Answer
a) $9\times10^{-6}\;\rm N/m^2$
b) $9\times10^{-3}\;\rm m/s^2$
c) $1.25\times10^{4}\;\rm m^2$
Work Step by Step
a) We know that the radiation pressure, with total reflection, is given by
$$P=\dfrac{2I_{avg}}{c}$$
Plugging the known;
$$P=\dfrac{2\cdot 1350}{3\times10^8}=\color{red}{\bf 9\times10^{-6}\;}\rm N/m^2$$
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b) We know that the force due to pressure is given by $P=\dfrac{F}{A}$, so
$$F=PA$$
and according to Newton's second law, the net force exerted on an object is given by $F=ma$, so
$$ma=PA\tag 1$$
Solving for $a$;
$$a=\dfrac{PA}{m}=\dfrac{\frac{PA}{A}}{\frac{m}{A}}=\dfrac{P }{\frac{m}{A}}$$
Plugging the known;
$$a=\dfrac{9\times10^{-6}}{1\times 10^-3 }=\color{red}{\bf9\times10^{-3}}\;\rm m/s^2$$
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c) We need to find the area of the solar sail, so we need to use the pressure law we used above in (1).
$$ma=PA$$
Solving for $A$;
$$A=\dfrac{ma}{P}=\dfrac{(m_{s}+m_p)a}{P}$$
Noting that the mass $m_s$ is the mass of the solar sail and the mass $m_p$ is the mass of the payload.
where $m_p=100$ kg and $m_s=10^{-3}A $
$$A= \dfrac{(10^{-3}A+100) a}{P}$$
Thus,
$$PA= 10^{-3}Aa+100a $$
$$PA- 10^{-3}Aa=100a $$
$$A\left[P - 10^{-3} a\right]=100a $$
$$A =\dfrac{100a }{P - 10^{-3} a} $$
Plugging the known;
$$A =\dfrac{100a }{P - 10^{-3} a} $$
Plugging the known;
$$A =\dfrac{100\cdot 1\times10^{-3} }{(9\times10^{-6}) - (10^{-3} \cdot 1\times10^{-3})} $$
$$A =\color{red}{\bf 1.25\times10^{4}}\;\rm m^2$$
