Can we reach Mars using thrusters and maintain a constant velocity?

[BobG]

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.

 

[remcook]

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]:<br /> <br /> \bar{r}_{12} = \bar{r}_2 - \bar{r}_1<br /> <br /> r1 and r2 are the positions of the two masses in an inertial reference frame.<br /> <br /> Acceleration of mass m_2 is given by Newton&#039;s F=ma (in vector notation)<br /> <br /> m_2 \bar{a}_2 = -G \frac{m_1 m_2}{r^2} \frac{\bar{r}_{12}}{r}<br /> <br /> r is magnitude of vector r12. You see that m_2 indeed drops.<br /> <br /> similarly:<br /> <br /> \bar{a}_1 = G \frac{m_2}{r^2} \frac{\bar{r}_{12}}{r}<br /> <br /> 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 <i>relative</i> to mass m1 is:<br /> <br /> \bar{a}_{12} = \bar{a}_2 - \bar{a}_1 = -G \frac{m_1 + m_2}{r^3} \bar{r}_{12}<br /> <br /> often written as:<br /> <br /> \bar{a}_{12} = - \frac{\mu}{r^3} \bar{r}_{12}<br /> <br /> for m_2 \ll m_1 , indeed \mu \approx G m_2.<br /> <br /> 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.

 

[remcook]

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?

 

[BobG]

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.)

 

[remcook]

What I'm also saying is that for all practical purposes, the two definitions of \mu are close to identical, and so the mass of a planet or any other body can be easily derived from the orbit without taking into account the mass of the spacecraft .
And mu IS GM, but only in an inertial reference frame.

and..nicole who?

 

[Jenab]

I worked out the theory of transfer orbits and put it in the thread entitled, "Transfer orbits for dummies: a hillbilly tutorial." I included an example problem of a spaceship departing from Vesta and arriving at Earth.

Jerry Abbott

 

[patrick_usc]

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 spaceship and applied some thrusters, you are pretty much changing the orientation and speed of your "free fall", right?

So, let's 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?

Yes you would maintain that speed given that space is a vacuum and nothing would slow you down (unless you get hit by some space debris)

 

[russ_watters]

Note, this is a 4 year old thread.