Question

To simulate artificial gravity, a spaceship accelerates through space at 9.8 m/s2 . a) If it starts from a stop, how long (in time) will it take to reach a speed of 0.1c where c is 3 x 108 m/s (the speed of light). b) How much space does it cover during this time?

Answer 1

Answer:

(a) Time taken to reach the speed will be $0.306\times 10^7sec$

(b) Space cover by spaceship $0.9183\times 10^{14}m$

Explanation:

We have given acceleration due to gravity $g=9.8m/sec^2$

As the spaceship starts from rest so initial velocity u = 0 m/sec

Final velocity v = 0.1 c, here c is speed of light

So final velocity $v=3\times 10^7m/sec$

(a) According to first law of motion we know that v=u+gt, here v is final velocity, u is initial velocity and t is time

So $3\times 10^7=0+9.8\times t$

$t=0.306\times 10^7sec$

(b) According to third law of motion

$h=ut+\frac{1}{2}gt^2=0\times .306\times 10^7+\frac{1}{2}\times 9.8\times (0.306\times 10^7)^2=0.9183\times 10^{14}m$

Answer 2

Answer:

The spaceship will take 850.34 hours to reach a speed of 0.1c and it will travel 45 918 367 346.9 km before reaching that speed.

Explanation:

The spaceship is traveling at a constant acceleration, then we can use the kinematic equations for that kind of movement to calculate the time it would take to reach a given speed.

$s=s_0+at$

Where $s_0$ is the initial speed, $a$ is the acceleration rate and $s$ is the speed the spaceship will have at a time $t$.

If the spaceship starts from a stop, then $s_0=0$. So:

$s=at$

$t=\frac{s}{a}=\frac{0.1c}{9.8\frac{m}{s^2}}=\frac{0.1(3*10^8\frac{m}{s})}{9.8\frac{m}{s^2}}=3061224.48 s$

$t=3061224.48s*\frac{1 min}{60 s} *\frac{1 hour}{60 min} =850.34hours$

Then, the spaceship will take 850.34 hours to reach a speed of 0.1c

To calculate the displacement of the spaceship during that time, we use the following equation:

$x=s_0t+\frac{1}{2}at^2$

Where $s_0$ is again the initial speed (which is zero), $a$ is the acceleration and $t$ is the travel time (which we've calculated is the previous step).

$x=s_0t+\frac{1}{2}at^2=\frac{1}{2}at^2=\frac{1}{2}(9.8 \frac{m}{s^2})(3061224.48s)^2=4.59*{10}^{13} m$

$x=4.59*{10}^{13} m * \frac{1 km}{1000m}=45918367346.9km$

Then, the spaceship will travel 45 918 367 346.9 km before reaching a 0.1c speed.