Determining density of spherical cloud by dropping probe and measuring velocity

  • Thread starter pirland
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  • #1
pirland
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This is a problem that has been giving me trouble, wondering if I could get some input:

A space probe of mass m is dropped into a previously unexplored
spherical cloud of gas and dust, and accelerates toward
the center of the cloud under the influence of the cloud’s gravity.
Measurements of its velocity allow its potential energy, [tex]U[/tex], to be
determined as a function of the distance [tex]r[/tex] from the cloud’s center.
The mass in the cloud is distributed in a spherically symmetric way,
so its density, [tex]\rho(r)[/tex], depends only on [tex]r[/tex] and not on the angular coordinates.
Show that by finding [tex]U(r)[/tex], one can infer [tex]\rho(r)[/tex] as follows:

[tex]\rho(r)=\frac{1}{4 \pi Gmr^2} \frac{d}{dr}(r^2 \frac{dU}{dr}) [/tex]

The first problem I am running into is finding the Equation for U. I assume it as a lot to do with the equation for work=[tex]GMm(\frac{1}{r_{2}}-\frac{1}{r_{1}}) [/tex] but I am getting tripped up over the value of M, or the mass of the cloud itself. I believe that this would change as r decreased, and less and less of the cloud would be exerting force on the probe. I assume I need to find a value for M in terms of r, but I am seemingly unable to come up with anything that works with later parts of the problem. If anyone could give me a push in the right direction it would be appreciated.
 

Answers and Replies

  • #2
HallsofIvy
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Assuming that the cloud is spherical and that density is a function of r only, yes, only the portion of the cloud closer to the center than the center than the probe has a net force on the probe (because any "chunk" of cloud outside the probe has an offsetting "chunk" diametrically opposite). You can find the force function, as well as the potential function, by integrating the from the center of the cloud to r.
 
  • #3
pirland
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Ok, so this is what I currently have for U:
If I solve Newtons law of gravitation for [tex]M[/tex] I get [tex] M=\frac {Fr^2}{Gm}[/tex]. I then substitute this value for [tex]M[/tex] in the equation [tex]Work=GMm(\frac{1}{r_{2}}-\frac{1}{r_{1}})[/tex] I get [tex]Work=F(r_{2}-r_{1})[/tex]
Is this a valid equation for work in the situation described in my first post? It seems almost to simple, as it is basically the same thing as [tex]W=Fd[/tex], which I thought was only really applicable for situations involving constant force.
 
  • #4
lightgrav
Homework Helper
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The clue is that, in your "Work Equation",
there is only ONE (1) M, but two (2) r's.
Obviously that equation is only valid if
the same amount of mass is within r2 as r1.

Do you recognize 4 pi r^2 rho(r) dr ?
in your case, dM is not zero, it's the key!

That is, yours is not an algebra question in disguise,
it is an honest-to-goodness calculus question.
 
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