Problem 70 - Calculus Single Variable

Question

qThe interior temperature \(T_{I}\) (in degrees \(K\) ) of a cooling white dwarf (star) satisfies the differential equation $$ \frac{d T_{I}}{d t}=-k\left(\frac{T_{I}}{7 \times 10^{7^{\circ}} \mathrm{K}}\right)^{7 / 2} $$ Here \(k\) is a constant with units degrees \(K\) per year, and \(t\) represents time in years. Solve for \(T_{I}\). If \(k=6^{\circ} \mathrm{K} / \mathrm{yr}\) and if \(T_{I}(0)=10^{8 \circ} \mathrm{K},\) then in how many years will the star \(\mathrm{cool}\) to \(10^{4} \mathrm{~K}\) ?

Step-by-Step Solution

Answer

The white dwarf will take approximately 23,146,667 years to cool to 10,000 K.

Step 1: Separate Variables

To solve the given differential equation, we start by separating the variables on both sides. We have: \[ \frac{d T_I}{dt} = -k \left( \frac{T_I}{7 \times 10^7} \right)^{7/2} \] We can rewrite this as: \[ \frac{dT_I}{\left(\frac{T_I}{7 \times 10^7}\right)^{7/2}} = -k \, dt \] Bring \(T_I\) to one side and \(dt\) to the other. Further simplify the left side of the equation to: \[ \left(\frac{7 \times 10^7}{T_I}\right)^{7/2} \, dT_I = -k \, dt \] Now, both variables \(T_I\) and \(t\) are separated.

Step 2: Integrate Both Sides

Integrate both sides of the equation from Step 1. The left side, with respect to \(T_I\), becomes: \[ \int \left(\frac{7 \times 10^7}{T_I}\right)^{7/2} \, dT_I \] The right side becomes: \[-k \int dt \] Perform the integrals: The integral for the left-hand side yields: \[ -\frac{2}{5} \times \frac{7 \times 10^7}{T_I^{5/2}} \] The integral for the right side is straightforward: \[ -kt + C \] where \(C\) is the integration constant.

Step 3: Simplify the Integrated Equation

After integration, we have: \[ -\frac{2}{5} \left(\frac{7 \times 10^7}{T_I}\right)^{5/2} = -kt + C \] Cancel the negative sign on both sides: \[ \frac{2}{5} \left(\frac{7 \times 10^7}{T_I}\right)^{5/2} = kt - C \] Simplify this equation to solve for \(T_I\).

Step 4: Determine the Constant of Integration

When \(t = 0\), we have \(T_I = 10^8 \) K. Substitute these values into the equation: \[ \frac{2}{5} \left(\frac{7 \times 10^7}{10^8}\right)^{5/2} = -C \] This results in: \[ \frac{2}{5} \left(0.7\right)^{5/2} = -C \] Calculate \(C\) (using a calculator if necessary): \[ C = \frac{2}{5} \left(0.7\right)^{5/2} \approx 0.1002 \] So the integration constant \(C \approx -0.1002\).

Step 5: Solve for Time \(t\) with Desired \(T_I\)

Now that we know the constant, the equation is: \[ \frac{2}{5} \left(\frac{7 \times 10^7}{T_I}\right)^{5/2} - 0.1002 = kt \] We want to find \(t\) when \(T_I = 10^4\) K. Plug in this value: \[ \frac{2}{5} \left(\frac{7 \times 10^7}{10^4}\right)^{5/2} - 0.1002 = 6t \] Calculate the left side: \[ \frac{2}{5} \times (7000)^{5/2} \approx 1.3888 \times 10^{11} \] Now: \[ 1.3888 \times 10^{11} - 0.1002 = 6t \] Finally, solve for \(t\) using: \[ t \approx \frac{1.3888 \times 10^{11}}{6} \approx 23146666.7 \text{ years}\]

Step 6: Conclusion

It will take approximately 23,146,667 years for the white dwarf to cool from its initial temperature of \(10^8 K\) to \(10^4 K\).

Key Concepts

Separation of Variables

The process of solving a differential equation often begins with the technique of 6lsquo;separation of variables6rsquo;. This method helps in simplifying the equation by rearranging it so that each variable and its derivatives appear on different sides of the equation. In our exercise with the cooling of a white dwarf, we were given the differential equation: \[ \frac{d T_I}{dt} = -k \left( \frac{T_I}{7 \times 10^7} \right)^{7/2} \] To separate the variables, we moved terms involving \(T_I\) to one side and terms involving \(t\) to the other. This results in: \[ \frac{dT_I}{\left(\frac{T_I}{7 \times 10^7}\right)^{7/2}} = -k \, dt \] By transforming the left side, we ended up with: \[ \left(\frac{7 \times 10^7}{T_I}\right)^{7/2} \, dT_I = -k \, dt \] This rearrangement makes it possible to integrate both sides independently, paving the way to find \(T_I\) as a function of \(t\). The focus is on creating a clear path for integration which is the next logical step.

Integration

Once the variables in a differential equation are separated, the next task is to integrate both sides of the equation. Integration helps in finding the antiderivatives and is a key step in solving differential equations analytically. For the cooling white dwarf problem, after separating the variables, we integrated:

• With respect to \(T_I\) on the left: \[ \int \left(\frac{7 \times 10^7}{T_I}\right)^{7/2} \, dT_I \]
• With respect to \(t\) on the right: \[-k \int dt \]

The integration of the left leads to the equation: \[ -\frac{2}{5} \times \left(\frac{7 \times 10^7}{T_I}\right)^{5/2} \] While integrating the right side yields a straightforward result: \[-kt + C\] where \(C\) represents the constant of integration. This constant appears because indefinite integrals include an unknown constant. It is crucial as it allows us to incorporate initial conditions, providing a complete solution to the differential equation.

Cooling of Stars

The cooling of stars, specifically in this context, involves the gradual decline in temperature of a white dwarf over time. Stars cool over an extended period as they radiate away heat. In our model for the white dwarf star, the star's interior temperature over time was governed by the differential equation discussed previously. This equation accounts for the physics of star cooling by incorporating a factor dependent on temperature raised to a fractional power, representing complex radiative heat transfer mechanisms. As the temperature of the star decreases, the rate of cooling changes, showing the non-linear relationship between temperature and time. In essence, determining how long a white dwarf takes to cool from a very high temperature to a significantly lower one requires solving such equations. The real-world application of these solutions helps astronomers understand stellar life cycles, especially the end stages like those of white dwarfs.

White Dwarf

A white dwarf represents an evolutionary stage of a star after it has exhausted its nuclear fuel and expelled its outer layers. Compared to living stars, white dwarfs are relatively small yet incredibly dense remnants composed predominantly of electron-degenerate matter. White dwarfs typically have a mass comparable to the Sun's yet occupy a volume similar to that of Earth. The cooling process of a white dwarf is incredibly slow due to its initial temperature and the nature of radiative heat emission. Understanding the cooling of white dwarfs allows scientists to deduce their ages, with the temperature of the white dwarf directly correlating to its age. The cooler a white dwarf, the older it is. Due to this cooling phenomenon, the white dwarf eventually becomes so cold and faint that it no longer emits significant radiation, evolving into a so-called "black dwarf", although such objects have not yet been observed, as the Universe is not old enough for them to exist.