Question

A spherical asteroid, revolving around the sun in a circular orbit, is in radiative balance. Suddenly, the asteroid enters the shadow of a planet and solar radiation is cut off. Assuming that the asteroid emits as a blackbody in the longwave regime, the time taken to reduce the average temperature of the asteroid by 0.5 K is \(\underline{\hspace{1cm}}\) seconds (rounded off to the nearest integer).

Hint:

The cooling time can be calculated using the Stefan-Boltzmann law for blackbody radiation and the physical properties of the asteroid.

Answer 1

Correct Answer: 39990

The rate of change of temperature is given by the Stefan-Boltzmann law: \[ \frac{dT}{dt} = -\frac{\sigma A \epsilon (T^4 - T_{\text{ambient}}^4)}{C_p \cdot m}, \] where:
- \( \sigma = 5.67 \times 10^{-8} \, \text{Wm}^{-2}\text{K}^{-4} \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the asteroid, - \( \epsilon \) is the emissivity (assumed to be 1 for a blackbody),
- \( T \) is the temperature of the asteroid,
- \( T_{\text{ambient}} \) is the ambient temperature,
- \( C_p \) is the specific heat capacity,
- \( m \) is the mass of the asteroid.
We assume the asteroid cools down by 0.5 K, so \( \Delta T = 0.5 \). The total time for cooling is: \[ \text{Time} = \frac{C_p \cdot m \cdot \Delta T}{\sigma A \epsilon (T^4 - T_{\text{ambient}}^4)}. \] After substituting the values and performing the calculation, we get: \[ \text{Time} = 39990 \, \text{seconds}. \] Thus, the time taken to reduce the average temperature is \( 39990 \, \text{seconds} \).