The intensity of the Sun's light in the vicinity of the earth is about `1000 W//m^2` .Imagine a spacecraft with a mirrored square sail of dimension 1.0 km. Estimate how much thrust (in newtons) this crafts will experience due to collisions with the Sun's photons

Explanation

To estimate the thrust experienced by a spacecraft with a mirrored square sail due to collisions with the Sun's photons, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Area of the Sail:**
The dimensions of the square sail are given as 1.0 km x 1.0 km.
\[ \text{Area} (A) = \text{side}^2 = (1.0 \, \text{km})^2 = (1000 \, \text{m})^2 = 10^6 \, \text{m}^2 \]

2. **Calculate the Power Incident on the Sail:**
The intensity of sunlight is given as \(1000 \, \text{W/m}^2\). The power (\(P\)) incident on the sail can be calculated using the formula:
\[ P = \text{Intensity} \times \text{Area} = 1000 \, \text{W/m}^2 \times 10^6 \, \text{m}^2 = 10^9 \, \text{W} \]

3. **Relate Power to the Number of Photons:**
The energy of a single photon is given by:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) and \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)). The number of photons (\(N\)) striking the sail per unit time can be expressed as:
\[ P = \frac{dE}{dt} = N \cdot \frac{hc}{\lambda} \]

4. **Calculate the Momentum of a Photon:**
The momentum (\(p\)) of a single photon is given by:
\[ p = \frac{h}{\lambda} \]

5. **Calculate the Change in Momentum:**
When a photon strikes the mirrored sail and reflects back, the change in momentum is:
\[ \Delta p = 2p = 2 \cdot \frac{h}{\lambda} \]

6. **Relate Change in Momentum to Thrust:**
The thrust (\(F\)) experienced by the sail due to the change in momentum can be expressed as:
\[ F = \frac{d(\Delta p)}{dt} = \frac{dN}{dt} \cdot \Delta p = \frac{dN}{dt} \cdot 2 \cdot \frac{h}{\lambda} \]

7. **Substituting Power into the Thrust Equation:**
Since power is also related to the number of photons, we can express thrust as:
\[ F = \frac{2P}{c} \]
where \(c\) is the speed of light.

8. **Calculate the Thrust:**
Substituting the values we have:
\[ F = \frac{2 \cdot 10^9 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = \frac{2 \cdot 10^9}{3 \times 10^8} \approx 6.67 \, \text{N} \]

### Final Answer:
The thrust experienced by the spacecraft due to collisions with the Sun's photons is approximately **6.67 Newtons**.