Respuesta :
Solution :
Given data:
Area of the solar sail, A = 16 [tex]$m^2$[/tex]
Mass (array + sail), m = [tex]$3 \times 10^{-3}$[/tex] kg
Power, P = 100 GW
= [tex]$10^{11}$[/tex] W
Time, t = 10 min
= 10 x 60 s
Distance, D = [tex]$4.3 \times 4 \times 10^{16}$[/tex] m
Kinetic energy, [tex]$KE=\frac{1}{2}mv^2=P\times t$[/tex]
[tex]$v=\sqrt{\frac{2Pt}{m}}$[/tex]
[tex]$v=\sqrt{\frac{2\times 10^{11}\times 600}{3 \times 10^{-3}}}$[/tex]
[tex]$=2 \times 10^8$[/tex] m/s
So, the acceleration is [tex]$a=\frac{v}{t}$[/tex]
[tex]$a=\frac{2 \times 10^8}{6 \times 10^2} \ m/s^2$[/tex]
[tex]$=333333.33 \ m/s^2$[/tex]
Therefore, force,
[tex]$F = ma$[/tex]
[tex]$=3 \times 10^{-3}\times 333333.33$[/tex]
= 1000 N
Pressure, [tex]$P=\frac{F}{A}$[/tex]
[tex]$=\frac{1000}{16}$[/tex]
[tex]$=62.5 \ N/m^2$[/tex]
Therefore, time taken is t
Now the distance is
[tex]$d_1=\frac{1}{2}at_1^2$[/tex]
[tex]$d_1=0.5 \times 333333.33 \times (600)^2$[/tex]
[tex]$=6 \times 10^{10} \ m$[/tex]
Now, the distance, [tex]$d_2 = D-d_1=v.t_2$[/tex]
Now, [tex]$t_2=\frac{(17.2 \times 10^{16})-(6 \times 10^{10})}{2 \times 10^8}$[/tex]
= 859999700 s
Therefore, total time is
[tex]$T=t_1+t_2$[/tex]
= 86000300 s
= 27.27 years
