Constant 1 g acceleration problem

[kelmcguir]

Multipart question:

1) How long would it take to travel to the moon from Earth at a constant 1 g acceleration?

2) How does fuel consumption work with constant 1 g acceleration or acceleration in general? I read that "it doesn't take a constant amount of energy to maintain a constant acceleration...it takes twice as much energy to go from 1 to 2 m/s as it does to go from 0 to 1 m/s". Is this true and why?

 

[PeroK]

Multipart question:

1) How long would it take to travel to the moon from Earth at a constant 1 g acceleration?

2) How does fuel consumption work with constant 1 g acceleration or acceleration in general? I read that "it doesn't take a constant amount of energy to maintain a constant acceleration...it takes twice as much energy to go from 1 to 2 m/s as it does to go from 0 to 1 m/s". Is this true and why?

I guess you are not familiar with the basic kinematic equations:
$$x = ut + \frac 1 2 at^2, \ \ \text{and} \ \ KE = \frac 1 2 mv^2$$
The first tells you how far you have traveled ($x$) in time $t$ with initial velocity $u$ and constant acceleration $a$. If the initial speed is zero ($u = 0$) then it's just $x = \frac 1 2 at^2$.

Could you use that to answer 1) for yourself? You could look up the value of $g$ and the distance to the Moon.

The second tells you the kinetic energy (KE) of an object of mass $m$ moving with speed $v$. You can see that KE increases with the square of the velocity. Hence it takes four times as much energy to go from $1$ to $2m/s$ as it does to go from $0$ to $1m/s$.

PS Although, in terms of space travel and rocket engines, things are not so simple.

 

[miltos]

How long would it take to travel to the moon from Earth at a constant 1 g acceleration?

About 2h and 25minutes.

Is that answer satisfying?

 

[Drakkith]

2) How does fuel consumption work with constant 1 g acceleration or acceleration in general?

Conventional space travel involves ejecting mass (propellant) out the back of a rocket engine, and you have to carry this mass with you. So a spacecraft would actually use less and less fuel to maintain a constant acceleration since its mass is falling as it expends propellant. Less mass requires less force to accelerate and thus less propellant expenditure.

 

[etotheipi]

I guess the $1\text{g}$ here is a coordinate acceleration, because I'm undergoing $1\text{g}$ g-force/proper acceleration right now and I'm not getting any closer to the moon

 

[A.T.]

1) How long would it take to travel to the moon from Earth at a constant 1 g acceleration?

If the constant 1g is measured by an acceloromter on board, then you never leave the launch pad.

 

[jbriggs444]

Conventional space travel involves ejecting mass (propellant) out the back of a rocket engine, and you have to carry this mass with you. So a spacecraft would actually use less and less fuel to maintain a constant acceleration since its mass is falling as it expends propellant. Less mass requires less force to accelerate and thus less propellant expenditure.

Well, there is good news and bad news. The good news is that it takes less and less force as you burn fuel. The bad news is that for a fixed payload, you need to have started with an exponentially large tank full.

 

[PeroK]

Well, there is good news and bad news. The good news is that it takes less and less force as you burn fuel. The bad news is that for a fixed payload, you need to have started with an exponentially large tank full.

The really bad news is that you'll crash into the Moon at about $90,000m/s$.

 

[kelmcguir]

Conventional space travel involves ejecting mass (propellant) out the back of a rocket engine, and you have to carry this mass with you. So a spacecraft would actually use less and less fuel to maintain a constant acceleration since its mass is falling as it expends propellant. Less mass requires less force to accelerate and thus less propellant expenditure.

But PeroK made an interesting point above. "The second tells you the kinetic energy (KE) of an object of mass m moving with speed v. You can see that KE increases with the square of the velocity. Hence it takes four times as much energy to go from 1 to 2m/s as it does to go from 0 to 1m/s." So, even though you are losing mass, wouldn't you still need more fuel for every increase in velocity gained during the acceleration because more energy is required for every step in velocity increase ?

 

[PeroK]

But PeroK made an interesting point above. "The second tells you the kinetic energy (KE) of an object of mass m moving with speed v. You can see that KE increases with the square of the velocity. Hence it takes four times as much energy to go from 1 to 2m/s as it does to go from 0 to 1m/s." So, even though you are losing mass, wouldn't you still need more fuel for every increase in velocity gained during the acceleration because more energy is required for every step in velocity increase ?

That's for a fixed mass. If you start ejecting mass and reducing the mass of the object you call your rocket, then it's not so simple.

 

[jack action]

2) How does fuel consumption work with constant 1 g acceleration or acceleration in general? I read that "it doesn't take a constant amount of energy to maintain a constant acceleration...it takes twice as much energy to go from 1 to 2 m/s as it does to go from 0 to 1 m/s". Is this true and why?

You are missing the concept of power somewhere. Let's imagine you are driving a car with negligible resistance (aerodynamic drag & rolling resistance) and with negligible fuel mass with respect to the car mass $m$. The car is going from point A to point B, separated by a distance $x$, at a constant acceleration $a$ and the trip lasts 60 minutes in total.

How much energy is required for the trip? That would be based on the required work:
$$E = max$$
Time, thus speed, has no say in this. Twice the acceleration means twice the required energy. And twice the distance traveled means twice the energy needed too. So half-way on our trip we spent half of our energy (i.e. fuel):
$$E_{@x_{\frac{1}{2}}} = ma\frac{x}{2}$$
But how much time does it take to travel half-way? We know it takes 60 minutes for the total trip, so this holds up:
$$t = \sqrt{\frac{2x}{a}}$$
and half-way, it will be:
$$t_1 = \sqrt{\frac{2\frac{x}{2}}{a}} = \sqrt{\frac{x}{a}} = \frac{t}{\sqrt{2}}$$
If $t$ is 60 minutes, then it will take 42 minutes to travel the first half ($t_1$) and 17 minutes to travel the second half ($t - t_1$). This is true because during the second half we are driving a lot faster.

What is not constant throughout our trip is the power $P$ needed which, for a certain period of time, averages to:
$$P = \frac{E}{t}$$
For both halves of the trip, we have the same value for $E$. But the time elapsed is different:
$$\frac{P_2}{P_1} = \frac{t_1}{t - t_1} = \frac{1}{\frac{t}{t_1} - 1} = \frac{1}{\sqrt{2} - 1} = 2.4$$
So, if you need, say, 100 hp on average during the first half of the trip, you will need an engine producing 240 hp on average during the second half of the trip. Of course, the actual power requirement is totally dependent on the speed of the car ($P = \frac{ma\Delta x}{\Delta t} = mav$).

In conclusion, distance-wise, you will need as much fuel to travel the first meter of the trip as for traveling the last one. But time-wise, you will need a lot more fuel to travel the last minute of the trip compared to what you will need for the first minute of your trip.

 

[Redbelly98]

The second tells you the kinetic energy (KE) of an object of mass $m$ moving with speed $v$. You can see that KE increases with the square of the velocity. Hence it takes four times as much energy to go from $1$ to $2m/s$ as it does to go from $0$ to $1m/s$.

Not to be nit-picky, but can we agree the factor of four would apply to going from zero m/s to 2 m/s, vs. 0 to 1 m/s?
:-)