Space Travel – How Fast Would a Spaceship Have to Go to Reach Alpha Centauri Within a Person’s Lifetime?

space-travelvelocity

Given the recent news about the discovery of an "Earthlike" planet orbiting Alpha Centauri (our nearest stellar neighbour) it got me wondering just how fast would spaceship have to travel to be able to reach Alpha Centauri within a person's lifetime (say 60 years)?

The reasoning is that even if we sent an unmanned probe – assuming that it would be a one way trip – the journey time would have to be sufficiently short to keep people's interest in the mission so that when it actually arrived and sent back data there would be somebody home to receive the data.

The added complication is that the craft would have to slow down sufficiently so that it could at least enter orbit around the star (I'm not going to suggest that it manages to orbit one of the planets!).

I suppose what I'm really interested in how this speed compares to the speeds we've managed to attain so far and thus get an idea of how much technology must advance for us to be able to even think of achieving this.

I know about relativistic effects due to travelling at a high percentage of the speed of light – but I'm not really interested in that here.

Best Answer

If you are not interested in relativistic effects, the answer to your question is easy to workout. According to Wikipedia, Alpha Centauri is 4.24 ly away (4.0114x$10^{16}\mathrm{m}$). So to get there in 60 years ($1892160000\mathrm{s}$).

So your non-relativistic answer is

$v = \frac{d}{t} = \frac{4.0114 \times 10^{16}}{1892160000} = 21200000 \mathrm{m}\,\mathrm{s}^{-1}$.

This is 21200 $\mathrm{km}\,\mathrm{s}^{−1}$. The fastest recored space flight was 24,791Mph which is around 11$\mathrm{km}\,\mathrm{s}^{−1}$ which is 0.05% of 21200$\mathrm{km}\,\mathrm{s}^{−1}$. This means we have to be able to get spaceships to travel 2,000 times faster than the fastest current spaceship.

Note, I believe satellites in geostationary orbits do $\approx 17\mathrm{km}\,\mathrm{s}^{−1}$.

Edit. The relativistic calculation can be found here.

Related Solutions

[Physics] When driving uphill why can’t I reach a velocity that I would have been able to maintain if I started with it

Short, short version: It's complicated.

Slightly longer version:

Internal combustion engines have at least two relevant performance characteristics: power and torque. Furthermore the maximum attainable values for both are functions of the current engine speed (RPM).

Acceleration will cease if the current requirement for either power or torque equal the engine's maximum value at the current speed.

Both the power and torque curves (as a function of RPM) start low, rise steadily and eventually turn over and drop off. The requirements for power and torque are both monotonically increasing, which means that there must be a speed where the power requirements curve crosses the power curve. At that speed acceleration drops to zero.

Likewise, there must be a place where the torque requirement crosses the available torque and again, you can't accelerate further from there.

These two crosses can come at different engine speeds.

The result is that you can may be able to maintain a speed that you can not accelerate up to.

Full version:

The full answer to your question would require knowing the relevant curves for your engine as well as the gross vehicular weight, the current effective gear ratio between the engine and the road, the slope of the road, the effective rolling friction and the car's drag coefficient. Which is why I'm not going to try to do the full version.

[Physics] How to graph acceleration

I am more familiar with graphing in Python than in the language you are using. The following snippet of code produces the graph you are asking for:

import numpy as np
import matplotlib.pyplot as plt

# sci fi velocity graph
D = 160934400000     # m
v_i = 3070           # m/s
total_time = 40*3600 # seconds: 40 hours

# to cover half the distance in half the time, we solve
# x(t) = x(0) + v_0*t + 0.5*a*t*t
# x(0) = 0
# x(t) = D/2
# t = total_time / 2 = th
th = total_time / 2
xt=D/2
a = 2.0*(xt/th-v_i)/th

print 'acceleration needed is %.2f m/s^2'%a
# this prints out 30.96 m/s^2 - encouraging

# calculate position during acceleration phase - every second:
tm = 20*3600 # time at mid point
x=[]
for t in range(0,tm+1):
    x.append(v_i*t + 0.5 * a * t * t)

xm = x[-1] # position at mid point

for t in range(tm+1, 2*tm):
    x.append((v_i + tm*a)*(t-tm) - 0.5 * a * (t - tm) * (t - tm) + xm)

time = np.arange(0,2*tm)/3600.0

plt.close('all')
plt.figure()
plt.plot(time, np.array(x)/1000)
plt.xlabel('time (hr)')
plt.ylabel('position (km)')
plt.show()

v_t = np.diff(x)
plt.figure()
plt.plot(np.array(x[:-1])/1000,v_t/1000)
plt.xlabel('position (km)')
plt.ylabel('velocity (km/s)')
plt.show()

plt.figure()
plt.plot(time[:-1], v_t/1000)
plt.xlabel('time (hr)')
plt.ylabel('velocity (km/s)')
plt.show()

This produces the following graph:

enter image description here

You can also plot velocity as a function of time:

enter image description here

Or as a function of position:

enter image description here

Best Answer

Related Solutions

[Physics] When driving uphill why can’t I reach a velocity that I would have been able to maintain if I started with it

[Physics] How to graph acceleration

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