Author Topic: Harder Physics Questions (Read 31196 times) Tweet Share

QuantumJG · « **Reply #15 on:** January 31, 2010, 10:08:01 pm »

Quote from: Cthulhu on January 31, 2010, 07:45:19 pm

Suppose that the radius of the Sun were increased to $5.90 * 10^{12}$ (the average radius of orbit for pluto), that the density of the expanded sun was uniform and that the planets revolved within it. Calculate earths orbital speed within the 'new' sun and what would be the earths new period of revolution.

To solve this I am going to first derive an expression for g

$\oint_{S} \bf{g} . d \bf{A} = - 4 \pi G M$

since r_Earth < r_Sun, then let:

M_Sun = $\dfrac{4}{3} \pi \rho_{Sun} r^{3}$

$\therefore \oint_{S} \bf{g} . d \bf{A} = - 4 \pi G \dfrac{4}{3} \pi \rho_{Sun} r^{3}$

$\therefore - \bf{g} . 4 \pi r^{2} = - 4 \pi G \dfrac{4}{3} \pi \rho_{Sun} r^{3}$

$\therefore \bf{g} = \dfrac{4}{3} G \pi \rho_{Sun} r$

I love using Guass's law on spheres!

Edit: Now, r = 1.5x10⁹m, $\rho_{Sun}$ = 1.41x10³ kg/m³

implying g = 591 m/s²

Now,

v_Earth = 941 km/s (31x faster than what speed it orbits at)

$g = \dfrac{ 4 \pi ^{2} r}{T^{2} }$

$\therefore T = \sqrt{ \dfrac{4 \pi^{2} r}{g} }$

implying Edit: T = 1.00x10⁴ s = 2 hours 47 minutes!

So if this were to happen, in theory a year would only be about 2 hours 47 minutes!

Am I right?

appianway · « **Reply #16 on:** January 31, 2010, 10:13:40 pm »

Hmm, that's essentially what I think I described earlier, but I think you have to consider the effect of the buoyancy force (although I could be mistaken, seeing as I've only had it mentioned for around 20 seconds in one lecture), as earth essentially displaces a portion of the sun.

Cthulhu · « **Reply #17 on:** January 31, 2010, 10:38:36 pm »

Sorry quantumJG, way off

and the question is to find the orbital velocity of earth... >.>

QuantumJG · « **Reply #18 on:** January 31, 2010, 11:15:47 pm »

Quote from: appianway on January 31, 2010, 10:13:40 pm

Hmm, that's essentially what I think I described earlier, but I think you have to consider the effect of the buoyancy force (although I could be mistaken, seeing as I've only had it mentioned for around 20 seconds in one lecture), as earth essentially displaces a portion of the sun.

hmm... I never considered that!

I calculated the buoyancy force to be 9.02x10²⁶ N and it's weight to be 3.53x10²⁷ N, implying a force of 2.63x10²⁷ N is pulling the Earth to the Sun's centre.

So

v_Earth = 812km/s

T = 11,601 s = 3 hours 13 minutes

Quote from: Cthulhu on January 31, 2010, 10:38:36 pm

Sorry quantumJG, way off and the question is to find the orbital velocity of earth... >.>

What is the answer?

Can I have a hint?

Cthulhu · « **Reply #19 on:** February 01, 2010, 11:15:26 am »

The earths orbital radius still comes into play here.
I suppose I could tell you that there is of course a centripetal force acting on the planet within in the new expanded sun and that gravitational force may still play a part in all of it.

QuantumJG · « **Reply #20 on:** February 01, 2010, 01:15:47 pm »

Quote from: Cthulhu on February 01, 2010, 11:15:26 am

The earths orbital radius still comes into play here.
I suppose I could tell you that there is of course a centripetal force acting on the planet within in the new expanded sun and that gravitational force may still play a part in all of it.

I'm really confused?

QuantumJG · « **Reply #21 on:** February 01, 2010, 02:35:05 pm »

I don't understand where I have gone wrong?

Cthulhu · « **Reply #22 on:** February 01, 2010, 06:58:38 pm »

IF YOU DON'T WANT TO SEE THE ANSWER DO NOT READ THIS POST
fuck this forums lack of spoiler tags
Sorry. It's a bit rushed. It might not make sense but its the answer none the less.
You're on your own to find the period of orbit though. This is just the major part of the problem.

First lets define some variables:
$R = 5.90 \times 10^{12} m$
$r = radius\ of\ earth\ orbit = 1.50 \times 10^{11} m$
$M = mass\ of\ sun\ within\ earths\ orbit$
$M_s = original\ mass\ of\ sun = 1.988435 \times 10^{30} kg$
$V = \frac{4}{3}\pi r^{3}$
$V_{orig} = 1.412 \times 10^{27} m^3$
$V_{new} = 8.603 \times 10^{38} m^3$
Since the density of the expanded sun is uniform we can use $\rho = \frac{m}{V}$ to find the mass of the sun within earths orbit. $\rho_{in}$ is the density of the sun when the radius is the orbital radius of earth
or $r$ while $\rho_{out}$ is the density of the sun at radius $R$
$ \rho_{in} = \rho_{out} $
$\frac{3M}{4 \pi r^3} = \frac{3M_s}{4 \pi R^3}$
solving for $M$ we get $M = \left( \frac{r}{R} \right)^3 M_s$
and so:
$ M = \left(\frac{1.50 \times 10^{11} m}{5.90 \times 10^{12} m}\right) \times 1.988435 \times 10^{30} kg = 3.27 \times 10^{25}kg $
Now using the equations $F = G \frac{Mm}{r^2}$ and using $F = \frac{mv^2}{r}$ because the earth is experiencing a centripetal force. setting these equal to each other we can solve for $v$ :
$v = \sqrt \frac{GM}{r}$ and therefore
$ v = \sqrt \frac{\left(6.67 \times 10^{-11}\right)\left(3.27 \times 10^{25}\right)}{1.50 \times 10^{11}} = 1.21 \times 10^{2} m/s $

QuantumJG · « **Reply #23 on:** February 01, 2010, 07:59:30 pm »

Quote from: Cthulhu on February 01, 2010, 06:58:38 pm

IF YOU DON'T WANT TO SEE THE ANSWER DO NOT READ THIS POST
fuck this forums lack of spoiler tags
Sorry. It's a bit rushed. It might not make sense but its the answer none the less.
You're on your own to find the period of orbit though. This is just the major part of the problem.

First lets define some variables:
$R = 5.90 \times 10^{12} m$
$r = radius\ of\ earth\ orbit = 1.50 \times 10^{11} m$
$M = mass\ of\ sun\ within\ earths\ orbit$
$M_s = original\ mass\ of\ sun = 1.988435 \times 10^{30} kg$
$V = \frac{4}{3}\pi r^{3}$
$V_{orig} = 1.412 \times 10^{27} m^3$
$V_{new} = 8.603 \times 10^{38} m^3$
Since the density of the expanded sun is uniform we can use $\rho = \frac{m}{V}$ to find the mass of the sun within earths orbit. $\rho_{in}$ is the density of the sun when the radius is the orbital radius of earth
or $r$ while $\rho_{out}$ is the density of the sun at radius $R$
$ \rho_{in} = \rho_{out} $
$\frac{3M}{4 \pi r^3} = \frac{3M_s}{4 \pi R^3}$
solving for $M$ we get $M = \left( \frac{r}{R} \right)^3 M_s$
and so:
$ \bf{M = \left(\frac{1.50 \times 10^{11} m}{5.90 \times 10^{12} m}\right) \times 1.988435 \times 10^{30} kg = 3.27 \times 10^{25}kg } $
Now using the equations $F = G \frac{Mm}{r^2}$ and using $F = \frac{mv^2}{r}$ because the earth is experiencing a centripetal force. setting these equal to each other we can solve for $v$ :
$v = \sqrt \frac{GM}{r}$ and therefore
$ v = \sqrt \frac{\left(6.67 \times 10^{-11}\right)\left(3.27 \times 10^{25}\right)}{1.50 \times 10^{11}} = 1.21 \times 10^{2} m/s $

I) There is a contradiction here! You have stated that the original Sun (I.e. Our real Sun) was not being expanded (I.e. the Sun's expanding it's volume and mass).

M = 1.99x10³¹ kg

r_{Earth's orbit} = 1.5x10⁹ m

$\therefore$ v_{Earth's orbit} = 941km/s

I rest my case!

Cthulhu · « **Reply #24 on:** February 01, 2010, 08:08:13 pm »

You have no case though. Your first two answers were both different to that and the first answer didn't even answer the question I asked. I worded the question poorly and I apologise; I will re-word it for future generations.

QuantumJG · « **Reply #25 on:** February 01, 2010, 08:49:31 pm »

Quote from: Cthulhu on February 01, 2010, 08:08:13 pm

You have no case though. Your first two answers were both different to that and the first answer didn't even answer the question I asked. I worded the question poorly and I apologise; I will re-word it for future generations.

Okay please word the question so we know our data and explain specifically what you want from us.

Okay first I made some mistakes, but fixed them up. Then I investigated appianway's buoyancy idea. It's almost like you want us to fail at this question.

I'm obsessed with getting on my soapbox

Cthulhu · « **Reply #26 on:** February 01, 2010, 09:10:27 pm »

I don't want you guys to fail. I actually thought it was a pretty good question had I worded it right I'm sure it would have all made sense. I apologise for that and next time I write a question I'll word it carefully to make sure you guys know what I want.

appianway · « **Reply #27 on:** February 01, 2010, 09:27:01 pm »

If the buoyancy force isn't considered (hence the only force causing circular motion being the gravitational force), can't it just be found using g like I originally described (equating the two g expressions, although I admittedly didn't know that the density was constant so used the GM/r^2 to find the mass in terms of the radius, but it's still the same general idea *coughcough*).

Good question though. Wording's always a bit tricky. It'd also be interesting to consider any drag forces acting on the body...

Cthulhu · « **Reply #28 on:** February 01, 2010, 09:36:44 pm »

I might as well do a survey while I'm here.
Are there any particular areas people are interested in getting questions in?

appianway · « **Reply #29 on:** February 01, 2010, 09:37:58 pm »

Thermodynamics, physical optics, special relativity, electromagnetism.

Oh, and how can I forget... quantum.

I probably need practise in geometrical optics too, seeing as it's my least favourite area...

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