Are These Astrophysics Calculations Correct?

In summary: ENDED ANSWER:So, you should have:U = \frac{4}{3}G\pi^2(\rho_1^2R_1^5 + \rho_2^2(R^5-R_1^5))3. Okay, this question has a little more depth to it...An asteroid with an iron core (density 8000kg/m^3) and a silicate mantle (density 3400kg/m^3) that is 20% of the radius of the asteroid. Ignoring the heat capacity of the mantle, I'm told to find the radius for which the cooling rate of the asteroid is about 1K per 10^6 years. The core has a constant
  • #1
Rob Hal
13
0
Hi all,

I'm looking for some suggestions for a few problems I'm working on.

1. A very simple problem with a comet of orbital period 100 years and perihelion distance from the center of the Sun that is twice the distance of the solar radius. Now, I was looking for the velocity of the comet at perihelion, and I calculated it using Kepler's 3rd and Vis Viva, and I came to 436km/s. Now, I'm not very familiar with comets so I just want to know if this makes sense. It seems a little fast, but my calculations look right.

2. Another simple question... I have two models of the structure of the Earth, one with uniform density and one with a single divide between two layers of different density. I'm asked to calculate the gravitational potential energy of the Earth in each model. Now I'm wondering if this is a typo... if what it really wants is something like the gravitational binding energy... because it asks me to compare two numerical answers. The PE of an object is it's energy due to its presence in a gravitational field so how can I do it for the Earth. (unless I take a test particle, but then I still get function that depends on radius) So am I interpreting this wrong or what?

3. Okay, this question has a little more depth to it...
An asteroid with an iron core (density [tex]8000kg/m^3[/tex]) and a silicate mantle (density [tex]3400kg/m^3[/tex]) that is 20% of the radius of the asteroid. Ignoring the heat capacity of the mantle, I'm told to find the radius for which the cooling rate of the asteroid is about 1K per 10^6 years. The core has a constant internal temperature throughout of [tex]T_i = 600K[/tex], the thermal conductivity of the mantle is 2W/mK and the surface temperature is 200 K. I'm also given that the thermal energy of the core is [tex]3kT_i[/tex] per atom.

So I've considered the heat loss to the surface to be
[tex]\frac{\Delta Q}{\Delta t} = \frac{(4\pi R^2)(2W/mK)(600K-200K)}{R}[/tex]
(So... area of surface of heat loss * temperature gradient * thermal conductivity of mantle...)

And the thermal energy of the core is just [tex]E_t = 3kT_iN[/tex], where N is the number of atoms in the core. And I found a function for N which depends on the radius of the core.

Now I don't really know what I should be doing here. I've tried a few things, but I can't really be sure I'm on the right track. I figured that the time for heat to flow to the surface would be something like the thermal energy of the core divided by the heat loss to the surface, but I'm not exactly sure if this is a completely correct assumption, and if it is, where to go from there...

So any advice is greatly appreciated.
Thank you.
 
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  • #2
Rob Hal said:
1. A very simple problem with a comet of orbital period 100 years and perihelion distance from the center of the Sun that is twice the distance of the solar radius. Now, I was looking for the velocity of the comet at perihelion, and I calculated it using Kepler's 3rd and Vis Viva, and I came to 436km/s. Now, I'm not very familiar with comets so I just want to know if this makes sense. It seems a little fast, but my calculations look right.
First of all, anybody that comes within a solar radius of the sun's surface has a period of infinity. It is never going to return.

But, assuming that this comet should miraculously survive to reach the perihelion, its speed is:

[tex]v = \sqrt{GM/R}[/tex]

Using M = 2e30 kg and R = 1.4e9m and G = 6.67e-11 Nm^2/kg^2,

v = 3.09e5 m/sec = 309 km/sec

2. Another simple question... I have two models of the structure of the Earth, one with uniform density and one with a single divide between two layers of different density. I'm asked to calculate the gravitational potential energy of the Earth in each model.
This is a measure of the work required to peel off the Earth layer by layer and move it to infinity.

If density is uniform, the potential of the first layer of thickness dr and mass [itex]\rho 4\pi r^2dr[/itex] is:

[tex]dU = \frac{GM\rho 4\pi r^2dr}{r}[/tex]

M is a decreasing function of r, as you peel off layers:

[tex]M = \rho \frac{4}{3}\pi r^3[/tex]

So:

[tex]U = \int_0^R \frac{G\rho^2 16\pi^2 r^4}{3}dr[/tex]

Work out the two integrals for an Earth with two different densities (ie. [itex] \rho_1[/itex] from 0 to R1 and [itex]\rho_2[/itex] from R1 to R) and compare to this expression. (The densities and value of R1 must be such that the total mass is the same as in the uniform density case).

AM
 
  • #3


Hi there,

1. Your calculation for the velocity of the comet at perihelion seems reasonable. Comets can have very high velocities when they are close to the Sun due to the strong gravitational pull. In fact, some comets can reach speeds of up to 800 km/s at perihelion. So 436 km/s is definitely within the realm of possibility.

2. It does seem like there might be a typo in the question. The gravitational potential energy of the Earth would be an extremely large number and not very meaningful to compare between models. It is possible that they meant the gravitational binding energy, which is the amount of energy needed to completely separate all the particles in an object. This would be a more meaningful comparison between the two models.

3. For this question, you are on the right track. The time for heat to flow to the surface would be the thermal energy of the core divided by the heat loss to the surface. This is because the thermal energy of the core is the amount of energy that needs to be transferred to the surface to maintain a constant temperature. From there, you can use the function for N to find the radius at which the cooling rate is 1K per 10^6 years. This would be the equilibrium point where the heat loss to the surface is equal to the thermal energy of the core.

I hope this helps and good luck with your problems!
 

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