How Does Solar Radiation Pressure Affect an Object in Space?

In summary, the conversation discusses the calculation of radiation pressure exerted on an object at a distance of 1 million kilometers from the surface of the sun. The formula for calculating pressure is provided, along with the explanation that the flux decreases as 1/R^2. The temperature and radius of the sun are also mentioned, as they are needed for an alternative formula to calculate flux using black body radiation. The conversation ends with the suggestion to use the formula P = 2*W/c to calculate force.
  • #1
mithil03
9
0
The question I have is:
Assume a 100 kg reflective load with 1m2 area at a distance of 1 million kilometers from
the surface of the sun. If the sun’s temperature is 6000K and radius is 695,000 km, find
the radiation pressure exerted on the object. Find how far the load would be from its
initial position after 10 days assuming it started at rest.

I went through the internet and found that
p= W (1+p)/c
Where p – pressure
c – speed of light, 3 ·108 m/sec
W – radiation flux (near the Earth ~1400 W/m2, so-called solar constant
p – overall surface reflectance (from 0 to 1)

But i couldn't relate the Flux value for other distances, and Why would they give the Temperature and radius of the sun ?
Help please..
 
Physics news on Phys.org
  • #2
Ok, flux drops as 1/R^2, where R is the distance from the sun. So in order to find flux at some distance R, you can take flux at Earth, WE, and use this formula:

[tex]W_R = W_E\frac{R_E^2}{R^2}[/tex]

Where RE is distance of Earth from the Sun.

The reason they give temperature and radius of the Sun is because there is another way to find the flux using black body radiation.

[tex]W_R = \sigma T^4 \frac{R_S^2}{R^2}[/tex]

Where T is temperature of the Sun, RS is radius of the Sun, R is the distance at which you want to find flux, and sigma is Stefan-Boltzman constant:

[tex]\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}[/tex]
 
Last edited:
  • #3
Thanks a lot K^2. Understood this now. Could you help me with the next part of the problem please..
 
  • #4
You already have the formula. Force is pressure times area of the reflector. I'm guessing reflectivity is assumed to be close to 1, so your formula reads P = 2*W/c. That's all you need.
 
  • #5


The sun's temperature and radius are important factors in calculating solar radiation pressure. This is because the intensity of solar radiation decreases with distance from the sun due to the inverse square law. This means that the further away an object is from the sun, the less intense the solar radiation will be. The temperature and radius of the sun are needed to calculate the radiation flux, which is the amount of solar radiation that reaches a given area at a certain distance from the sun.

To calculate the radiation pressure exerted on the object, we can use the formula p = W(1+p)/c, where p is the pressure, c is the speed of light, W is the radiation flux, and p is the overall surface reflectance. In this case, we know that the object has a mass of 100 kg and a surface area of 1m2, and is located at a distance of 1 million kilometers from the sun's surface. We also know that the solar constant, or radiation flux near Earth, is approximately 1400 W/m2.

Using these values, we can calculate the radiation pressure exerted on the object as follows:

p = (100 kg)(1400 W/m2)(1+1)/ (3 x 108 m/s)
= 0.0000004667 N/m2

This is a very small amount of pressure, but it is still significant enough to cause movement in the object over time.

To calculate how far the load will be from its initial position after 10 days, we can use the equation d = 0.5at2, where d is the distance, a is the acceleration (in this case, the radiation pressure), and t is the time. We know that the object starts at rest, so the initial velocity is 0.

Plugging in the values, we get:

d = 0.5(0.0000004667 N/m2)(10 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute)2
= 0.0006024 meters

Therefore, after 10 days, the object will have moved approximately 0.0006 meters away from its initial position due to solar radiation pressure. This may seem like a small distance, but it is important to consider that this is only for a 100 kg object with a 1m2 surface area. For larger objects or objects closer to the sun, the effect of solar radiation pressure
 

Related to How Does Solar Radiation Pressure Affect an Object in Space?

1. What is solar radiation pressure?

Solar radiation pressure is the force exerted by sunlight on an object. It is caused by the transfer of momentum from the photons in sunlight to the surface of the object.

2. How does solar radiation pressure affect space objects?

Solar radiation pressure can have a significant impact on the orbits and trajectories of space objects. It can cause changes in orbital speed, altitude, and orientation, and can even lead to the loss of control of a space object.

3. Can solar radiation pressure be used for propulsion?

Yes, solar radiation pressure can be harnessed for propulsion by using large, reflective sails to capture and reflect sunlight. This can provide a continuous thrust, allowing for long-distance space travel without the need for traditional rocket engines.

4. How is solar radiation pressure measured?

Solar radiation pressure is typically measured using a device called a radiometer, which detects the amount of radiation being absorbed and reflected by an object. It can also be calculated using the object's surface area, reflectivity, and the intensity of sunlight at its location.

5. How can solar radiation pressure be mitigated?

There are several techniques that can be used to mitigate the effects of solar radiation pressure on space objects. These include using special coatings to reduce the object's reflectivity, using attitude control systems to maintain a specific orientation, and designing objects with stronger structural materials to withstand the force of solar radiation pressure.

Similar threads

  • Introductory Physics Homework Help
2
Replies
35
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
4K
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top