Calculating Gravity of a New Planet with a Gas Atmosphere - Homework Solution"

[cmcd]

Homework Statement

A space-probe discovers a planet with a gas atmosphere, detects a non-zero value for speed of sound 60km relative to the surface of the planet. It measures a surface temperature of -5 degrees C and an average atmospheric molecular mass of 38 x 10 ^ -27kg.

Find G_p the gravity of the planet.

Homework Equations

barometric height formula:
rho(h) = rho(0) . e ^ - (rho(0) g h/ p(0))
where rho = density, p = pressure, g gravity, h height above surface...
index 0 = planet conditions

Taylor series/ maclaurin series for e^x = sum of (x^n)/n! from n=0 to n= infinity.

Kb value given.

The Attempt at a Solution

-

Any help appreciate it

 

[BvU]

Still need an equation for the speed of sound (which i hope is 60 km/h, not 60 km ? )...

 

[cmcd]

The question doesn't give one, just said non zero value at 60km

 

[BvU]

Sorry, I misread completely. So there is atmosphere at 60 km height. Probably that's an upper limit where $\rho(h)\approx 0$ ?
Still think an equation more is needed. What about the average molecule mass ?

 

[cmcd]

Maybe something to do with Kb. I've tried using pv=N.K_b . T. I think it has something to do with the fact that the taylor/maclaurin series for e^x is given.

 

[cmcd]

And subbed V(h) = (m_ave ) / rho(h)

 

[BvU]

the taylor/maclaurin series for e^x is given.

Oh, I found it under relevant equations. What is the complete problem statement ?
And the values, variables, given/known data etc. Which of the relevant eqns did you contribute ?At first I thought the K was $K_{bulk}$ for the speed of sound equation $c = \sqrt{\frac{K}{\rho}}$ :( (that's is what I get for accepting an incomplete problem statement). Boltzmann constant is usually written $k_B$, so I read something else. Could have been prevented if I'd asked for the given value first, so my mistake. Goes to show that a complete problem statement, all variables, given/known data really speeds up the assistance process, thereby allowing others some time too...

Anyway, it seems we probably need the ideal gas law listed under 2. relevant equations.

And as icing on the cake, PF rules require something under 3. too. So show some more...

 

[cmcd]

Yeah, that's all the question gave though. Should have said Boltzmann constant soz.
If you can't help don't sweat

 

[berkeman]

Homework Statement

A space-probe discovers a planet with a gas atmosphere, detects a non-zero value for speed of sound 60km relative to the surface of the planet. It measures a surface temperature of -5 degrees C and an average atmospheric molecular mass of 38 x 10 ^ -27kg.

Find G_p the gravity of the planet.

Homework Equations

barometric height formula:
rho(h) = rho(0) . e ^ - (rho(0) g h/ p(0))
where rho = density, p = pressure, g gravity, h height above surface...
index 0 = planet conditions

Taylor series/ maclaurin series for e^x = sum of (x^n)/n! from n=0 to n= infinity.

Kb value given.

The Attempt at a Solution

-

Any help appreciate it

Yeah, that's all the question gave though. Should have said Boltzmann constant soz.
If you can't help don't sweat

Please check your private messages. We require that you show your Attempt at a Solution when posting schoolwork questions here. Please post your detailed work so far, based on the hints you have been given in this thread. We do not do your schoolwork problems for you. We are happy to help it you show your detailed calculations and efforts.

 

[cmcd]

The full question is:

A space-probe from Earth has been sent out of Earth's solar system and eventually reahes a neighbouring solar system. The space - probe has discovered a planet with a gas atmosphere in this solar system and approches for landing. During the landing process the space probe measures the velocity od sound. The space -probe found that from $$h_s = 60km$$ relative to the planetary surface it is able to detect a non - zero value for the velocity of sound. On the surface of the planet the space-probe measures a temperature of $$T=-5\text{ degrees celsius}$$ and an average molecular mass $$m_a$$ of about $$38*10^{-27}kg$$ for a gas-molecule of the atmosphere. What is the gravitational acceleration $$g_p$$ of the planet?

Hint : The barometric height - formula is given by
$\rho(h)=\rho_0\exp^-\frac{\rho_0g_ph}{p_0}$

where $$\rho$$ is the density, $$p$$ is the pressure, $$h$$ is the height, $$g$$ is the gravitational acceleration, and the index 0 denotes the conditions on the planetary surface.

The exponential function can always be written as a sum:

$\exp^x = \sum\limits_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{1}{2}x^2 + ...$

(Boltzmann - constant $$k_B = 1.38 x 10^{-27} J/K$$)

Relevant Equations:
$\rho(h)=\rho_0\exp^{-\frac{\rho_0g_ph}{p_0}}$

Ideal Gas Law using $$N=1$$ and $$K_B$$ :

$$p(h)V(h) = NK_BT$$

I guess the expansion is relevant too but I don't know how to use it!

Attempt at solution:

$$V(h) = \frac {M_a}{\rho(h)}$$

surface temp:

$$T=-5\text{ degrees celsius}$$

$$p_0\frac {M_a}{\rho_0} = K_B(273.15 -5)$$

$$p_0 = \frac{K_B(268.15)}{\frac {M_a}{\rho_0}}$$

$\rho(h)=\rho_0\exp^{-\frac{\rho_0g_ph}{\frac{K_B(268.15)}{\frac {M_a}{\rho_0}}}}$

$$\rho(h)=\rho_0\exp^{-\frac{m_a*g_p*h}{268.15K_B}}$$

Here is where I think I need to use the taylor series for e^x and/or a suitable assumption but I'm confused. Thanks, Colin

 

[haruspex]

I don't understand how one is supposed to use the 60km information. Seems like it should imply a particular threshold of density. Is there anything in your course notes that sheds light on that?

 

[cmcd]

Thanks for the quick reply;
I don't think we did get notes for it, it's a question from a past exam
taken to get a scholarship so they're a little tougher than the course
content.
The expansion for $$exp^x$$ is the bit that confuses me the most.
Thanks though

 

[haruspex]

Thanks for the quick reply;
I don't think we did get notes for it, it's a question from a past exam
taken to get a scholarship so they're a little tougher than the course
content.
The expansion for $$exp^x$$ is the bit that confuses me the most.
Thanks though

I don't know whether you're confused about that because you don't know how to do it or because you don't know how it's useful.
Doesn't seem to me that it is particularly useful. What you need to do with your equation is plug in values for h and $\rho(h)$ and solve for g. This is where the 60km info comes in, but what I don't get is what you use for $\rho(60km)$.
All the expansion of exp allows you to do is solve for g without taking logarithms. Probably just the first two terms of the expansion are good enough: exp(x) ~ 1 + x.