19 Jan '10 04:29>
Originally posted by joe shmoI know what it means. What I meant is what is the value of t.
time, the same time that is on the other side of the equation
We know a. We don't know v or t, so your equation doesn't help.
Originally posted by AThousandYoungOr, if you prefer:
Pt = K
(m d^2/t^3)(t) = (m d^2/2t^2)
(m d^2/t^3)(t) = 1/2 m(at)^2
2(m d^2/t^3)(t)/(at)^2 = m
2P/ta^2 = m
See? The mass depends on the time given constant power and acceleration.
Originally posted by AThousandYoung"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)"
Or, if you prefer:
Pt = K
Pt = mv^2/2
P/t = mv^2/2t^2
P/t = ma^2/2
2P/ta^2 = m
Originally posted by AThousandYoungyes yours is the same as mine.
Or, if you prefer:
Pt = K
Pt = mv^2/2
P/t = mv^2/2t^2
P/t = ma^2/2
2P/ta^2 = m
Originally posted by AThousandYoungNo
Here's an interesting question.
Since P, a and m cannot all be constant, if P and m are constant the rate of acceleration will slow down proportionate to the square root of the velocity.
Is there an asymptote? Is there a point at which there is no acceleration without more power? What value is it?
Originally posted by wolfgang59T=square root of 2S/A, that says how much time it takes to go a certain distance under a certain acceleration. You don't need mass in this version. This is just using the famous S=(A*(T)^2)/2 where S= distance in whatever units, T is time in seconds, A is acceleration in the same units as S and solving for T as in the first equation. Simple stuff really.
No
If there is a power you are putting energy into the system.
in other words 1/2 m v^2 is increasing
therefore v is increasing
therefore there is acceleration no matter what the mass
or what the power
or what the velocity
(Ignoring relativistic effects as previously stated)
Perhaps an interesting problem is what is the equation of motion
(as a function of t) for a constant mass with a constant power supply?
Originally posted by sonhouseYou are assuming constant acceleration and deceleration which as you point out is a trivial problem.
T=square root of 2S/A, that says how much time it takes to go a certain distance under a certain acceleration. You don't need mass in this version. This is just using the famous S=(A*(T)^2)/2 where S= distance in whatever units, T is time in seconds, A is acceleration in the same units as S and solving for T as in the first equation. Simple stuff really.
...[text shortened]... d. And you arrive at zero relative velocity at Mars so you can actually think about landing🙂
Originally posted by wolfgang59That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.
You are assuming constant acceleration and deceleration which as you point out is a trivial problem.
My question was
[b]Perhaps an interesting problem is what is the equation of motion
(as a function of t) for a constant mass with a constant power supply?
We have from previous posts that v^2 = 2P/mt
so v ={2P/m}^0.5 * t^-0.5
define K={2p/m ...[text shortened]... 5
v of course is dx/dt so to X as a function of t we need to integrate the above
Any takers?[/b]
Originally posted by sonhouseI'll give it to you again:
That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.
Originally posted by sonhouseI gave you the answer. So did several others. We cannot give you a single numerical value because it varies with t somehow.
That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.
Originally posted by sonhouseWhy do you think a constant power supply on a constant mass will give a constant acceleration? Besides "it's obvious" or "it seems to make sense"?
That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.