Calculating cosmic velocities using different frames of reference

[etotheipi]

Homework Statement

Determine the cosmic velocity required for a trajectory into the sun

Relevant Equations

N/A

This problem is conventionally solved using the Earth frame of reference. We require that the hyperbolic excess velocity w.r.t. the Earth has the same magnitude as the speed of the Earth around the sun, so that we zero the velocity in the heliocentric frame. Energy conservation per unit mass in the Earth frame looks like$$-\frac{\mu}{r} + \frac{1}{2}v^2 = \frac{1}{2}v_{\infty}^2$$ $$v^2 = v_{\infty}^2 + v_e^2 \implies v = \sqrt{v_{\infty}^2 + v_e^2} = \sqrt{v_E^2 + v_e^2}$$using $v_e^2 = \frac{2\mu}{r}$. But now as a sanity check I tried to perform the same calculation in the heliocentric frame. In the heliocentric frame gravitational forces will do work on the Earth which we must take into account. We transform our velocities into the heliocentric frame by adding back $\vec{v}_E(t)$, the velocity of the Earth, so that energy conservation now looks like$$-\frac{\mu}{r} + \frac{1}{2}(v-v_E)^2 + \frac{1}{2} \frac{M_E}{m} v_E^2 = \frac{1}{2}\frac{M_E}{m} {v_E'}^2$$ $$(v-v_E)^2 = v_e^2 + \frac{M_E}{m}({v_E'}^2 - v_E^2)$$We can safely apply momentum conservation in this frame, which yields that$$v_E' = \frac{M_Ev_E - m(v-v_E)}{M_E} = v_E - \frac{m}{M}(v-v_E)$$ $${v_E'}^2 = v_E^2 - \frac{2m}{M_E}(v-v_E)v_E + \left (\frac{m}{M_E} \right)^2 (v-v_E)^2$$ $$\implies \frac{M_E}{m}({v_E'}^2 - v_E^2) = \frac{M_E}{m} \left(- \frac{2m}{M_E}(v-v_E)v_E + \left (\frac{m}{M_E} \right)^2 (v-v_E)^2 \right) = -2v_E (v - v_E) + \frac{m}{M_E}(v-v_E)^2$$we put this back into our energy equation,$$(v-v_E)^2 = v_e^2 -2v_E (v - v_E) + \frac{m}{M_E}(v-v_E)^2 = v_e^2 -2v v_E+ 2v_E^2 + \frac{m}{M_E}(v-v_E)^2$$ $$v^2 = v_e^2 + v_E^2 + \frac{m}{M_E}(v-v_E)^2$$It definitely reduces to the first case in the limit $\frac{m}{M_E} \ll 1$, but I have never seen this latter equation written so I hope I have not made a mistake? Surely the error in the first approach is that it fails to take into account the change in the velocity of the Earth when nullifying the heliocentric velocity?

 

[pbuk]

I'm not familiar with the term "cosmic velocity" in this context, are you referring to the orbital escape velocity for an Earth satellite in the Earth+Sun gravity field?

I think that if $\neg( \frac{m}{M_E} \ll 1)$ then you cannot make the assumption that $r$ can be measured to the centre of the Earth and not the barycentre of the Earth-Satellite system and all these approximations fail.

 

[etotheipi]

I think that if $\neg( \frac{m}{M_E} \ll 1)$ then you cannot make the assumption that $r$ can be measured to the centre of the Earth and not the barycentre of the Earth-Satellite system and all these approximations fail.

$r$ is defined as the distance between the satellite and the centre of the Earth, so it is not an assumption.

I'm not familiar with the term "cosmic velocity" in this context, are you referring to the orbital escape velocity for an Earth satellite in the Earth+Sun gravity field?

I refer to the minimum launch velocity required for a trajectory into the sun

 

[pbuk]

$r$ is defined as the distance between the satellite and the centre of the Earth, so it is not an assumption.

Well yes, but this is no longer the orbital radius for $v_e^2 = \frac{2\mu}{r}$

 

[etotheipi]

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

That relation is also exact by definition, it is the escape velocity

 

[pbuk]

That relation is also exact by definition, it is the escape velocity

It's not defined, it's derived. And I think you'll find that you have to neglect some terms $O(\frac{m}{m_E})$ along the way.

 

[pbuk]

Oh OK, I was considering barycentric coordinates so you have to take the Earth's PE into account.

I think we are looking at the same thing from different angles: you are saying

Surely the error in the first approach is that it fails to take into account the change in the velocity of the Earth when nullifying the heliocentric velocity?

and I am saying

Surely the error in the first approach is that it fails to take into account the change in the position of the Earth/object barycentre when moving the object to infinity.

Does it make a difference how you look at it, the first approach ignores some terms $O(\frac{m}{m_E})$ so why are you surprised that your more accurate calculation includes such a term?

Or am I missing something subtle?

 

[etotheipi]

Oh OK, I was considering barycentric coordinates so you have to take the Earth's PE into account.

I will note that potential energy is a property of a system of particles, not a single body, and that it is classically a coordinate independent quantity, but I trust you are aware of this in which case...

Or am I missing something subtle?

no, that should be all. I just look for confirmation that the formula $v_e^2 + v_{\infty}^2 = v^2$ that is often quoted in orbital mechanics texts is an approximation.

 

[pbuk]

I will note that potential energy is a property of a system of particles, not a single body, and that it is classically a coordinate independent quantity,

Yes, exactly that - and we neglect this when we derive the simple escape velocity equation because we only look at part of the system.

no, that should be all. I just look for confirmation that the formula $v_e^2 + v_{\infty}^2 = v^2$ that is often quoted in orbital mechanics texts is an approximation.

Sure is, as a result of that neglect.

 

[pbuk]

I am not a fan of quoting from Wikipedia but it's the only thing I can find and I'm not going to do the calcs myself: "For a mass equal to a Saturn V rocket, the escape velocity relative to the launch pad is 253.5 am/s (8 nanometers per year) faster than the escape velocity relative to the mutual center of mass".

Who knew?

 

[etotheipi]

Yes you are correct the formula is not exact, I try just now to transform into the frame of the heavy object and write the equations of motion of the motion in reverse$$-\frac{G(m_1 + m_2)}{r^2} = -\frac{\mu}{r^2} = \ddot{r} = v_r \frac{dv_r}{dr}$$with $\mu = G(m_1 + m_2)$, for brevity, then$$-\int \frac{\mu}{r^2} dr = \int v_r dv_r \implies \frac{\mu}{r} = \frac{1}{2}v_r^2 + 0$$ $$v_r ^2= \frac{2\mu}{r} = \frac{2G(m_1 + m_2)}{r} \approx \frac{2Gm_1}{r} \quad \text{if}\quad m_1 \gg m_2$$If we do the analysis in the non-inertial Earth frame then a fictitious force also acts on the body $m_2$ whilst it escapes, doing negative work, so the initial $v_r$ must be larger. If we do the analysis in the inertial space frame, where the Earth is free to move, then the smaller mass does positive work on the Earth (which starts from rest) and accounting for final kinetic energy of the Earth makes $v_r$ larger.

Thank you for your help ☺

 

[PeroK]

If you get stumped by the maths you could always try this:

 

[etotheipi]

If you get stumped by the maths you could always try this:

And if that fails, I think we could all learn a thing or two from famous astrophysicist Ali G,

Ali G, to Buzz Aldrin: “When is man going to walk on da sun?”
Buzz Aldrin: “It’s much too hot on the sun.”
Ali G: “We could go in da winter, when it’s colder.”