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An idea to explore the nearest Earth-sized planet in a habitable zone to our Sun is to attach a small sensor array to a 4 m by 4 m solar sail and shine a 100 GW laser on the solar sail for 10 minutes. This will get the spaceship up to speed after which it will coast the rest of the way at the speed attained after the 10 minutes of acceleration. Assume the sensor array and solar sail has a mass of 3 grams. The nearest Earth-sized planet in a habitable zone orbits the Alpha Centauri system at 4.3 Lightyears. 1 GW = 1*109 W. 1 Light year = 4*1016 m. Assume a perfectly reflecting solar sail and assume all 100 GW of power are intercepted by the solar sail.

a. What is the pressure on the Solar Sail from the laser?
b. What is the force on the Solar Sail from the laser?
c. What speed will the spaceship attain after 10 minutes assuming it started from rest and assuming a constant acceleration?
d. How long will it take the spaceship to reach Alpha Centauri in years?

User MeltingDog
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1 Answer

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Solution :

Given data:

Area of the solar sail, A = 16
$m^2$

Mass (array + sail), m =
$3 * 10^(-3)$ kg

Power, P = 100 GW

=
$10^(11)$ W

Time, t = 10 min

= 10 x 60 s

Distance, D =
$4.3 * 4 * 10^(16)$ m

Kinetic energy,
$KE=(1)/(2)mv^2=P* t$


$v=\sqrt{(2Pt)/(m)}$


$v=\sqrt{(2* 10^(11)* 600)/(3 * 10^(-3))}$


$=2 * 10^8$ m/s

So, the acceleration is
$a=(v)/(t)$


$a=(2 * 10^8)/(6 * 10^2) \ m/s^2$


$=333333.33 \ m/s^2$

Therefore, force,


$F = ma$


$=3 * 10^(-3)* 333333.33$

= 1000 N

Pressure,
$P=(F)/(A)$


$=(1000)/(16)$


$=62.5 \ N/m^2$

Therefore, time taken is t

Now the distance is


$d_1=(1)/(2)at_1^2$


$d_1=0.5 * 333333.33 * (600)^2$


$=6 * 10^(10) \ m$

Now, the distance,
$d_2 = D-d_1=v.t_2$

Now,
$t_2=((17.2 * 10^(16))-(6 * 10^(10)))/(2 * 10^8)$

= 859999700 s

Therefore, total time is


$T=t_1+t_2$

= 86000300 s

= 27.27 years

User Pbies
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