# Physics of space travel

1. Jan 27, 2004

### Nibles

It's my understanding that everything in space is in some sort of "free fall", which is pretty much what an orbit is. So, if you were in a space ship and applied some thrusters, you are pretty much changing the orientation and speed of your "free fall", right?

So, lets say I wanted to get to Mars. If I turned my thrusters on full blast would I keep constantly accelerating (as long as the thrusters are active)? Would the gravitational pull of Earth cause a slight deceleration? Because I'm closer to Earth, Earth wants to pull me back, but when I cross over the 50% distance mark then Mars's pull will be more influential and will start pulling me in. Of course this 'threshold' wouldn't be 50% because Mars and Earth are not exactly the same size, and would not have the same GP.

My main question is: Could you simply apply thrusters to reach a certain velocity, and then turn them off (to save fuel) and maintain that velocity until you reached your destination?

2. Jan 27, 2004

### chroot

Staff Emeritus
If the only two bodies in the universe were the Earth and Mars, yes. All you'd have to do is cross the first Lagrange point (where the gravitational pull of both bodies is equal, just in opposite directions), and you'd free-fall until you crash into your destination body.

In the solar system, however, the dominant body is the sun, so it's not that simple. For the majority of the ride from the Earth to Mars, the Sun is the only source of gravitational forces that you really need to be concerned with.

To travel from Earth to Mars, you have to put yourself in a so-called "transfer" orbit -- an elliptical orbit (around the Sun) that crosses both Earth's orbit and Mars's orbit at carefully calculated times.

- Warren

3. Jan 27, 2004

### enigma

Staff Emeritus
To expand on Warren's post, this is exactly what you do for conventional rockets.

You apply a big thrust at the start, coast for a few minutes/hours/days/months, then apply a second thrust to stabilize the orbit.

You usually apply the thrusters several times during the transit to keep yourself on course (trying to aim straight at a dime in New York when you're starting in California is damn near impossible).

One exception to this is if you are using electric propulsion systems like 'ion drives'. The thrust from those engines is so small (about the weight of a sheet of paper), that you need to keep them on for a long time to get up to speed. The reason you'd want to use them is because they use less fuel for the same amount of 'oomph'.

Last edited by a moderator: Apr 20, 2017
4. Mar 12, 2004

### yowda

All these sci-fy shows like star trek and star wars say they use things like warp drives where anti matter and matter are reacted i guess... Is this theory of anti matter and matter reacting to produce incredible ammount of thrust even slightly realistic in any way??? Why would anti matter and matter react??? Would it not be like an acid and a base reacting in a neutralization or am i totally wrong???

5. Mar 13, 2004

### Nereid

Staff Emeritus
Welcome to Physics Forums yowda!

Matter + anti-matter = energy.

IF the energy produced by 'burning' anti-matter could be harnessed to produce thrust, it would be the most efficient rocket fuel possible, in terms of thrust per unit of mass.

The first (or last?) difficulty is that when matter and anti-matter annihilate, the energy produced is hard gamma rays, which aren't easily converted to thrust

Then there's the small matter (oops, is that a pun?) of storing anti-matter, not to mention producing it. Even Bill Gates may not be able to afford more than a picogram or two.

6. Mar 13, 2004

### Staff: Mentor

Acid/base reactions do result in a neutralized solution, but the reaction is generally extremely exothermic (energetic).

7. Mar 13, 2004

### Janitor

Another tidbit that may be useful to consider

A general rule of thumb in rocket dynamics is to burn as much of your propellant low in a gravitational potential well as possible. You can think of it this way: at any given time in the ascent, the propellant mass that is still in the tanks had to be lifted against the gravitational gradient at the expense of propellant already burned in order to get it to the altitude it is at.

Of course, on a Mars trip, there are at least three gravity wells to take into consideration: those of Earth, the Sun, and Mars.

8. Mar 13, 2004

### enigma

Staff Emeritus
This is sortof right.

The difference in gravity between two altitudes in the well is not that large unless you're talking about BIG differences in altitude.

For example, at 250km altitude, g is still more than 9 m/s^2.

The biggest losses come from not utilizing the curvature of the Earth to your advantage by going so fast that it gets "out of the way".

I'm working on a rocket trajectory program for a project right now. The first cut went straight up to get it to 250km, and then burned tangentially. The final velocity was around 1.5km/s. By following a sinusoidal trajectory the typical final velocities are around 4km/s... not enough (I'm still tweaking), but much better.

Last edited: Mar 13, 2004
9. Mar 13, 2004

### Janitor

Imagine two modest sized solid-fuel rockets. Let us say their performance is not great enough to achieve an altitude where we have to worry about gravitational acceleration changing to a lower value, i.e. we can take g to be constant. Let us agree to ignore air drag effects as well. (If you want to think of this experiment taking place in the vacuum on the moon, go ahead and do so.) Let us suppose that the total impulse of both rockets is the same, but one rocket has twice the thrust of the other rocket, and only is able to fire for half as long of a time. Does enigma agree with me that the higher-thrust rocket will achieve greater apogee?

If so, then taking things to a somewhat absurd limit, it should also follow that the best performance of all would come if all of the propellant could be used up at the instant of launch, in one big bang so to speak. I.e., burn all of the propellant up at zero altitude. Here I am making the simplifying assumption that the structure of the rocket would not have to be beefed up to stand the ultra-high acceleration of such a quick burn of the propellant, of course.

10. Mar 13, 2004

### enigma

Staff Emeritus
No. Purely theoretically, with all else being equal, they'll go to the same apogee.

If the one rocket has twice the thrust but twice the mass flow, the specific impulses are the same.

Isp = Thrust/(mass flow*g0)

To calculate the theoretical velocity after the burn, you follow the ideal rocket equation:

delta V = -Isp * g0 * ln (final mass/initial mass)

In real life, the rocket with the higher thrust will be heavier because for the greater thrust, you'll need a bigger chamber pressure which means heavier structure. This brings down the 'final/inital' fraction (inert mass fraction), which brings down the velocity after the burn, which brings down the energy of the orbit, which brings down the apogee.

11. Mar 13, 2004

### Janitor

I'm off to the gym to exercise soon, but I will think more about this.

To further pin down my claims, I am saying that if we ignore issues of the structure having to be more massive to withstand greater thrust, then out in field-free interstellar space rockets with the same total impulse and the same structure mass will achieve the same speed upon burnout. I am pretty sure you agree with me on that. But I still think that in a gravitational field, a rocket with higher thrust (even though it have the same total impulse and again ignoring structural beefiness issues so that it has the same structure mass) will achieve a higher apogee. I could be wrong, though, and I will think more about it later tonight.

Last edited: Mar 13, 2004
12. Mar 13, 2004

### Janitor

Enigma,

Your equation gives the increase in speed for a rocket out in free space, with no gravitational gradient.

For the case where the rocket is climbing vertically against a gravitational gradient, the equation becomes (and we can take initial velocity to be zero, since the rocket starts at rest on its launch pad) the velocity at any time t < t_co (with "co" referring to engine cutoff) is:

v = -I_sp * g * ln ({initital mass-mass burned}/initial mass) - g * t.

This matches your equation, except that there is the added term due to the constant acceleration g of gravity working against the rocket.

The height obtained at any time t < t_co before cutoff is:

z = I_sp g {[(M_i - m_r t)/m_r] ln [1-(m_r/m_0)] - t} - (1/2) g t^2.

where m_r is the mass flow rate of the propellant (if I have done the integration correctly).

For values of time t greater than t_co, we obviously just have a ballastics problem.

I could go on to calculate the apogee, and then to compare what happens when you double F and m_r and halve t_co. But it will get ugly. I haven't yet learned TEX very well.

Would you like me to continue, or do you grant that under the idealization that thrust could be doubled without increasing structure mass, the apogee will be higher if thrust is greater while I_sp remains the same and the propellant is used up correspondingly sooner?

Last edited: Mar 13, 2004
13. Mar 14, 2004

### enigma

Staff Emeritus
OK,

but the direction of final velocity is not in the direction which gravity pulls, but is perpendicular to it. You're thinking about this like an infinite flat Earth model.

Since the gravity gradient is miniscule:

EDIT: fixed error
$$-2*\mu*r^{-3}*\frac{dr}{dt}$$

mu=398600.4 km^3/s^2
r at sea level is 6378km
->or -6.14e-6 * dr/dt

over the burn duration of a typical rocket, you are not really not gaining much by increasing the thrust to get 'up' early in the burn.

BTW:
apogee can be found by manipulating the Vis-Viva equation:

$$V=\sqrt{\mu*(\frac{2}{r}-\frac{1}{a})}$$

Apogee will be 2*a - r, where r is the altitude after completion of the burn.

Last edited: Mar 14, 2004
14. Mar 14, 2004

### enigma

Staff Emeritus
On thinking on this a little more, you are right that the gravity losses will be less with a higher thrust rocket. Not because you go 'up' quicker, however, but because you gain orbital tangential velocity quicker. The effect is also not a super huge effect unless you're talking about really big durations staying at low speeds.

15. Mar 14, 2004

### Janitor

Thanks, enigma.

If you have a chance, could you explain more to me about your mu term?

Analogies are dangerous because they can mislead so easily. So take this one with a good dose of skepticism: A person steps into an elevator (in which the doors are permanently wedged open) on the ground floor and finds a pile of rocks sitting on the elevator floor. He hits the button to take him to the tenth floor. On the way up, he steadily throws the rocks out the door until the last one is pitched out just as he gets to the tenth floor. It is clear that the energy needed by the elevator motor to do that task is more than if the man pushed all the rocks out by the time he reached the second floor, and then rode the rest of the way up to the tenth floor without any rocks along for the ride. That is sort of how I am thinking about the rocket, flying straight up and straight down in a constant gravity field.

16. Mar 14, 2004

### enigma

Staff Emeritus
Sorry: mu is the gravitational parameter, or G*mass of the attracting body.

G*M comes up often enough in Celestial mechanics that the two are just combined to a constant and tabulated by celestial body.

The problem with your analogy is that you're ignoring the fact that in a rocket, it IS the rocks which provide the upwards velocity, not the elevator engine.

So-long as you are considering only +-z motion (which would be a really short flight), assume rocket one takes 100s to discharge its fuel, and rocket 2 takes 50s. In either case, the fuel will be gone long before the rocket gets much higher than 100 km. The difference in gravitational acceleration between the burnout points is on the order of 10^-1 to 10^-2 m/s^2. Really not that much...

17. Mar 14, 2004

### Janitor

Thanks for taking so much time to discuss this with me, Enigma. It makes sense that a constant would be defined for the product G*M_body, now that I think about it. And yes, my analogy certainly has shortcomings!

The thought that keeps coming to me is that if that last pound of propellant to be consumed immediately prior to burnout had to be lifted all the way up to 10 miles (just as an example, for some modest rocket going straight up from the launch pad), the propellant already consumed prior to that had to do extra work to lift it there, compared to if the same total amount of propellant had been burned quicker, such that the last pound was only lifted to 7 miles, or whatever. But I realize my intuition is not a reliable guide, so I will drop the argument--unless I happen to come across something in the literature that supports one side of the argument or the other.

18. Mar 14, 2004

### enigma

Staff Emeritus
There's no reason to drop it if I can explain it better.

Alright, the exhaust velocity of the rocket fuel is constant for a specific nozzle shape and propellant combination.

We'll assume that we're dealing with the same rocket parameters, only with a larger sized throat (which increases the mass flow, and thus the thrust).

By Newton's third law, the momentum given to the propellants is equal but opposite to the momentum given the rocket.

$$\frac{d}{dt} mV = F = ma$$

$$\frac{dm}{dt}V +m\frac{dV}{dt} = ma$$

Since we're assuming that the exhaust flow is constant, dV/dt is zero. Since we're solving for the final velocity of the rocket by looking at the velocity of the fuel, V will be equal to -Ve

$$-V_e \frac{dm}{dt}=ma$$

$$-V_e \frac{1}{m} \frac{dm}{dt}=a$$

$$\int_{t_i}^{t_f} \frac{-V_e}{m} \frac{dm}{dt}dt = \int_{t_i}^{t_f}a dt$$

$$-V_e \int_{t_i}^{t_f} \frac{dm}{m} = V(t_f)-V(t_i)$$

$$-V_e*(ln(m_f)-ln(m_i)) = \Delta V$$

$$\Delta V = -V_e*ln(\frac{m_f}{m_i})$$

Which is the ideal rocket equation given earlier.

The last bit of rocket fuel needs to get accelerated up to full velocity regardless of whether it's done in 60 seconds or 100. By integrating Newton's second law, you can see that the mass flow doesn't even come into play.

In real life, it will matter a little, because as you pointed out, you have gravity which is a fraction higher lower in the gravity well. Still, the differences in the gravity field over the range of a typical rocket's altitude profile during ascent is really small.

PS: Check your private messages

Last edited: Mar 14, 2004
19. Mar 15, 2004

### Janitor

Yes, I agree completely with your equations in your last post regarding rocket performance in a gravity-free region. The ma term would become m(a+g) if we took a constant gravity field into account, or m[a+g(z)] if we took a gravity field varying with altitude z into account. (Here I am assuming a one-dimensional problem, with the rocket going straight up, so that the g vector points along the rocket axis. For trajectories which are not vertical, one would have to use vectors explicitly since g and a would not lie along the same line.)

I tried my pm button for the first time. It took me to a box where I can apparently send a message to someone, but I see no way of telling if anyone has sent anything to me, so I am still in the dark about your PS. Any advice?

Last edited: Mar 15, 2004
20. Mar 15, 2004

### enigma

Staff Emeritus
At the bottom of the main PF page, under the "current users" section is your private message overview. If you click on "private messages" it'll take you to your mailbox.

What you did by clicking PM under your post was to try to send a PM to yourself.