Radiation Pressure Formula
Last Updated : 04 Feb, 2024
It is important to know the term radiation so that we can better comprehend what radiation pressure is. Radiation is defined as a source of energy or particles that move through the atmosphere or other media. Ionizing and non-ionizing radiation are two different forms of radiation. The several types of radiation include electromagnetic radiation, acoustic radiation, particle radiation, and gravitational radiation. Photons, or quanta, are discrete units of radiant energy that makeup radiation. An electric and magnetic disturbance called electromagnetic radiation moves through space at the speed of light. In addition to radio waves, electromagnetic waves also include X-rays, infrared, visible light, microwaves, gamma rays, and ultraviolet. The electromagnetic spectrum is made up of all these wavelengths. Learn about the Radiation Pressure Formula now.
Radiation Pressure
The momentum transfer between the electromagnetic field and the object gives birth to the idea of radiation pressure. Since electromagnetic waves carry transport momentum, we may observe this force.
The mechanical pressure that is exerted to any surface as a result of the exchange of momentum between an item and an electromagnetic field is known as radiation pressure.
Photons striking the surface of the object in this instance cause a momentum shift. Any surface's susceptibility to radiation pressure depends on its makeup and the amount of light being utilized. It can be proven that the size of the total momentum provided to a surface (for complete absorption) is if the total energy transmitted to a surface is U in time "t", p = U/c.
When the sun's rays hit our hands, they become heated. This is a straightforward illustration of radiation pressure. Since the amount of momentum conveyed by electromagnetic waves is relatively small compared to the size of c, which causes your hands to warm up, you cannot feel the pressure. This occurs as a result of the hands' surface absorbing energy from the electromagnetic waves. The radiation pressure of visible light was measured in 1903 by American researchers Nicols and Hull. The outcome was 7 × 10-6 Nm-2. Therefore, based on the surface area, it is 10-2 Radiation-related force is only approximately 7 × 10-9 N.
Radiation Pressure Formula
Radiation pressure has an impact on astronomical objects. Even though we may not always feel pressure, astronomical objects like stars are very sensitive to photons since they emit vast numbers of them as radiation. When a star is in the blackbody condition, the radiation pressure is proportional to the temperature raised by the fourth power and is given by the equation:
Inside a star, the radiation pressure formula is:
p=\frac{4\sigma}{3c}T^4
Where,
- p = Radiation Pressure,
- σ = Stefan-Boltzmann constant (σ = 5.6704×10-8 W/(m2-K4),
- T = Temperature,
- c = Speed of light.
Outside a star, the radiation pressure formula is:
p=\frac{xLcos^2(\alpha)}{4\pi R^2c}
Where,
- p = Radiation Pressure,
- x = Surface type,
- L = Star luminosity,
- α = angle between a surface's surface that is either reflecting or absorbing and a light beam,
- R = Distance of Star,
- c = Speed of light.
Remember that, The formula provides Radiation Pressure,
P_{R}=\frac{(1+\alpha)I}{c}
Where,
- α = surface reflection coefficient.
Radiation Pressure is independent of incident light wavelength and depends on the type of surface onto which light is reflected.
Derivation
A fully reflecting surface experiences radiation pressure when an electromagnetic wave hits it.
Force = Rate of change of momentum
F = ΔP/Δt
F = ΔE/Δtc …(P = E/c)
F = 1/c × ΔE/Δt
∴ ΔE = FcΔt
Radiation Intensity,
I = ΔE / AΔt
I = FcΔt / AΔt
∴ I = Fc/A
∴ F/A = I/c
∴ PR = I/c (Pressure absorption)
For Totally Reflection,
PR = 2I/c
∴ PR = energy density = u
Applications of Radiation Pressure
There are some actual or at least proposed applications for radiation pressure, despite the fact that it often has a very small effect:
- Various techniques for cooling lasers use the effects of light on atomic particles.
- High-power lasers can be used to propel spacecraft using beams of energy. However, such propulsion would only offer weak forces and use a tremendous amount of electricity. As opposed to rocket propulsion, for instance, the generated momentum per kg of spent fuel (such as nuclear fuel) may be far higher. Therefore, such a strategy has been recommended for exceptionally far-off trips.
- The radiation pressure effect may measure a high-power laser beam's optical power. Despite the very low mechanical forces that must be measured, a benefit is that no significant heat power must be lost, and the optical power that must be measured does not need to be absorbed, keeping it ready for use.
Sample Questions
Question 1: Luminosity is 2 solar, distance is 4 au, and internal temperature is 4500000 K if the star's surface is opaque. Calculate the radiation pressure.
Solution:
Given: L = 2 solar, R = 4 au, T = 4500000 K, x = 1 (Surface is opaque)
Outside sun radiation Pressure,
p=\frac{xLcos^2(\alpha)}{4\pi R^2c}
p=\frac{1\times2\times cos^2(0)}{4\times3.14\times4^2\times299792458}
∴ p = 5.70
Inside sun radiation Pressure,
p=\frac{4\sigma}{3c}T^4
p=\frac{4\times5.670367\times10^-8\times(4500000)^4}{3\times299792458}
∴ p = 103 G Pa
Question 2: If the star's surface is reflective, the luminosity is 100 W, the distance is 423 m, and the internal temperature is 3000000 K. Determine the pressure of radiation outside.
Solution:
Given: L = 100 W, R = 423 m, T = 3000000 K, x = 2 (Surface is reflective)
p=\frac{xLcos^2(\alpha)}{4\pi R^2c}
p=\frac{2\times100\times cos^2(0)}{4\times3.14\times(423)^2\times299792458}
∴ p = 2.96
Question 3: The star has a luminosity of 3 solars, a distance of 4 au, and a temperature of 2200000 K if its surface is opaque. How do you calculate the internal radiation pressure?
Solution:
Given : L = 3 solar, R = 4 au, T = 2200000 K, x = 1 (Surface is opaque)
p=\frac{4\sigma}{3c}T^4
p=\frac{4\times5.670367\times10^-8\times(2200000)^4}{3\times299792458}
∴ p = 590 G Pa
Question 4: The star has an outside Radiation pressure of 80 Pa on a reflective surface, a distance is 220 m. Then calculate the luminosity.
Solution:
Given: p = 80 pa, R = 220 m, x = 2 (Surface is reflective)
L=\frac{4\pi pR^2c}{xcos^2(\alpha)}
L=\frac{4\times3.14\times80\times(220)^2\times299792458}{2\times cos^2(0)}
∴ L = 7292123 W
Question 5: If the star's surface is opaque, the luminosity is 90 W, and the distance is 300 m. Determine the pressure of radiation outside.
Solution:
Given: L = 90 W, R = 300 m, x = 1 (Surface is opaque)
p=\frac{xLcos^2(\alpha)}{4\pi R^2c}
p=\frac{1\times90\times cos^2(0)}{4\times3.14\times(300)^2\times299792458}
∴ p = 2.65
Question 6: Calculate the distance of the star when the outside radiation pressure is 139 Pa, and the luminosity is 320 W.
Solution:
Given: p = 139 Pa, L = 320 W
R=\sqrt{\frac{Lxcos^2(\alpha)}{4p\pi c}}
R=\sqrt{\frac{320\times1\times cos^2(0)}{4\times139\times3.14\times299792458}}
∴ R = 2.472
Question 7: What is a star's radiation pressure?
Answer:
Everything that comes into contact with electromagnetic radiation experiences a small force. The most massive stars are supported largely by radiation pressure against gravity, which ultimately defines the maximum mass that a star may have.
Question 8: What technique is used to gauge radiation intensity?
Answer:
You can gauge the strength of x-rays or gamma rays by counting the number of ions they produce. The amount of ionization in the air that the radiation has created is known as exposure. In science, an exposure is measured in roentgens (R or r).