Center of Infinite Mass?

Classical Mechanics Level 5

Consider a planet with radius $r_{0}$ located somewhere far away in the universe. Interestingly, it is a perfect sphere of uniform density. There is an infinitely tall cylinder on the planet. It is filled with a mixture of hydrogen and helium such that the molar mass of the mixture is $M$ . The pressure , acceleration due to gravity and the temperature at the bottom of the cylinder, i.e., at the surface of the planet are $P_{0}$ , $g_{0}$ and $T_{0}$ respectively.
Also the temperature of the mixture in the cylinder varies according the observed formula given by,
$T = \frac{T_{0}}{(1 + \frac{h}{r_{0}})}$
where h is the height from the surface of the planet i.e., the base of the cylinder.

Find the value of $\frac{dg}{dT}$ at the center of mass of the cylinder.

Details and Assumptions:

The cylinder is only affected by the planet's gravitational field and not by field of any other celestial body.

The area of cross section of the cylinder is too small to be affected by any gravitational aspect.

$r_{0} = \frac{4}{3} \times 10^3 km$

$M = 2.5 g/mol$

$T_{0} = 1000 K$

$g_{0} = 10 m/s^2$

$R = \frac{25}{3} J/K-mol$ , where $R$ is the universal gas constant.

The answer is 0.01.

2 solutions

Jatin Yadav
Feb 25, 2014

$T = \frac{T_{0} r_{0}}{z}$ , where $z$ is the distance from center to an element of thickness $dz$ .

We know that :

$dP = -\rho g dz$ , where $g = \frac{g_{0} r_{0}^2}{z^2}$

$PM = \rho RT$

Divide them to get :

$\displaystyle \frac{dP}{P} =-\frac{Mg}{RT} dz$

$\Rightarrow \displaystyle \frac{dP}{P} = -\frac{Mg_{0}r_{0}}{T_{0} R} \frac{dz}{z}$

$\Rightarrow \displaystyle \frac{dP}{P} = -4 \frac{dz}{z}$

$\Rightarrow \displaystyle P = P_{0} \bigg(\frac{z}{r_{0}} \bigg)^{-4}$

Now, $\displaystyle \rho(z) = \frac{PM}{RT} = \frac{P_{0} M}{RT_{0}} \bigg(\frac{z}{r_{0}}\bigg)^{-3}$

Now, $\displaystyle z_{cm} = \frac{ \int_{r_{0}}^{\infty} z \rho(z) dz}{\int_{r_{0}}^{\infty} \rho(z) dz}$

= $2 r_{0}$ (after very easy calculations).

Now $\displaystyle \frac{dg}{dT} = \frac{\frac{dg}{dz}}{\frac{dT}{dz}} = \frac{g_{0}{r_{0}}^2 (-2/z^3)}{T_{0} r_{0} (-1/z^2)} = \frac{2 g_{0} r_{0}}{T_{0}}$

At $z = z_{cm} = 2r_{0} , \frac{dg}{dT} = \frac{g_{0}}{T_{0}} = 0.01 m/s^2/K$

I have done in the same way.

jinay patel - 7 years, 2 months ago

So the trick was to put the value of the variables there in the working itself. I attempted to solve the entire thing in variables and got stuck in the integral. Great solution!!

Sanat Anand - 6 years, 11 months ago

John Samuel
Mar 29, 2014

I did the same way as well

Center of Infinite Mass?

The answer is 0.01.

2 solutions

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