet of unknown dust. The sheet has a radius of 106km and a density of 7×1011kg/m2. The spaceship and the dust attract each other due to the gravitational force. a) Find the initial acceleration of the spaceship. b) Find the acceleration of the spaceship right before the spaceship hits the dust. c) What will the astronauts in the spaceship experience right before the spaceship hits the dust?

2. Relevant equations
F=(GMm)/r2

3. The attempt at a solution
I realized that the distance from the spaceship to any point on the sheet (except the center) would constantly be changing as the spaceship moves close and closer. This prompted me to set up an integral of the form F=∫(Gm*(2Mr/R2)*h)/(r2+h2)3/2 dr with integrating factors from h2 to( R2 + h2) using U-sub and moving the constants out I obtained the answer ((1/2)*GMm)/R2)(1-h)/(r2+h2)

then use F=ma and solving for a a=F/m I obtained a=(((1/2)GM/r2)(1-(h/sqrt(r2+h2)

this seems right, but does not make any sense for the acceleration right before the spaceship hits the cloud. I would assume this value would be zero.

2. Oct 2, 2016

### Simon Bridge

Why would the distance to the centre of the sheet not change as the spacecraft gets closer?
$$F=\int_{h^2}^{R^2+h^2} Gm\frac{2Mr}{R^2}h\left(\frac{1}{(r^2+h^2)^{3/2}}\right)\; dr$$ ... but you neglected to define your variables or explain your reasoning, so I have no way to evaluate your approach.

Because.........??? Perhaps because you would expect that, with the spacecraft dead center in a uniform cloud the force would be equal in all directions ... thus zero net force and so no acceleration? Is that how you are reasoning?

3. Oct 2, 2016

### Dusty912

Oh, yea I guess that distance would be changing.

Let me elaborate. But is this integral correct?

Consider the circular sheet (total mass M and radius R) as being composed of infinitesimally thin rings. One such ring, between radius r and r+dr, has an infinitesimal mass given by

dm = M/(pi R^2) * 2 pi r dr = 2 M r/R^2 dr

The distance d from this ring to the spaceship (of mass m) at height h from the center of the ring is given by Pythagoras' theorem:

d^2 = r^2 + h ^2

So this infinitesimal ring will exert an infinitesimal force

dF = G m (2M/R^2) r dr / (r^2+h^2)

But by symmetry, all the components of force perpendicular to the axis will cancel, so the component of force along the axes survives. This brings an additional geometrical factor h/sqrt(r^2+h^2). Hence, the force from the ring along the axes is

dF// = G m (2M/R^2) r h dr / (r^2+h^2)^(3/2)

The total force on the spaceship at height h then follows by adding (integrating) all contributions from the concentric infinitesimally thin rings:

4. Oct 2, 2016

### Dusty912

and yes that is my reasoning for why there would be no acceleration immediatly before it hits the cloud

5. Oct 3, 2016

### Simon Bridge

The problem statement gives you the surface mass density $\sigma$ of the cloud already so there is no need to calculate it. Thus $dm = 2\pi \sigma r \;dr$, well done. You do not need M.

Yep.

This statement is not correct ... however you go on:

... which is looking good.

Just a note on notation - you cannot use $m$ for the mass of the ship and $dm$ for the mass of a ring because, by definition $m=\int\;dm$ so you are using the same variable for two things.
... if you want the total mass of the ring to be M, then an element of that mass would be dM.
Your first step should have been: $dF = [Gm/(d^2)][h/d]\;dM$ then you can sub for d and dM and integrate over r below...

... looks good to me. Since the integral is over r, what are the limits of r?

Aside:
* "integrating factor" is a technical term in mathematics - the a and b in $\int_a^b$ are more usually called "the limits of the integration".
* "cylindrical polar coordinates" would have helped you here.
* suggest you learn LaTeX ... it will make your equations easier to read (and write).
See: https://www.physicsforums.com/help/latexhelp/

6. Oct 3, 2016

### Dusty912

are you saying that my work looks fine? or do I need to change something?

and yes I should have used latex. my apologies

7. Oct 3, 2016

### Dusty912

and yes, I know it's limits of integration. I'm leaning about integrating factors in my differential equation class and mixed up the terms

8. Oct 3, 2016

### Simon Bridge

Doing good so far ... you need to make some changes to your notation if this is to be presented as a long answer.
Next step is to justify the limits of the integration.