Q29E
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?
Step-by-Step Solution
VerifiedThe total power is, \(9.48 \times {10^{27}}\;{\rm{W}}\).
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, \(I = 15.4\;{{\rm{W}} \mathord{\left/
- {\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.
- \kern-\nulldelimiterspace} {{{\rm{m}}^{\rm{2}}}}}\).
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.
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, \(9.48 \times {10^{27}}\;{\rm{W}}\).