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Spherical distribution with homogeneous density

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A galaxy can be described as a sphere with homogeneous distributed density. The mass at any point of the space is $M(r)$, that is described as:

$M(r)$= $kr^3$ when $r\le R$
$M(r)$= $k'r$ when $r > R$

a) Find the velocity of a body in a circular orbit around a more massive one.
b) Find the circular velocity of a particle around the galaxy at a distance $r$ from the center at any point.
c) Relate $k$ and $k'$
d) Draw the fuction of the circular velocity in function of the distance to the center for all $r$.

These are my solutions.

a) The centripetal force on a circular orbital path equals to the gravitational force. Therefore:

$$ \frac{mv^2}{r} = \frac{GMm}{r^2}$$

$$ v=\sqrt{ \frac{GM}{r}} $$

Which is the circular velocity of the smaller body around the more massive one.

b) As the galaxy has a spherical distribution with homogeneous density, the sphere (galaxy) can be regarded as a point mass (point particle) placed at the origin of coordinates (galaxy's center).
We have two different cases:

Case 1: when $r\le R$, we have $M(r)$= $kr^3$. Then:

$$ \frac{mv^2}{r} = \frac{Gkr^3m}{r^2}$$

$$ v=\sqrt{Gkr^2} $$

Case 2: when $r > R$, we have $M(r)$= $k'r$. Then:

$$ v=\sqrt{Gk'} $$

c) We can relate both $k$ and $k'$ at the limit case $r=R$. Therefore:

$$k'=R^2k$$

$$or$$

$$k'=r^2k$$

d) Plugging $k'=R^2k$ into $ v=\sqrt{Gkr^2} $ and $ v=\sqrt{Gk'} $,
we obtain $ v=r\sqrt{Gk} $ and $ v=R\sqrt{Gk} $ respectively.

I represented it graphically as follows:

At c) I had doubts if both $$k'=R^2k$$ $$and$$ $$k'=r^2k$$ were right. I gave a thought to the matter, and I concluded $$k'=R^2k$$ is more suitable as R is constant and r is not.

At d), I did the representation in function of $\sqrt{Gk}$, instead of $\sqrt{Gk'}$. I guess it does not matter.

What do you think about these results?

Thanks

asked Apr 30, 2018 in Physics Problems by Jorge Daniel (606 points)
Your answers look ok to me.
These are my worked solutions. This problem is not coming from a book or anything like that (from an unsolved exam). I just want to check if my answers are 100% right. About a) and b) I don’t have doubts, just wrote it down so as to checking with you. About c) and d) my doubts are exposed in the question
Questions asking for someone to check your work (to make sure it is 100% right) are not very interesting. If the result contradicts something else you know, or you've been told you've got the wrong answer but you don't understand why, then there is something to explain, a lesson to be learnt. Otherwise, the answer is just "yes that is correct" and there is no lesson for anyone. Unless you see a difficulty, you need to have confidence in your own abilities.
I agree with you, since now I will just ask what I am unsure about. Not more checking :)

1 Answer

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The definition of $M(r)$ is quite confusing. It is not the mass at a point, because there are an infinite number of points so there is an infinite amount of mass. It could be the density at a radius $r$. But the simplest interpretation is the one you used - that $M(r)$ is the mass inside a sphere of radius $r$.

(c) The function $M(r)$ must be continuous at the boundary. The mass inside the sphere of radius $r$ cannot suddenly increase as you cross a boundary. But it is not necessarily a smoothly varying function.

$M(R_{-}) = M(R_{+})$

$k'r=kr^3$ when $r=R$ but not necessarily at any other radius.

(d) The question only asks you to relate $k$ and $k'$, not to express $k'$ in terms of $k$. So you could use either $k$ or $k'$ in your answer.

answered May 1, 2018 by sammy gerbil (28,448 points)
selected May 5, 2018 by Jorge Daniel
Then the statement 'The mass at any point of the space is $ M(r) $' would be technically wrong?  It is written down like that in the test.
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