Accelerating into space for a day.

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I wish to accelarete out into space at 9.8m/s/s for 1 day. The reason I want to accelerate at 9.8m/s/s is because I want to feel the normal effects of gravity as I travel. After accelerateing for one day, I want to look out my rocket window, and take a snap shot of our sun, then I want to turn my rocket around and accelerate back towards earth for one day.

I would like to know if such a trip is possible?

I have heard that the faster you travel the heavier you become so hence the more energy you need to maintain a constant accelaration. Does this mean that it would be impossible to maintain a constant accelaration in a straight line for one day.

If it is possible, what would my picture of the sun look like? Thanks
 

Answers and Replies

  • #2
berkeman
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Well, you would need to feel greater than 9.8m/s^2 acceleration at first, so that you could lift off and get away from the Earth. Say you use 2g acceleration until you get a few 100km away, and then throttle back to 1g for the rest of the day.

Use the kinematic equation <<Totally incorect equation subsequently EDITed out by berkeman>> to calculate what your velocity will be versus time. What does your max speed turn out to be? What fraction of the speed of light is it? Unless you are moving at a large fraction of the speed of light, you don't need to worry about mass changing. Quiz question -- how long do you have to accelerate at 1g to get to 0.5c?
 
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  • #3
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Well, i thought after 1 day you would be travelling much faster than the speed of light.

v = 4.9 * Seconds in one day^2
= 4.9 * 86400^2
= 36,578,304,000 m/s >> c = 299,792,458 m / s

The above equation would imply that you would reach the speed of light after 2 and a bit hours.
 
  • #4
cristo
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Use the kinematic equation [tex]v = v_0 + \frac{1}{2}a t^2[/tex] to calculate what your velocity will be versus time...
Is the equation not [tex]v = v_0 + a t[/tex]? The kinematic eqn for displacement has (1/2)at^2 in.
 
  • #5
D H
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v = 4.9 * Seconds in one day^2
You are applying the wrong equation, and the "right" equation is wrong as well. You are using the equation for distance, not velocity. Velocity, as pointed out in the previous post, is [itex]v = v_0 + a*t[/itex]. But even that equation is "wrong"; it violates the speed of light. The answer is in relativity.

Assuming you have an incredibly capable spacecraft equipped with an accelerometer and a thrust controller that keeps the accelerometer pegged at 9.8 m/s^2. You can operate your spacecraft forever, constantly feeling a 1G acceleration, and yet any observer will see you as traveling slower than the speed of light.
 
  • #6
berkeman
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Is the equation not [tex]v = v_0 + a t[/tex]? The kinematic eqn for displacement has (1/2)at^2 in.
Oopsies. I have to be more careful. Sorry about that!!:blushing:
 
  • #7
berkeman
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Thanks for catching my error. I was wondering why it took such a short time to get up to relativistic speeds.

Using the correct equation and assuming just a constant 1g accleration in free space, you'd get up to 0.1c in about 8300 hours. Assuming I didn't get the math wrong too.
 
  • #8
Janus
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You are applying the wrong equation, and the "right" equation is wrong as well. You are using the equation for distance, not velocity. Velocity, as pointed out in the previous post, is [itex]v = v_0 + a*t[/itex]. But even that equation is "wrong"; it violates the speed of light. The answer is in relativity.
But in this case, the difference between the correct relativistic answer and the "wrong" answer only differ by about 2 meters/sec.
 
  • #9
DaveC426913
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I just did this calc a couple of days ago. I got 90% of c after 312 days of 1g acceleration.
 
  • #10
russ_watters
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Well, i thought after 1 day you would be travelling much faster than the speed of light.

v = 4.9 * Seconds in one day^2
= 4.9 * 86400^2
= 36,578,304,000 m/s >> c = 299,792,458 m / s

The above equation would imply that you would reach the speed of light after 2 and a bit hours.
v=a*t, not a*t^2. After 1 day, you are going 800,000 m/s - less than 1/300th the speed of light. At that low speed, relativity doesn't have a noticeable impact.
 
  • #11
D H
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But in this case, the difference between the correct relativistic answer and the "wrong" answer only differ by about 2 meters/sec.
Yes, but I intuited that the OP posed the question assuming that uniform acceleration would somehow result in a speed that exceeds that of light if the uniform acceleration applied for a long enough time.

The relevant equation is the relativistic rocket equation,

[tex]\frac v c = \tanh^{-1}(\frac {a \tau}c)[/tex]

where [itex]\tau[/itex] is the proper time of the time interval over which the acceleration occurs.

It takes 520.7 days (as measured on the rocket) to get to 0.9c accelerating constantly at 9.80665 m/s^2.
 
  • #12
George Jones
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The relevant equation is the relativistic rocket equation,

[tex]\frac v c = \tanh^{-1}(\frac {a \tau}c)[/tex]

where [itex]\tau[/itex] is the proper time of the time interval over which the acceleration occurs.

It takes 520.7 days (as measured on the rocket) to get to 0.9c accelerating constantly at 9.80665 m/s^2.
I think that you didn't mean to use an inverse.
 
  • #13
D H
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You're right; no inverse.
 
  • #14
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DH- Thanks, thats an interesting equation.
Is the proper time, the time elasped for a person in the rocket, or for a person on earth?

So if I accelerated straight out into space, for 2 years at 1g, then turned around and came back traveling at 1 g, then how could you calculate how many days had elapsed on earth?
 
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  • #15
George Jones
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DH- Thanks, thats an interesting equation.
Is the proper time, the time elasped for a person in the rocket, or for a person on earth?
A person on Earth.

So if I accelerated straight out into space, for 2 years at 1g, then turned around and came back traveling at 1 g, then how could you calculate how many days had elapsed on earth?
For these types of scenarios, I like to use use ([itex]y[/itex]) as the unit of time and lightyears ([itex]ly[/itex]) as the unit of distance. Then, the speed of light is [itex]c = 1 ly/y[/itex], and [itex]g = 1.032 ly/y^2 [/itex].

Then, in the frame of the Earth (with, for convenience, the Earth at x = 1 ly) the, the ship's position [itex]x[/itex] and time [itex]t[/itex] as functions of the ship's clock's (proper) time [itex]T[/itex] are given by

[tex]t = a^{-1} \mathrm{sinh}(aT)[/tex]

and

[tex]x = a^{-1} \mathrm{cosh}(aT).[/tex]

D H's equation for speed is just [itex]dx/dt[/itex], which can be calculated by differentiating the expressions for [itex]x[/itex] and [itex]t[/itex] with respect to [itex]T[/itex] and then using the chain rule.

For a round-trip voyage, the ship should, at the point halfway to the destination, fire engines so as to "decelerate" at [itex]g[/itex]. The ship will momentarily stop at the destination, and then start accelerating towards Earth. At the hafway point, the ship should again reverse firing direction, so that it will stop when it reaches Earth.

Then, the relationship between [itex]t[/itex] and [itex]T[/itex] for the entire trip is given by

[tex]t = 4 a^{-1} \mathrm{sinh}(aT/4).[/tex]
 
  • #16
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damoclark said:
DH- Thanks, thats an interesting equation.
Is the proper time, the time elasped for a person in the rocket, or for a person on earth?
A person on Earth.
Isn't this a little simplistic?

Proper time between "person on Earth starts clock" and "person on Earth stops clock" is the time elapsed for the person on Earth.

But for the person (or clock) in the rocket, proper time is time elapsed for a person (or clock) in the rocket.

(from http://en.wikipedia.org/wiki/Proper_time" [Broken]: "In relativity, proper time is time measured by a single clock between events that occur at the same place as the clock. It depends not only on the events but also on the motion of the clock between the events.")

Isn't it?

cheers,

neopolitan
 
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  • #17
DaveC426913
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2 1/2 year necroposty. Cool.
 

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