Q53E - University Physics with Modern Physics

Question

Magnetar. On December 27, 2004, astronomers observed the greatest flash of light ever recorded from outside the solar system. It came from the highly magnetic neutron star SGR 1806-20 (a magnetar). During $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{\text{ s}}$, this star released as much energy as our sun does in  $\mathbf{250}\mathbf{,}\mathbf{000}\mathbf{\text{ years}}$. If $\mathit{P}$ is the average power output of our sun, what was the average power output (in terms of $\mathit{P}$) of this magnetar?

Step-by-Step Solution

Answer

The average output of this magnetar is$\left(3.9\times {10}^{13}\right)P$.

Step 1: Power:

Power is defined as the rate of energy. The expression of the energy is,

$\mathit{P}\mathbf{=}\frac{\mathbf{E}}{\mathbf{t}}$

Here, $\mathit{P}$ is the power, $\mathit{E}$ is the energy, and $\mathit{t}$ is the time.

Step 2: Time is second:

The time for the sun in the second is defined as below.

 $\begin{array}{rcl}{t}_{sun}& =& \left(250,000\text{ years}\left(\frac{\text{365 days}}{\text{1 year}}\right)\left(\frac{\text{24 h}}{\text{1 day}}\right)\left(\frac{\text{60 min}}{\text{1 h}}\right)\left(\frac{\text{60 s}}{\text{1 min}}\right)\right)\\ & =& 7.884\times {10}^{12}\text{ s}\end{array}$

The time for star,  $t_{star}=0.20\text{ s}$

Step 3: The average output of this magnetar:

The power $P_{\text{sun}}$ for the sun is,

$P_{\text{sun}}=\frac{E_{\text{sun}}}{t_{\text{sun}}}$

Here, $E_{\text{sun}}$ is the energy released by the sun and $t_{\text{sun}}$ is the time for the sun.

 

The power $P_{\text{star}}$ for star is,

$P_{\text{star}}=\frac{E_{\text{star}}}{t_{\text{star}}}$

Here, $E_{\text{star}}$ is the energy released by the star and $t_{\text{star}}$ is the time for the star.

Now, divide the equations of $P_{\text{star}}$ and  $P_{\text{sun}}$.

$\frac{P_{\text{star}}}{P_{\text{sun}}}=\frac{{\displaystyle \frac{E_{{}_{\text{star}}}}{t_{{}_{\text{star}}}}}}{{\displaystyle \frac{E_{\text{sun}}}{t_{\text{sun}}}}}$

Since the released energy by the sun and the star is equal. Substitute $E_{\text{star}}$ for $E_{\text{sun}}$ in the above equation, and you get,

$\begin{array}{rcl}\frac{P_{\text{star}}}{P_{sun}}& =& \frac{{\displaystyle \frac{E_{\text{star}}}{t_{\text{star}}}}}{{\displaystyle \frac{E_{\text{sun}}}{t_{\text{sun}}}}}\\ & =& \frac{t_{\text{sun}}}{t_{\text{star}}}\end{array}$

 

Substitute $0.20\text{ s}$ for $t_{\text{star}}$ and $7.884\times {10}^{12}\text{ s}$ for $t_{\text{sun}}$ in the above equation.

$\begin{array}{rcl}\frac{P_{\text{star}}}{P_{\text{sun}}}& =& \frac{7.884\times {10}^{12}\text{ s}}{0.20\text{ s}}\\ & =& 3.942\times {10}^{13}\\ & \approx & 3.9\times {10}^{13}\end{array}$

 

Substitute $P$ for $P_{\text{sun}}$ and solve for $P_{\text{star}}$.

$$\begin{gathered} \frac{P_{\text{star}}}{P}=3.9\times {10}^{13} \\ {P}_{\text{star}}=\left(3.9\times {10}^{13}\right)P \end{gathered}$$

Therefore, the average output of this magnetar is, $P_{\text{star}}=\left(3.9\times {10}^{13}\right)P$.