Recognitions:
Gold Member
Staff Emeritus

## Physics of space travel

 Originally posted by Janitor 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.
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.

 Recognitions: Science Advisor 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?
 Recognitions: Gold Member Science Advisor 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.
 I know that there are all these confusing formulas out there but when do you guys/girls think that 'space travel' can become realistic. I dont mean going to space going vroom vroom and coming back to earth. When do u think ppl will be on mars??? and do u think mars will be cononized and if so when??? Im just curious what u guys/girls think...[a)]
 O ya i forgot one more question... how??? with what type of travel??? anti matter matter reactions???
 Recognitions: Science Advisor My wild guess is it will be in another 30 years, and it won't be done with anything as exotic as matter/antimatter.
 I have a question (more of a request, really). I am just a college student who is writing a research paper for english class. I don't know much about physics so what you all are posting is basically greek to me! My research topic is space travel and the possibilities / limitations for now and the future. If anyone has ANY good - credible- sources that I could quote or use for info could you PLEASE e-mail me? Any websites (preferably .edu or .org) and science journals / magazines / recent papers would be great. Thank you!
 Also, if anyone is a professional (I mean, has crudentials that I can include in my references) and you don't mind my quoting you, I can gladly use your own views in my paper.
 I am still REALLY in need of some good research tips. Help!
 Recognitions: Gold Member Science Advisor Staff Emeritus Hi Rachel, sorry, I missed your posts earlier. It may or may not help you, but the book which helped me make up my mind to go into aerospace is called Mining the Sky by John S. Lewis. He goes into great detail into what resources are out in space and what would be needed to go get them. It's not "space travel" persay, but it does cover what's needed to establish a lasting presence in space.

 Quote by enigma Sorry: mu is the gravitational parameter, or G*mass of the attracting body.
Couldn't help myself: mu is actually G*(m1+m2), but for most problems (the ones that involve artificial satellites, planets revolving around the Sun,...)
one of the masses is much smaller than the other and is thus neglected. Hence the G*M.

BTW: how do you get those fancy LaTeX formulas in your posts?

 Rachel, a fun book about the possibilities of space travel is Robert Zubrin's "Entering Space". Zubrin is a well-known (and a tad arrogant) aerospace engineer.
 BTW2: $$\frac{n^e V_e r}{Mi_n d}$$ about the Latex :)

Recognitions:
Gold Member
Homework Help
 Quote by remcook Couldn't help myself: mu is actually G*(m1+m2), but for most problems (the ones that involve artificial satellites, planets revolving around the Sun,...) one of the masses is much smaller than the other and is thus neglected. Hence the G*M.
Actually, $$\mu$$ is just GM.

To calculate the force on an object due to gravity, you still have to include the mass of the object in your equations.

$$F=\frac{\mu m_2}{r^2}$$ for example

To calculate the effect the force has, (i.e. - the resulting acceleration), you not only have to consider the mass in the force, but also in the object's resistance to being moved by the force - inertia. Thus, two objects with a different mass are pulled toward the Earth by a different amount of force, but accelerate towards the Earth at the same rate.

$$a_g = \frac{\mu}{r^2}$$

Since most of your orbit equations are only concerned about the motions of objects, you see quite a few equations with the mass scaled out - hence, angular momentum of an orbit is actually 'the specific angular momentum per unit of mass', the specific energy the 'specific energy per unit of mass', etc.

Why not? If you calculate the orbit of one piece of the satellite, and all the pieces are connected, then they all must follow the same orbit, right? (Well, not quite right, since now you get into torques on the spacecraft and attitude control issues, but close enough for government work).

But, you're definitely right about the relative size of the objects being important. In actuality, both objects are accelerated towards their combined center of mass. With two binary stars, for example, each star is orbiting around a point in space that could be considered to have the combined mass of both stars. In the case of artificial satellites around the Earth, which has a mass of around $$6 x 10^{24} kg$$ even a huge satellite of around $$4.5 x 10^5 kg$$ (mass of the ISS) won't change the combined mass enough for you to see on a calculator and the combined center of mass will be so close to the center of the Earth that you won't be able to measure the difference.

 In your example - yes But not always (often not actually, expecially when satellite orbits are concerned). ! This is because most equations are described relative to the centre of mass of one of the masses, and not in the barycentre itself (or any other inertial reference frame). Allow me to demonstrate: The acceleration of a particle of mass $$m_2$$ due to gravity of mass $$m_1$$ is, of course: $$F=G \frac{m_1 m_2}{r^2}$$ Imagine the vector from $$m_1$$ to $$m_2[\tex]: [tex]\bar{r}_{12} = \bar{r}_2 - \bar{r}_1$$ r1 and r2 are the positions of the two masses in an inertial reference frame. Acceleration of mass $$m_2$$ is given by Newton's F=ma (in vector notation) $$m_2 \bar{a}_2 = -G \frac{m_1 m_2}{r^2} \frac{\bar{r}_{12}}{r}$$ r is magnitude of vector r12. You see that $$m_2$$ indeed drops. similarly: $$\bar{a}_1 = G \frac{m_2}{r^2} \frac{\bar{r}_{12}}{r}$$ BUT...this equation is still set up in an inertial reference system! A reference frame that is placed at the centre of mass of the large body is not an inertial one (it is accelerating). the acceleration relative to mass m1 is: $$\bar{a}_{12} = \bar{a}_2 - \bar{a}_1 = -G \frac{m_1 + m_2}{r^3} \bar{r}_{12}$$ often written as: $$\bar{a}_{12} = - \frac{\mu}{r^3} \bar{r}_{12}$$ for $$m_2 \ll m_1$$ , indeed $$\mu \approx G m_2$$. It is also true that the barycentre is nearly at the centre of mass of the large mass when the other one is very small, and that the large mass is hardly accelerating due to the small mass, but that does not mean $$\mu = G m_2$$ exactly.

hmm ... it refuses my edit of the tex (there's a minus-sign missing for a12)
and there's still "tex" somewhere
and the last statement is of course Gm1 (twice! aargh)

another remark:

 Thus, two objects with a different mass are pulled toward the Earth by a different amount of force, but accelerate towards the Earth at the same rate.
You meant the other way around right?

 Recognitions: Gold Member Homework Help Science Advisor I see what you're saying now. I started from a definitions point of view. $$\mu$$ is a defined constant. Of course, it is its value that's 'defined', since that's been determined from reliable observations. From the 'defined' value (plus a few observations of other orbiting systems) you can derive a couple of other constants: the universal gravitational constant and the mass of the Earth. While $$\mu$$ may normally be defined as the product of the two constants derived from it, that's a bit of circular logic, at best. What you're also saying is that even the 'defined' value of $$\mu$$ can't really represent the 'true' gravitational potential since the 'true' gravitational potential can't be a constant. The defined value ignores the mass of one of the objects and the location of the center of mass in order to provide a number you can remember well enough to plug into equations. Fair enough, but, personally, I think you're secretly in love with Nicole Capitaine. (I wonder what she looks like, anyway.)