Solar Powered Spaceship

Firstly,
Let us calculate the power that is received by the sheet at that distance.

$P_{received} = \frac{P_{sun}}{4\pi d^2}\cdot$ Area of the sheet

Thus,
$P_{received} = \frac{4\times10^{26}}{4\pi(1.5\times10^{11})^2} (10^3)^2 = 1,414,710,605$ Watts

Now,
Since the Power is received in the form of photons ,
$P_{received} = \eta\times$ Energy of each Photon ,
where $\eta$ represents the number of photons received by the sheet per unit time

Therefore,
$\eta \cdot \frac{hc}{\lambda} = 1,414,710,605$

Also, note that the Change in Momentum of each photon as it reflects off the sheet is $2p$

Hence,
Force $= \frac{dp_{total}}{dt} = \eta \times 2p = \eta \times \frac{2h}{\lambda}$

$\Rightarrow$ Force $= 1,414,710,605 \times \frac{\lambda}{hc} \times \frac{2h}{\lambda}$

$\Rightarrow$ Force $= \frac{1,414,710,605\times2}{3\times10^8} \approx \boxed{9.43N}$

Notice that the particle absorbs the radiation, then reflecting it back. Thus the change in momentum is $\Delta p = \frac{2 \Delta U}{c}.$ Since the change in energy is equal $P\Delta t$ , where $P$ is the power of the sun, we get $\Delta U = AI\Delta t,$ where A and I denote area of the sail and the intensity of the sun. We know $F\Delta t = \Delta p$ , consequently, $F = \frac{2AI}{c}.$

The solar power per unit area when it reach Earth and the solar sail:

$\Delta P_s = \cfrac{P_s}{4\pi R^2} = \cfrac{4\times 10^{26}}{4\pi (1.5\times 10^{11})^2}=1414.710605W/m^2$

The radiation power collected by the 1-km square solar sail:

$P_{sail} = \Delta P_sA = 1414.710605 \times 10^6 = 1414710605W$

Since all the radiation power is converted to mechanical power and that $E=\cfrac{hc}{\lambda}=cp \Rightarrow$ momentum $p=\cfrac{E}{c}$ , then the force:

$F = \cfrac{dp}{dt} = 2\times \cfrac{P_{sail}}{c}= 2\times \cfrac{1414710605}{300000000}= \boxed{9.43N}$