How many days in the robotic probe's time frame is needed to stop in Proxima Centauri from Earth?

Classical Mechanics Level 2

This problem's question: **How many days in the robotic probe's time frame is needed to stop in Proxima Centauri from Earth?" The time unit is days (of 86400 seconds).

A number of assumptions will be made that make this just a mathematical problem.

The speed of light is $299792458\, m/s$ . This is, in fact, the current definition.

The robotic probe is capable of immediately changing acceleration quantities without any other effects. This is not possible, The probe either is not accelerating or is accelerating at $1 km/s^2$ . The energy source that makes this possible is just assumed. The acceleration can be maintained without a lime limit. This is also just assumed.

The distance to Proxima Centauri is assumed to be exactly 4.244 lightyears (converted to kilometers: $40\,151\,340\,125\,632.9$ ).

It is assumed that Accelerated motion and special relativity transformations by A A Ketsaris is correct. The formula needed is within this paper at equation (3). You will need to apply a correction similar in spirit to the "+ constant" of an indefinite integral so that no distance is traveled at robotic time frame 0 seconds. The formula can be used to develop that correction.

It is assumed that correct answer is twice the time to travel half of the distance. Please, feel free to comment about this assumption against my offered solution. Remember the answer is in the robotic probe's frame and not Earth's where the answer remains about four years and three months.

The answer is 42.3765480425859.

1 solution

A Former Brilliant Member
Jun 2, 2019

The Rindler distance function is $r:\,(a,t)\to \frac{c^2}{a} \cosh{\frac{a\,t}{c}}$ .

The correction factor is $r(0\,\text{s},1\,\frac{\text{km}}{\text{s}}) \Rightarrow 8.98755178736818\times10^{13}\,\text{m}$ .

The Rindler distance function with the correction applied is:

$x:\,(a,t)\to r(a,t)- r(0\, \text{s},a)$

A numerical search for the solution for the robotic probe time for one half the distance:

$\text{FindRoot}[\text{QuantityMagnitude}[\text{UnitConvert}[x(t\,\text{days},a),\text{LightYears}]]=2.122,\{\text{days},7,60\}] \Rightarrow t\to 21.1882740212929$ .

The Wolfram Mathematica procedure FindRoot works on numbers and not quantities. Therefore the output of the $x$ function needs to be converted to lightyears and the number extracted from the quantity.

$2\times 21.18827402129294 \Rightarrow 42.3765480425859$ days.

How many days in the robotic probe's time frame is needed to stop in Proxima Centauri from Earth?

The answer is 42.3765480425859.

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