<p>Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness.</p>
<h3>Orbital Distance</h3>
<p>We're told the period is <span class="math-container">$T=5.5\:\mathrm{years}$</span>. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, <span class="math-container">$a$</span>). Since we're talking about a comet orbiting the Sun, we can simply use:</p>
<p><span class="math-container">$$T^2 = a^3$$</span></p>
<p>where <span class="math-container">$T$</span> is the orbital period in units of <span class="math-container">$\mathrm{years}$</span> and <span class="math-container">$a$</span> is the semi-major axis in units of <span class="math-container">$\mathrm{AU}$</span>. I find that <span class="math-container">$a = 3.116\:\mathrm{AU} = 4.66\times 10^{11}\:\mathrm{m}$</span>. Good so far.</p>
<h3>Volume of Evaporated Shell</h3>
<p>We're told that the radius of the comet decreases by <span class="math-container">$\Delta R = 20\:\mathrm{cm}$</span>. We can assume the comet is perfectly spherical and calculate the volume of the evaporated shell which should depend on both <span class="math-container">$R$</span>, the radius of the comet, and <span class="math-container">$\Delta R$</span>. This will be necessary since we need to know the total amount of ice that was evaporated. The volume of this shell is given by</p>
<p><span class="math-container">$$V = \frac{4}{3}\pi(R^3-(R-\Delta R)^3)$$</span>
<span class="math-container">$$V = \frac{4}{3}\pi(R^3-R^3+3R^2\Delta R-3R\Delta R^2 + \Delta R^3)$$</span>
<span class="math-container">$$V = \frac{4}{3}\pi(3R^2\Delta R-3R\Delta R^2)$$</span>
<span class="math-container">$$V = 4\pi R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$</span></p>
<p>Note that I've made a specific choice here. I've dropped the <span class="math-container">$\Delta R^3$</span> term from the second line. The reason being that it is a third order term and <span class="math-container">$\Delta R^3 \ll R\Delta R^2$</span>. You could argue that I could drop the second order term, <span class="math-container">$R\Delta R^2$</span> since <span class="math-container">$R\Delta R^2 \ll R^2\Delta R$</span>, but I'm choosing to keep this second order term around so we wind up with an <span class="math-container">$R$</span> to solve for in the final answer.</p>
<h3>Power Input</h3>
<p>The next step is to find the total energy input per second to this comet, e.g., the power. You've basically already defined this part. The flux on this comet is defined as</p>
<p><span class="math-container">$$F = \frac{L_\odot}{4\pi a^2}$$</span></p>
<p>The power input is simply the flux times the area of the comet.</p>
<p><span class="math-container">$$P_{\mathrm{in}} = FA = \frac{L_\odot}{4\pi a^2} \pi R^2 = \frac{1}{4}L_\odot \frac{R^2}{a^2}$$</span></p>
<p>Note that this assumes the comet is in a circular orbit and thus always at the orbital radius of <span class="math-container">$a$</span>. If the comet's orbit had any sort of eccentricity, then <span class="math-container">$F$</span> would be a function of radius and you'd have a much harder time.</p>
<h3>Evaporation Energy</h3>
<p>Now we need to calculate the total energy necessary to evaporate the evaporated shell of volume <span class="math-container">$V$</span> from above. In order to evaporate a solid ice, you have to go through four stages of heating. First, you raise the ice's temperature to the melting point. The energy input for this is defined by the specific heat capacity for ice, <span class="math-container">$c_{\mathrm{ice}}$</span>. Then you add energy to convert the ice into water. The energy input for this is defined by the latent heat of fusion, <span class="math-container">$\ell_f$</span>. Now you can raise the temperature of the water until it reaches the next stage. This is defined by the specific heat of water, <span class="math-container">$c_{\mathrm{water}}$</span>. Finally, you add energy to convert the water to gas, defined by the latent heat of vaporization, <span class="math-container">$\ell_v$</span>.</p>
<p>All of these can be put together into a single equation.</p>
<p><span class="math-container">$$E_{\mathrm{evap}} = c_{\mathrm{ice}}(m_{\mathrm{shell}}\Delta T_1) + \ell_fm_{\mathrm{shell}}+c_{\mathrm{water}}(m_{\mathrm{shell}}\Delta T_2) + \ell_vm_{\mathrm{shell}}$$</span></p>
<p><span class="math-container">$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)m_{\mathrm{shell}}$$</span></p>
<p>Each one of the terms in this equation accounts for the energy input of one of the stages I described above. Note that <span class="math-container">$\Delta T_1$</span> is the temperature change from the initial temperature to the melting point (<span class="math-container">$273.15\:\mathrm{K}$</span>). A reasonable temperature change might be <span class="math-container">$73.15\:\mathrm{K}$</span> (assuming an initial temperature of <span class="math-container">$200\:\mathrm{K}$</span>), based on the <a href="http://www.astronomy.com/news/2014/08/rosetta-takes-comets-temperature" rel="nofollow noreferrer">temperature of comet 67P as determined by Rosetta</a>. The <span class="math-container">$\Delta T_2$</span> is the temperature change from the melting point to the boiling point which is necessarily <span class="math-container">$100\:\mathrm{K}$</span>.</p>
<p>You can look in a table somewhere and find that <span class="math-container">$c_{\mathrm{ice}} = 2.108\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$</span> and <span class="math-container">$c_{\mathrm{water}} = 4.187\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$</span>.</p>
<p>Lastly, you need to define the mass of the evaporated shell, <span class="math-container">$m_{\mathrm{shell}}$</span>. This is simply the volume already determined, multiplied by the density of ice/water. Technically those densities will be different, but they're close enough that we can just use <span class="math-container">$\rho_{shell} = 1000\:\mathrm{kg\:m^3}$</span>. So ultimately we have:</p>
<p><span class="math-container">$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)V \rho_{\mathrm{shell}}$$</span></p>
<p>For simplicity's sake, I'm going to define</p>
<p><span class="math-container">$$\eta \equiv (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)$$</span></p>
<p>so that</p>
<p><span class="math-container">$$E_{\mathrm{evap}} = 4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$</span></p>
<h3>Putting it all together</h3>
<p>We now know the total energy necessary to evaporate the comet's <span class="math-container">$20\:\mathrm{cm}$</span> shell as well as the energy input per second. We know that it receives this energy input per second for one orbital period of <span class="math-container">$5.5\:\mathrm{years}$</span> which means we can say:</p>
<p><span class="math-container">$$T = \frac{E_{\mathrm{evap}}}{P_{\mathrm{in}}}$$</span></p>
<p><span class="math-container">$$T = \frac{4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)}{\frac{1}{4}L_\odot \frac{R^2}{a^2}}$$</span></p>
<p><span class="math-container">$$T = 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot} \left(1-\frac{\Delta R}{R}\right)$$</span></p>
<p>Again, for simplicity, I'll define</p>
<p><span class="math-container">$$\xi \equiv 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot}$$</span></p>
<p>so that</p>
<p><span class="math-container">$$T = \xi \left(1-\frac{\Delta R}{R}\right)$$</span></p>
<p>It should be pretty easy to see now that</p>
<p><span class="math-container">$$\boxed{R = \frac{\Delta R}{1-T/\xi}}$$</span></p>
<p>The rest is just plugging everything in.</p>

CC BY-SA 3.0
2017-07-11T18:36:39.903Z
2017-07-11T20:08:21.873Z
2020-06-17T09:47:40.653Z
zephyr
<p>A Comet is orbiting around the sun with period of $T= 5.5$ years. The comet has radius $R$. After 5.5 years (A period) we understand that its radius has decreased about $20cm$. How can we calculate the R. We know that comet is approximately a sphere and it is made of Ice. We have the $G = 6.67 * 10^{-11}$ SI_Unit, $M_{sun} = 1.99 * 10^{30} Kg$, $\sigma = 5.67 *10^{-8}$ SI_Unit (Stefan-Boltzmann constant) , $L_{sun}=3.85 * 10^{26} w$, $l_f = 333 KJ/Kg$ , $l_v = 2260 KJ/Kg$. It would be fine that if the solution doesn't need the numerical value of density of ice but if it can't be solved without density of ice, you are allowed to use it.</p>

<p>I can calculate the distance to sun ($a$) with formula:
$T^2 = 4 *\pi^2 a^3 /(GM_{Sun})$.</p>

<p>$a= 4.66* 10^{11} m$</p>

<p>I also can calculate $F= L_{sun} / (4 * \pi * a^2)$.</p>

<p>I also approximated the power that the comet gives is: $P = F * Area$ (think ice don't reflect any light)</p>

<p>But I don't know how to calculate the R and formulate how it decreases every moment (because it affects the area).</p>

CC BY-SA 3.0
2017-07-10T18:00:38.213Z
As a hint: You need to use the latent heat of fusion, $l_f$, and vaporization, $l_v$, which, in part, tells you the energy input to convert the ice on the comet to gas such that it leaves and the comet shrinks.You might also have to assume an initial temperature of the comet and look up the specific heat of ice/water. Basically, calculate the total energy necessary to turn the 20 cm shell of the comet into a gas.
zephyr
CC BY-SA 3.0
2017-07-10T18:25:10.657Z
@zephyr I can calculate the heat with formula like $\rho * \Delta V * (l_f + l_v + c*100)$ where $\Delta V = 4/3 * \pi * ((R)^3 -(R - 0.2)^3) $  . (with init temp = 0 Celsius) my problem is about how I can calculate the heat It gains in different times. because its radius changes, so its area and the heat it gains, change.
titansarus
CC BY-SA 3.0
2017-07-10T18:44:21.287Z
The problem doesn't specify anything about the orbit of the comet other than its period. Without further information, its probably safe to assume it has a circular orbit and thus receives constant heating throughout its orbit. If you had to account for differences in heating with distance, you'd need to know something like the eccentricity of the orbit and you'd have to integrate the heat input along the path of the orbit. It sounds like assuming a circular orbit is plenty sufficient for this problem.
zephyr
CC BY-SA 3.0
2017-07-10T19:43:15.567Z
@zephyr assume it is circular. How I must integrate? (What must I integrate?) I am not very familiar with integration in physics.
titansarus
CC BY-SA 3.0
2017-07-10T20:33:53.710Z
My point was that you don't need to integrate. If you assume a circular orbit, you receive a constant flux input from the star so no integration is necessary. You would only need to integrate if your orbit wasn't circular, but a circular orbit simplifies the math significantly and means no integration is necessary.
zephyr
CC BY-SA 3.0
2017-07-11T16:48:07.760Z
@zephyr My problem is that due to change in radius, hence the area and $F* Area$ changes. We also don't know anything about how large the comet is and so, we don't know if this change has great impact or not. So how can I calculate correctly due to this change in area?
titansarus
<astrophysics><comets><radiation>
A question about calculating the radius of a comet
titansarus
# A question about calculating the radius of a comet

(Asked by: username_0 on 2017-07-10)

A Comet is orbiting around the sun with period of $T= 5.5$ years. The comet has radius $R$. After 5.5 years (A period) we understand that its radius has decreased about $20cm$. How can we calculate the R. We know that comet is approximately a sphere and it is made of Ice. We have the $G = 6.67 \* 10^{-11}$ SI\_Unit, $M\_{sun} = 1.99 \* 10^{30} Kg$, $\sigma = 5.67 \*10^{-8}$ SI\_Unit (Stefan-Boltzmann constant) , $L\_{sun}=3.85 \* 10^{26} w$, $l\_f = 333 KJ/Kg$ , $l\_v = 2260 KJ/Kg$. It would be fine that if the solution doesn't need the numerical value of density of ice but if it can't be solved without density of ice, you are allowed to use it.

I can calculate the distance to sun ($a$) with formula:
$T^2 = 4 \*\pi^2 a^3 /(GM\_{Sun})$.

$a= 4.66\* 10^{11} m$

I also can calculate $F= L\_{sun} / (4 \* \pi \* a^2)$.

I also approximated the power that the comet gives is: $P = F \* Area$ (think ice don't reflect any light)

But I don't know how to calculate the R and formulate how it decreases every moment (because it affects the area).

**Tags:** astrophysics, comets, radiation

> **Comments:**
> **username_1**: As a hint: You need to use the latent heat of fusion, $l_f$, and vaporization, $l_v$, which, in part, tells you the energy input to convert the ice on the comet to gas such that it leaves and the comet shrinks.You might also have to assume an initial temperature of the comet and look up the specific heat of ice/water. Basically, calculate the total energy necessary to turn the 20 cm shell of the comet into a gas.  
> **username_0**: @username_1 I can calculate the heat with formula like $\rho * \Delta V * (l_f + l_v + c*100)$ where $\Delta V = 4/3 * \pi * ((R)^3 -(R - 0.2)^3) $  . (with init temp = 0 Celsius) my problem is about how I can calculate the heat It gains in different times. because its radius changes, so its area and the heat it gains, change.  
> **username_1**: The problem doesn't specify anything about the orbit of the comet other than its period. Without further information, its probably safe to assume it has a circular orbit and thus receives constant heating throughout its orbit. If you had to account for differences in heating with distance, you'd need to know something like the eccentricity of the orbit and you'd have to integrate the heat input along the path of the orbit. It sounds like assuming a circular orbit is plenty sufficient for this problem.  
> **username_0**: @username_1 assume it is circular. How I must integrate? (What must I integrate?) I am not very familiar with integration in physics.  
> **username_1**: My point was that you don't need to integrate. If you assume a circular orbit, you receive a constant flux input from the star so no integration is necessary. You would only need to integrate if your orbit wasn't circular, but a circular orbit simplifies the math significantly and means no integration is necessary.  
> **username_0**: @username_1 My problem is that due to change in radius, hence the area and $F* Area$ changes. We also don't know anything about how large the comet is and so, we don't know if this change has great impact or not. So how can I calculate correctly due to this change in area?  

---

## 1 answers

### Answer by username_1 on 2017-07-11. Score: 4 | Accepted answer

Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness.

### Orbital Distance

We're told the period is $T=5.5\:\mathrm{years}$. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, $a$). Since we're talking about a comet orbiting the Sun, we can simply use:

$$T^2 = a^3$$

where $T$ is the orbital period in units of $\mathrm{years}$ and $a$ is the semi-major axis in units of $\mathrm{AU}$. I find that $a = 3.116\:\mathrm{AU} = 4.66\times 10^{11}\:\mathrm{m}$. Good so far.

### Volume of Evaporated Shell

We're told that the radius of the comet decreases by $\Delta R = 20\:\mathrm{cm}$. We can assume the comet is perfectly spherical and calculate the volume of the evaporated shell which should depend on both $R$, the radius of the comet, and $\Delta R$. This will be necessary since we need to know the total amount of ice that was evaporated. The volume of this shell is given by

$$V = \frac{4}{3}\pi(R^3-(R-\Delta R)^3)$$
$$V = \frac{4}{3}\pi(R^3-R^3+3R^2\Delta R-3R\Delta R^2 + \Delta R^3)$$
$$V = \frac{4}{3}\pi(3R^2\Delta R-3R\Delta R^2)$$
$$V = 4\pi R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$

Note that I've made a specific choice here. I've dropped the $\Delta R^3$ term from the second line. The reason being that it is a third order term and $\Delta R^3 \ll R\Delta R^2$. You could argue that I could drop the second order term, $R\Delta R^2$ since $R\Delta R^2 \ll R^2\Delta R$, but I'm choosing to keep this second order term around so we wind up with an $R$ to solve for in the final answer.

### Power Input

The next step is to find the total energy input per second to this comet, e.g., the power. You've basically already defined this part. The flux on this comet is defined as

$$F = \frac{L\_\odot}{4\pi a^2}$$

The power input is simply the flux times the area of the comet.

$$P\_{\mathrm{in}} = FA = \frac{L\_\odot}{4\pi a^2} \pi R^2 = \frac{1}{4}L\_\odot \frac{R^2}{a^2}$$

Note that this assumes the comet is in a circular orbit and thus always at the orbital radius of $a$. If the comet's orbit had any sort of eccentricity, then $F$ would be a function of radius and you'd have a much harder time.

### Evaporation Energy

Now we need to calculate the total energy necessary to evaporate the evaporated shell of volume $V$ from above. In order to evaporate a solid ice, you have to go through four stages of heating. First, you raise the ice's temperature to the melting point. The energy input for this is defined by the specific heat capacity for ice, $c\_{\mathrm{ice}}$. Then you add energy to convert the ice into water. The energy input for this is defined by the latent heat of fusion, $\ell\_f$. Now you can raise the temperature of the water until it reaches the next stage. This is defined by the specific heat of water, $c\_{\mathrm{water}}$. Finally, you add energy to convert the water to gas, defined by the latent heat of vaporization, $\ell\_v$.

All of these can be put together into a single equation.

$$E\_{\mathrm{evap}} = c\_{\mathrm{ice}}(m\_{\mathrm{shell}}\Delta T\_1) + \ell\_fm\_{\mathrm{shell}}+c\_{\mathrm{water}}(m\_{\mathrm{shell}}\Delta T\_2) + \ell\_vm\_{\mathrm{shell}}$$

$$E\_{\mathrm{evap}} = (c\_{\mathrm{ice}}\Delta T\_1 + \ell\_f +c\_{\mathrm{water}}\Delta T\_2 + \ell\_v)m\_{\mathrm{shell}}$$

Each one of the terms in this equation accounts for the energy input of one of the stages I described above. Note that $\Delta T\_1$ is the temperature change from the initial temperature to the melting point ($273.15\:\mathrm{K}$). A reasonable temperature change might be $73.15\:\mathrm{K}$ (assuming an initial temperature of $200\:\mathrm{K}$), based on the [temperature of comet 67P as determined by Rosetta](http://www.astronomy.com/news/2014/08/rosetta-takes-comets-temperature). The $\Delta T\_2$ is the temperature change from the melting point to the boiling point which is necessarily $100\:\mathrm{K}$.

You can look in a table somewhere and find that $c\_{\mathrm{ice}} = 2.108\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$ and $c\_{\mathrm{water}} = 4.187\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$.

Lastly, you need to define the mass of the evaporated shell, $m\_{\mathrm{shell}}$. This is simply the volume already determined, multiplied by the density of ice/water. Technically those densities will be different, but they're close enough that we can just use $\rho\_{shell} = 1000\:\mathrm{kg\:m^3}$. So ultimately we have:

$$E\_{\mathrm{evap}} = (c\_{\mathrm{ice}}\Delta T\_1 + \ell\_f +c\_{\mathrm{water}}\Delta T\_2 + \ell\_v)V \rho\_{\mathrm{shell}}$$

For simplicity's sake, I'm going to define

$$\eta \equiv (c\_{\mathrm{ice}}\Delta T\_1 + \ell\_f +c\_{\mathrm{water}}\Delta T\_2 + \ell\_v)$$

so that

$$E\_{\mathrm{evap}} = 4\pi \eta \rho\_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$

### Putting it all together

We now know the total energy necessary to evaporate the comet's $20\:\mathrm{cm}$ shell as well as the energy input per second. We know that it receives this energy input per second for one orbital period of $5.5\:\mathrm{years}$ which means we can say:

$$T = \frac{E\_{\mathrm{evap}}}{P\_{\mathrm{in}}}$$

$$T = \frac{4\pi \eta \rho\_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)}{\frac{1}{4}L\_\odot \frac{R^2}{a^2}}$$

$$T = 8\pi \eta \rho\_{\mathrm{shell}} \Delta R \frac{a^2}{L\_\odot} \left(1-\frac{\Delta R}{R}\right)$$

Again, for simplicity, I'll define

$$\xi \equiv 8\pi \eta \rho\_{\mathrm{shell}} \Delta R \frac{a^2}{L\_\odot}$$

so that

$$T = \xi \left(1-\frac{\Delta R}{R}\right)$$

It should be pretty easy to see now that

$$\boxed{R = \frac{\Delta R}{1-T/\xi}}$$

The rest is just plugging everything in.


---

