Question

In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500W/m2 at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.)

A) Find the average radiation pressure (in pascals) on a totally absorbing section of the floor.

B) Find the average radiation pressure (in atmospheres) on a totally absorbing section of the floor.

C) Find the average radiation pressure (in pascals) on a totally reflecting section of the floor.

D) Find the average radiation pressure (in atmospheres) on a totally reflecting section of the floor.

Answer 1

Concepts and reason

The concept used in this problem is Average Radiation Pressure.

Initially, calculate the average radiation pressure on a totally absorbing section of the floor by using the concept of average radiation Pressure and then calculate the average radiation pressure for a totally reflecting section of the floor by using the same concept.

Fundamentals

Average Radiation Pressure:

Radiation pressure is defined by the pressure applied beginning with any surface due to the interchange of momentum between the Electromagnetic field and the object. This comprises the electromagnetic radiation or the momentum of light of a different wavelength which is emitted by the matter, reflected or absorbed.

For Perfect Absorber:

The expression for the average radiation pressure for perfect absorber is given by,

$P = \frac{I}{c}$

Here, $P$ is the radiation pressure, $I$ is the incident energy and $c$ is the speed of light.

For Perfect Reflector:

The expression for the average radiation pressure for perfect absorber is given by,

$P = \frac{{2I}}{c}$

(A)

The expression for the average radiation pressure for perfect absorber is given by,

$P = \frac{I}{c}$

Substitute $2500{\rm{ W/}}{{\rm{m}}^2}$ for $I$ and $3 \times {10^8}{\rm{ m/s}}$ for c in the above equation.

$\begin{array}{c}\\P = \frac{{2500{\rm{ W/}}{{\rm{m}}^2}}}{{3 \times {{10}^8}{\rm{ m/s}}}}\\\\ = 8.3 \times {10^{ - 6}}{\rm{ Pa}}\\\end{array}$

(B)

The expression for the average radiation pressure for perfect absorber is given by,

$P = \frac{I}{c}$

$1{\rm{ atm }} = {\rm{ }}1.01325 \times {10^5}\;{\rm{Pa}}$

The average pressure in terms of atmosphere is given by,

$\begin{array}{c}\\P = 8.3 \times {10^{ - 6}}{\rm{ Pa}}\left( {\frac{{1{\rm{ atm}}}}{{1.01325 \times {{10}^5}\;{\rm{Pa}}}}} \right)\\\\ = 8.224 \times {10^{ - 11}}{\rm{ atm}}\\\end{array}$

(C)

The expression for the average radiation pressure for the perfect reflector is given by,

$P = \frac{{2I}}{c}$

Substitute $2500{\rm{ W/}}{{\rm{m}}^2}$ for $I$ and $3 \times {10^8}{\rm{ m/s}}$ for c in the above equation.

$\begin{array}{c}\\P = \frac{{2\left( {2500{\rm{ W/}}{{\rm{m}}^2}} \right)}}{{3 \times {{10}^8}{\rm{ m/s}}}}\\\\ = 16.6 \times {10^{ - 6}}{\rm{ Pa}}\\\end{array}$

(D)

The expression for the average radiation pressure for the perfect reflector is given by,

$P = \frac{{2I}}{c}$

$1{\rm{ atm }} = {\rm{ }}1.01325 \times {10^5}\;{\rm{Pa}}$

The average pressure in terms of atmosphere is given by,

$\begin{array}{c}\\P = 16.6 \times {10^{ - 6}}{\rm{ Pa}}\left( {\frac{{1{\rm{ atm}}}}{{1.01325 \times {{10}^5}\;{\rm{Pa}}}}} \right)\\\\ = 16.448 \times {10^{ - 11}}{\rm{ atm}}\\\end{array}$

Ans: Part A

The average radiation pressure for a total absorbing section of the floor in Pascal is ${\bf{8}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ Pa}}$ .

Part B

The average radiation for a total absorbing section of the floor in atmospheres is ${\bf{8}}{\bf{.224 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{\bf{ atm}}$ .

Part C

The average radiation pressure for a totally reflecting section of the floor in Pascal is ${\bf{16}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ Pa}}$ .

Part D

The average radiation pressure for a totally reflecting section of the floor in atmospheres is ${\bf{16}}{\bf{.448 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{\bf{ atm}}$ .