An object of mass 1 kg at rest experiences a force of 1 N in space with no resistance force. What will be the speed of the object after 20 years?

If I go with the high school physics. F=ma ; gives the acceleration of 1 m/s^2. And v = u + at; gives me v = 0 + 1 * 20 * 365 * 24 * 60 * 60 = 6.3 x 10^8 m/s (which is ~2 times speed of light) I am sure something is wrong here, but I am not able to find the theory behind it.

  • $\begingroup$ If I go with the high school physics. F=ma ; gives the acceleration of 1 m/s^2. And v = u + at; gives me v = 0 + 1 * 20 * 365 * 24 * 60 * 60 = 6.3 x 10^8 m/s (which is ~2 times speed of light) I am sure something is wrong here, but I am not able to find the theory behind it. $\endgroup$ – Raj Feb 20 at 1:20
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    $\begingroup$ The theory is the theory of Special Relativity. That is the theory used to calculate my answer below. $\endgroup$ – Dale Feb 20 at 1:28
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  • $\begingroup$ You should check out the article on the Relativistic Rocket that John Rennie mentions in the question linked by Fellow Traveller. $\endgroup$ – PM 2Ring Feb 20 at 7:19

Assuming that the force and the time are as measured by the object then a 1 m/s^2 proper acceleration for 20 years of proper time gives a velocity of 0.78 c in the inertial frame where the object was initially at rest.

No matter the force nor the duration you will never get a velocity greater than c.

See here for a convenient calculator: https://www.omnicalculator.com/physics/space-travel

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This is because Newtonian physics, such as $F=ma$, is only accurate for small masses and low speeds.

As mass and/or speed approaches astronomical scales, Length Contraction (altering perceived lengths) and Time Dilation (altering perceived time) become significant factors in any calculations involving distances and times, and the observed conditions from the static reference frame and from the moving reference frame begin to diverge.

Looking at this situation while taking only time dilation into account is a relatively simple way of approaching the question. Assume that your one-kilogram object is an analogue clock with a seconds-hand, and you are a static observer watching the clock with a telescope as it accelerates away from you. From your perspective, the clock's hands will appear to slow down so that the seconds-hand takes more and more time to reach the next tic mark. The apparent increase in duration can be calculated with the function

$$ Dilation \ Factor = \frac{1}{\sqrt {1 - \frac{v²}{c²}}} $$ where $c$ is the speed of light and $v$ is the velocity as a percentage of $c$.

For simplicity, set the speed of light to $1$ and let $v$ be any value between $0$ (inclusive) and $1$ (exclusive). When you observe the clock to be moving at 10% the speed of light, $v$ is $0.1$ and the apparent time dilation is $\approx 1.005$ ⁠— that is, you will have to wait ≈1.005 of your seconds for the clock to show one second. To get a dilation effect of 1.1, the clock needs to move 41.66% of lightspeed.

If you measure the moving clock's acceleration using the time displayed on the moving clock itself, the calculated acceleration will always stay constant. But if you measure it using the time displayed on a static clock next to you, the acceleration will appear to decrease as the inverse of the time dilation factor.

At 41.66% of lightspeed, you will see the clock take 1.1 seconds to accelerate the same amount as it did when it first began accelerating.
At 74.55% it will take 1.5 seconds.
At 86.6% it will take 2 seconds, appearing to accelerate at only half of the initial rate.
At 99.98611%, it will take a full minute - an apparent acceleration of one sixtieth of the initial rate.
At 99.9998%, one second on the moving clock will take five hundred seconds by your clock.

You can probably see where this is going.

The faster the clock is moving, relative to a static reference frame, the more time it takes to gain another meter-per-second of speed. As its apparent speed becomes infinitely close to lightspeed, its apparent acceleration becomes infinitely close to zero, and the lightspeed barriar will never be broken.

You can reach the same end result by observing time dilation from the clock's accelerating reference frame instead of the static observer's reference frame, or by observing length contraction from either reference frame. In all cases, one of the values in $acceleration = \frac {distance} {time^2}$ is being altered and one is held fixed, determining the value for the third.

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  • $\begingroup$ I edited it but made a mistake, the x should still be squared. $\endgroup$ – bemjanim Feb 20 at 12:26
  • $\begingroup$ The c should not be there $\endgroup$ – bemjanim Feb 20 at 12:58
  • $\begingroup$ @bemjanim I'm getting that equation from the book "Dynamics and Relativity" by Jeffrey Forshaw and Gavin Smith, section 6.1 "Time Dilation, Length Contraction, And Simultaneity" - books.google.com/… The $c$ goes away during simplification if you set it to 1 and $x$ is a percentage of lightspeed, but by including the $c$ term the equation also works if you set $c$ to 299,792,458 meters per second and $x$ is the object's speed in meters per second. $\endgroup$ – Lawton Feb 20 at 15:33
  • $\begingroup$ In the book it says $\frac{v^2}{c^2}$. By definition, $x=\frac{v}{c}$. It makes no sense to add a quantity with units to a pure number, so the c can't belong there $\endgroup$ – bemjanim Feb 20 at 15:53
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    $\begingroup$ That looks good, I'm sorry for making a fuss about a minor issue but your notation confused me. $\endgroup$ – bemjanim Feb 20 at 18:12

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