Change in acceleration

[julianwitkowski]

Does the change in acceleration along the path of a space flight to the moon, for example, get taken into account when planning initial velocity?

I would imagine that the initial velocity leaving Earth will be decelerating and the deceleration will increase the further it gets until another bodies gravitation compels it?

How is this change in acceleration taken into account? I am especially wondering if it is an average value of the acceleration function over the radius, or if an average doesn't work in this case?

Any input would be appreciated

Thanks!

 

[Matterwave]

The important quantity that we look at is the acceleration because that is the quantity which corresponds to forces through Newton's second law F=ma. Of course, when we work with planning any kinds of flight paths, we will look at an acceleration which will be generally a function of time a=a(t) and not a constant one. This is definitely taken into account.

 

[julianwitkowski]

The important quantity that we look at is the acceleration because that is the quantity which corresponds to forces through Newton's second law F=ma. Of course, when we work with planning any kinds of flight paths, we will look at an acceleration which will be generally a function of time a=a(t) and not a constant one. This is definitely taken into account.

Is it taken as an average when its increasing and decreasing throughout a trajectory path?

 

[Matterwave]

Is it taken as an average when its increasing and decreasing throughout a trajectory path?

No, when you are solving equations for these problems, you usually solve for r(t) which is the position as a function of time. This function determines uniquely all relevant quantities as functions of time. The velocity is v(t)=r'(t) and the acceleration a(t)=r''(t).

We rarely work with averaging since the instantaneous quantities tell us a great deal more information and it is very necessary to be precise.

 

[THE HARLEQUIN]

I would imagine that the initial velocity leaving Earth will be decelerating and the deceleration will increase the further it gets until another bodies gravitation compels it?

when a rocket leaves Earth it's not decelerating , rather it is accelerating with a decreasing acceleration ... and this acceleration has different equation regarding the path you choose and will be a function of time and mass

 

[julianwitkowski]

No, when you are solving equations for these problems, you usually solve for r(t) which is the position as a function of time. This function determines uniquely all relevant quantities as functions of time. The velocity is v(t)=r'(t) and the acceleration a(t)=r''(t).

We rarely work with averaging since the instantaneous quantities tell us a great deal more information and it is very necessary to be precise.

Ok, I was thinking an average like this

b
∫ G*M/r² * 1/(b-a)
a

So, in that case, how would you keep track of the change in acceleration between upper interval b and lower interval a? Do you use the interval as a function of time? and if so what equation?

 

[julianwitkowski]

when a rocket leaves Earth it's not decelerating , rather it is accelerating with a decreasing acceleration ... and this acceleration has different equation regarding the path you choose and will be a function of time and mass

Yes the rocket accelerates, but it accelerates against gravity accelerating back at it, so eventually you will come back down if you don't have escape velocity...

 

[Matterwave]

Ok, I was thinking an average like this

b
∫ G*M/r² * 1/(b-a)
a

So, in that case, how would you keep track of the change in acceleration between upper interval b and lower interval a? Do you use the interval as a function of time? and if so what equation?

That's quite a peculiar equation...I am pretty baffled why you think an "averaging" of this kind is desirable or necessary when doing calculations (I am assuming you are integrating there with respect to r?).

Simpler would be to look at the instantaneous forces on this rocket (definitely not a simple equation to solve though): $$\vec{F}_{net}(t)=\vec{F}_{thrust}(t)-\frac{GM_E m_r(t)}{|\vec{r}(t)|^2}\hat{r}(t)=m_r(t)a(t)$$

 

[THE HARLEQUIN]

averaging is not possible here ... its not a linear equation .. and accelaration changes every second and moreover , a rocket travels in a parabolic path amd when it's not so close to Earth the thrust equation also changes ( we have to take the gravitational force of moon into account )

 

[julianwitkowski]

averaging is not possible here ... its not a linear equation .. and acceleration changes every second and moreover , a rocket travels in a parabolic path amd when it's not so close to Earth the thrust equation also changes ( we have to take the gravitational force of moon into account )

I'm was specifically thinking about once its left earth, say the lower bound is the Karman line, and you need to propel your craft with enough velocity to the moon, if you make your main shot with enough velocity to get to the point where G ⋅M_e/r² = G ⋅M_m/r², the moon would pull you the rest of the way, so the main consideration would be to make sure that you don't decelerate and be attracted back to Earth before you reach where G ⋅M_m/r² > G ⋅M_e/r² So then, if this is the case, the v_i= v_f - a ⋅ t

 

[THE HARLEQUIN]

once you start a rocket that has enough fuel you will never decelerate until your fuel is over ... you can prove it with this

F⃗ net(t)=F⃗ thrust(t)−GMEmr(t)|r⃗ (t)|2r^(t)=mr(t)a(t)

 

[julianwitkowski]

once you start a rocket that has enough fuel you will never decelerate until your fuel is over ... you can prove it with this

But what about the poor man's space craft, which does not have a big enough tank to burn gas the whole way there, only one controlled burn to get a burst of velocity, the only thing that will slow you down is the gravity force attracting to earth, so if you get a good enough impulse from LEO in the right direction you will make TLI?

 

[THE HARLEQUIN]

But what about the poor man's space craft, which does not have a big enough tank to burn gas the whole way there, only one controlled burn to get a burst of velocity, the only thing that will slow you down is the gravity force attracting to earth, so if you get a good enough impulse from LEO in the right direction you will make TLI?

that poor man has to be richer than the rocket owner ... within a very little time he has to put in a great enough force to get that spacecraft the escape velocity ... that's an impossible case you are talking about... but yes if you can get that escape velocity you will leave the Earth's gravitational field

 

[Drakkith]

once you start a rocket that has enough fuel you will never decelerate until your fuel is over ... you can prove it with this

I'm not sure what you mean by this. You will certainly cease accelerating when you stop your rocket engine. Turning around and activating it again will cause you to decelerate.

 

[Drakkith]

But what about the poor man's space craft, which does not have a big enough tank to burn gas the whole way there, only one controlled burn to get a burst of velocity, the only thing that will slow you down is the gravity force attracting to earth, so if you get a good enough impulse from LEO in the right direction you will make TLI?

That's correct. That's how all of the Apollo missions worked. A short burn near Earth, and then they coasted the rest of the way, with perhaps a few course corrections if needed.

 

[THE HARLEQUIN]

I'm not sure what you mean by this. You will certainly cease accelerating when you stop your rocket engine. Turning around and activating it again will cause you to decelerate.

...i didn't mean it will still accelerate when the engine is stopped ... i just tried to say that when the engine is on it always accelerates

 

[julianwitkowski]

That's correct. That's how all of the Apollo missions worked. A short burn near Earth, and then they coasted the rest of the way, with perhaps a few course corrections if needed.

If centre to centre distance = d, and the distance from Earth where the moon's F_G > earth' F_G = a

How do you calculate the loss of velocity required to get to the point where → G ⋅M_e/ a² = G ⋅ M_m/ (d-a)²

I was thinking an average like this
b
∫ G*M/r² * 1/(b-a)
a

I thought if I summed the acceleration outcomes in the interval and averaged them like this it would work but I guess the values don't add up because they're based on time and not just the radius... Can you think of anything else. I don't see how the F_net(t)=F_thrust(t)... equation gets me what I'm looking for but I'm still learning about it so I need time on that.

I'm probably missing something, so if you can lead me in the right direction I'd be grateful.

 

[Drakkith]

How do you calculate the loss of velocity required to get to the point where → G ⋅Me/ a2 = G ⋅ Mm/ (d-a)2

Sorry, no idea! :s

 

[jbriggs444]

How do you calculate the loss of velocity required to get to the point where → G ⋅M_e/ a² = G ⋅ M_m/ (d-a)²

You could use potential energy to get to that number.

Let's take a simplified scenario where we imagine that Earth and Moon are stationary and that we launch with an initial burn and coast to the mid-point between Earth and Moon where their gravitational attraction cancels out.

The gravitational potential energy of the spacecraft at its initial location at the surface of the Earth will be the sum of its potential energy with respect to the Earth and its potential energy with respect to the moon. That's $\frac{-GmM_e}{R_e} + \frac{-GmM_m}{d}$

The gravitational potential energy of the spacecraft at its midpoint location at distance "a" from Earth is $\frac{-gmM_e}{a} + \frac{-GmM_m}{d-a}$

By conservation energy, the change in kinetic energy on a coasting trajectory is equal and opposite to the change in potential energy. Use $KE=\frac{1}{2}mv^2$ to work out the change in velocity required to match the change in potential energy computed as above.

 

[julianwitkowski]

That's quite a peculiar equation...I am pretty baffled why you think an "averaging" of this kind is desirable or necessary when doing calculations (I am assuming you are integrating there with respect to r?).

Simpler would be to look at the instantaneous forces on this rocket (definitely not a simple equation to solve though): $$\vec{F}_{net}(t)=\vec{F}_{thrust}(t)-\frac{GM_E m_r(t)}{|\vec{r}(t)|^2}\hat{r}(t)=m_r(t)a(t)$$

Yes sorry, the integral is with respect to r. Anyway's, your equation is a little above my paygrade atm, I wouldn't know where to start with the matrices I believe it requires and so far my thought problem is 1 dimensional. Why is this "simpler" than calculating the loss of velocity? Can you think of any other ways I could do this, maybe with a derivative for the rate of change?

 

[julianwitkowski]

You could use potential energy to get to that number.

The gravitational potential energy of the spacecraft at its initial location at the surface of the Earth will be the sum of its potential energy with respect to the Earth and its potential energy with respect to the moon. That's $\frac{-GmM_e}{R_e} + \frac{-GmM_m}{d}$

The gravitational potential energy of the spacecraft at its midpoint location at distance "a" from Earth is $\frac{-gmM_e}{a} + \frac{-GmM_m}{d-a}$

By conservation energy, the change in kinetic energy on a coasting trajectory is equal and opposite to the change in potential energy. Use $KE=\frac{1}{2}mv^2$ to work out the change in velocity required to match the change in potential energy computed as above.

I always forget about PE = KE... So much wow, much like. Thanks much.

 

[A.T.]

I always forget about PE = KE... So much wow, much like. Thanks much.

And if you want to account for Earth & Moon orbiting their common CoM, then you get a third PE-term in their common rest frame, associated with the inertial centrifugal force.

 

[julianwitkowski]

So is it $\frac{-gmM_e}{a} + \frac{-GmM_m}{d-a}$ = m ⋅ g ⋅ h when the object is directed towards Earth from point a or did I get that backwards?

thanks.

 

[jbriggs444]

So is it $\frac{-gmM_e}{a} + \frac{-GmM_m}{d-a}$ = m ⋅ g ⋅ h when the object is directed towards Earth from point a or did I get that backwards?

As you may know, potential energy is always relative to an arbitary position that one takes as a reference. Only differences in potential energy are relevant. So it makes no difference what reference point is chosen, as long as the choice is then adhered to consistently.

$PE = m \cdot g \cdot h$ is a formula for an object's gravitational potential energy with respect to Earth, taking a reference point at the surface of the earth. It is an approximation that is valid only for positions near enough to the Earth's surface that the Earth's gravity is approximately constant. If you get far enough so that the inverse square drop-off is significant, the approximation becomes inappropriate.

To account for the inverse square law, you can integrate the inverse square gravitational force over distance. $\int \frac{GmM_e}{r^2}dr$ yields $\frac{-GmM_e}{r}$. Leaving the constant of integration at zero is equivalent to selecting a reference point at infinity. That is the convention that is normally used.

Note that this is big G (Newton's universal gravitational constant), not little g (the acceleration of gravity near the Earth's surface).

[As many discussions in these forums point out, there are complications when computing potential energy for positions below the Earth's surface. Don't try to use this potential energy formula for burrowing satellites].

 

[julianwitkowski]

To account for the inverse square law, you can integrate the inverse square gravitational force over distance. $\int \frac{GmM_e}{r^2}dr$ yields $\frac{-GmM_e}{r}$.

I'm guessing that m before M_e is it the mass of the object?

If so, the result is -13874145771.424060200600472979367237404 ? What exactly is this number other than the area under x with an interval reference at infinity? Is it the PE? Oops, we went through that already... Energy Lost? ... Potential Lost = Kinetic Gained? 1.39e¹⁰ J = 1/2 m ⋅ v²... Or maybe I failed...

Thank you so much for taking the time to explain this for me :)

 

[jbriggs444]

I'm guessing that m before M_e is it the mass of the object?

Yes, m is the mass of the object.

If so, the result is -13874145771.424060200600472979367237404 ?

That's a lot of significant digits.

If you have $M_e$ in kilograms, $m$ in kilograms, $G$ in Newton meter²/kilogram² and r in meters then $\frac{-GmM_e}{r}$ will be in Joules.

It represents the minimum kinetic energy in Joules that the object would need to possesses so that its free fall trajectory would proceed infinitely far away from the Earth, if the Earth were alone in the universe. That is to say, it is the "escape energy" of the object with respect to the Earth at that distance.

If you were to take the object's escape velocity (at that orbital radius) v and plug it into $KE=\frac{1}{2}mv^2$ then you should get the same result.

 

[julianwitkowski]

That's a lot of significant digits.

Thank you again!

 

[julianwitkowski]

The gravitational potential energy of the spacecraft at its midpoint location at distance "a" from Earth is $\frac{-GmM_e}{a} + \frac{-GmM_m}{d-a}$

I was thinking about this again and I was just wondering why it isn't $\frac{-GmM_e}{a} - \frac{-GmM_e}{6.471e6}$ if its the potential energy of a fall from this said radius = a? I'm guessing maybe I just don't understand the math :(

I'm kinda confused, on second thought I would assume the integral to sum P_e is $\frac{-GmM_e}{r}$ and r max = a, and min = 6.471e⁶ for the initial escape from LEO. One thing is that I don't understand why the moon factors into it, if the integral here is the product of P_E from the a distance to where the initial acceleration from Earth is. Could you explain this a little?

 

[jbriggs444]

I was thinking about this again and I was just wondering why it isn't $\frac{-GmM_e}{a} - \frac{-GmM_e}{6.471e6}$ if its the potential energy of a fall from this said radius = a?

The above formula would give you the potential energy of the object with respect to the Earth only and would put the baseline "zero potential energy" position to the surface of the earth. It would ignore the potential energy of the object with respect to the Moon.

I'm kinda confused, on second thought I would assume the integral to sum P_e is $\frac{-GmM_e}{r}$ and r max = a, and min = 6.471e⁶ for the
initial escape from LEO.

You would be integrating $\frac{GmM_e}{r^2}$ to compute the potential energy with respect to the Earth and $\frac{GmM_m}{r^2}$ to compute the potential energy with respect to the Moon. If you want to put the reference point for zero potential energy at the surface of the Earth where r = 6.471x10⁶ meters, that is your choice.

One thing is that I don't understand why the moon factors into it, if the integral here is the product of P_E from the a distance to where the initial acceleration from Earth is. Could you explain this a little?

The integral is of the product of force times incremental distance, yielding potential energy. It is not the product of potential energy multiplied by something else. Possibly I have misunderstood your question.

If you want to use conservation of energy, you have to pay attention to all of the energy, not just part of it. Our goal as I recall involved computing an object's velocity to get to the point where the Earth's gravitational force and the Moon's gravitational force were equal and opposite. That means that the Moon's gravitational attraction is not insignificant and must be accounted for.

 

[julianwitkowski]

Thank you, this puts it in much better perspective!