Question

At a distance of \(7.00 \times {10^{12}}\;{\rm{m}}\) from a star, the intensity of the radiation from the star is \(15.4\;{{\rm{W}} \mathord{\left/

{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.

\kern-\nulldelimiterspace} {{{\rm{m}}^{\rm{2}}}}}\). Assuming that the star radiates uniformly in all directions, what is the total power output of the star?

Short answer

The total power is, \(9.48 \times {10^{27}}\;{\rm{W}}\).

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The distance of a star is, \(r = 7.0 \times {10^{12}}\;{\rm{m}}\).
• The intensity of the radiation is, \(I = 15.4\;{{\rm{W}} \mathord{\left/
• {\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.
• \kern-\nulldelimiterspace} {{{\rm{m}}^{\rm{2}}}}}\).

Significance of the intensity of the radiation

The power radiated by an object on which light rays are incident at a specific angle is known as the intensity of radiation. The rate of emissivity, temperature, and dimensions of the object all affect how much energy is radiated from its unit area.

Determination of the total power

The relation between the total power, intensity of the radiation and distance are expressed as,

\(\begin{aligned}{l}I = \frac{P}{{4\pi {r^2}}}\\P = \left( {4\pi {r^2}} \right)I\end{aligned}\)

Here \(P\) is the total power, \(I\) is the intensity of radiation and \(r\) is the distance of the star.

Substitute \(7.0 \times {10^{12}}\;{\rm{m}}\) for \(r\) and \(15.4\;{{\rm{W}} \mathord{\left/

{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.

\kern-\nulldelimiterspace} {{{\rm{m}}^{\rm{2}}}}}\) for \(I\) in the above equation.

\(\begin{aligned}{l}P = 4\pi {\left( {7.0 \times {{10}^{12}}\;{\rm{m}}} \right)^2} \times 15.4\;{{\rm{W}} \mathord{\left/

{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.

\kern-\nulldelimiterspace} {{{\rm{m}}^{\rm{2}}}}}\\P = 9.48 \times {10^{27}}\;{\rm{W}}\end{aligned}\)

Hence the total power is, \(9.48 \times {10^{27}}\;{\rm{W}}\).