Velocity required to escape the solar system

[theblazierbroom]

Homework Statement

The speed of earth in its orbit 18.5 $mi \, s^{-1}$. If it is desired to have the probe moving in a prescribed motion when it has escaped from the sun, what then is the maximum launching speed $v_{max}$ that could be required? (Problem 10.31 (b) from Exercises for the Feynman Lectures)

Relevant Equations

$$ F_g =G \frac{M \cdot m}{r^2} $$
$$ \Delta T = W = \int_A^B F \cdot d\textbf{r} $$

For the (a) portion of the problem, it asks to calculate the minimum speed a probe must be launched from earth to escape the solar system with residual speed of 10 $mi \, s^{-1}$ relative to the sun.

To find the minimum speed, I assumed the gravitational force affecting the probe by the sun is much smaller than the gravitational pull due to the Earth. Then, I assumed the distance when escaping the solar system is infinity. Then I integrated:

$$
\Delta T = \frac{1}{2} m \, v_{min}^{2} - \frac{1}{2} m \, (10 \frac{mi}{s})^{2} = \int_R^\infty F_g dr
$$

where $F_{g}$ = Gravity due to Earth, and $R$ = Radius of Earth. I got that the minimum speed required to escape the solar system is 12.1 $mi \, s^{-1}$, which the answer is 11.8 $mi \, s^{-1}$. Cool.

For (b), however, my initial assumption was that the maximum launching speed required is when the probe is launched in the opposite direction of the orbit at a given moment. Then, simply adding 12.1 $mi \, s^{-1}$ to the orbit speed of 18.5 $mi \, s^{-1}$ gets 30.67 $mi \, s^{-1}$. The answer is 47.1 $mi \, s^{-1}$.

I am lost on how to analyze the situation of this problem. Any hints are appreciated!

 

[haruspex]

For (b), however, my initial assumption was that the maximum launching speed required is when the probe is launched in the opposite direction of the orbit at a given moment. Then, simply adding 12.1 $mi \, s^{-1}$ to the orbit speed of 18.5 $mi \, s^{-1}$

That doesn’t seem quite to capture the difference between the two launch directions.

Btw, I could not follow your reasoning in part a). If you would like that checked, please clarify it.

 

[theblazierbroom]

Okay, so now I'm thinking, regardless of what launch direction it is, the launch probe must have with a minimum of 12.1 $mi \, s^{-1}$ to leave Earth's gravitational pull. The issue is breaking out of the orbit around the sun.

If launching in the direction of orbit, launch with 12.1+18.5 = 30.67 $mi \, s^{-1}$ to break out of both Earth's gravity and the orbit around the Sun.

If launching in the direction against the orbit, you launch with 12.1+18.5+18.5 = 49.1 $mi \, s^{-1}$ to accommodate for the initial momentum backwards.

But I think this reasoning is trying to fit into the solution rather than understanding the principles haha

 

[haruspex]

Okay, so now I'm thinking, regardless of what launch direction it is, the launch probe must have with a minimum of 12.1 $mi \, s^{-1}$ to leave Earth's gravitational pull. The issue is breaking out of the orbit around the sun.

If launching in the direction of orbit, launch with 12.1+18.5 = 30.67 $mi \, s^{-1}$ to break out of both Earth's gravity and the orbit around the Sun.

If launching in the direction against the orbit, you launch with 12.1+18.5+18.5 = 49.1 $mi \, s^{-1}$ to accommodate for the initial momentum backwards.

But I think this reasoning is trying to fit into the solution rather than understanding the principles haha

Throwing these numbers around does not help me understand your reasoning. Please use variables and define their meanings.
Be wary of adding velocities when perhaps you should be adding energies.

First, what energy is needed for a mass m to escape Earth's gravity, given that the Earth has radius Re and surface gravity g?

If the probe only has the energy to do that, what is then its motion relative to the Sun?

If an object is in circular orbit around a large mass, what is the relationship between its KE and the additional energy needed to escape to infinity?

 

[theblazierbroom]

First, what energy is needed for a mass m to escape Earth's gravity, given that the Earth has radius Re and surface gravity g?

Thanks for the outline!

The kinetic energy needed to escape Earth's gravity is:
$$
\frac{1}{2} m v_{escape}^{2} = m g R_{e}
$$

If the probe only has the energy to do that, what is then its motion relative to the Sun?

It will be in a circular orbit around the Sun, with relative velocity around sun = $\textbf{v}_{escape} + 18.5 \, mi \, s^{-1}$.

If an object is in circular orbit around a large mass, what is the relationship between its KE and the additional energy needed to escape to infinity?

Using the fact that Potential Energy is twice the Kinetic Energy in a circular orbit,

$$
KE + PE = \frac{1}{2} m v^{2} - m v^2 = -\frac{1}{2} m v^2 = Total \, Energy
$$

and in order to escape the orbit, the sum of the kinetic energy and potential energy must cancel out. Thus the additional kinetic energy needed is equal to the Total Energy.

If it launches in the direction of orbit of Earth, it will be,

$$
Total Energy = -\frac{1}{2} m (v_escape + 18.5 \, mi \, s^{-1})^{2}
$$

which will be the maximum total energy, since the speed of orbit around the sun is greatest.

I'm really having a tough time thinking through this problem. Is this the proper line of thinking?

 

[haruspex]

Thanks for the outline!

The kinetic energy needed to escape Earth's gravity is:
$$
\frac{1}{2} m v_{escape}^{2} = m g R_{e}
$$

Yes.

It will be in a circular orbit around the Sun, with relative velocity around sun = $\textbf{v}_{escape} + 18.5 \, mi \, s^{-1}$.

That's a contradiction. Yes, it is orbiting the sun at the same radius as the Earth, but that means it has the same velocity as the earth. It "used up" $v_{escape}$ to get there.

Thus the additional kinetic energy needed is equal to the Total Energy.

Equal to its magnitude, yes.

If it launches in the direction of orbit of Earth, it will be,

$$
Total Energy = -\frac{1}{2} m (v_escape + 18.5 \, mi \, s^{-1})^{2}
$$

Now you have lost me.

As hinted, you need to be adding the energies for each step:

• the energy to escape earth
• the energy to escape the sun, having escaped earth
• the energy needed to have the required final velocity having escaped both

 

[theblazierbroom]

That's a contradiction. Yes, it is orbiting the sun at the same radius as the Earth, but that means it has the same velocity as the earth. It "used up" to get there.

I see the contradiction and what's been throwing me off. Then, I have that:

$mgR_{e}$ ... The minimum energy required to escape Earth's gravitational pull
$\frac{1}{2}mv_{orbit}^{2}$ ... The minimum energy required to escape Sun's gravitational pull
$\frac{1}{2}mv_{f}^2$ ... The final energy of probe with residual speed $v_{f}$

I set it up as

$$
mgR_{e} + \frac{1}{2}mv_{orbit}^{2} + \frac{1}{2}mv_{f}^{2} = \frac{1}{2}mv_{0}^{2}
$$
$$
\sqrt{2gR_{e} + v_{orbit}^{2} + v_{f}^{2}} = v_{0}
$$

with $g = 6.1 \times 10^{-3} \, \rm{mi/s^2}$, $R_{e} = 3963.1 \, \rm{mi}$, $v_{orbit} = 18.5 \, \rm{mi}$, and $v_{f} = 10 \, \rm{mi/s^{-1}}$.

I get that $v_{0} = 22.15 \, \rm{mi/s}$. Unless I've done something wrong, I'm thinking this is the minimum velocity required for the probe to escape the sun at $10 \, \rm{mi/s}$, considering Earth's orbit. The question asks for the maximum velocity that may be required. Is there an angle dependence on the launch that can affect the required velocity?

 

[haruspex]

Unless I've done something wrong, I'm thinking this is the minimum velocity required for the probe to escape the sun at $10 \, \rm{mi/s}$, considering Earth's orbit. The question asks for the maximum velocity that may be required. Is there an angle dependence on the launch that can affect the required velocity?

What you have done looks ok to me. But I may have oversimplified it by thinking in terms of energy, as though a distinct burn can be performed at each stage. In the question, there is just the launch speed and angle, which limits the possible trajectories.
Having escaped Earth gravity, it has a residual velocity. To waste no energy, you want that to be parallel to the orbital velocity about the Sun. Maybe for part b you should waste the maximum by making it antiparallel.

 

[jbriggs444]

$mgR_{e}$ ... The minimum energy required to escape Earth's gravitational pull
$\frac{1}{2}mv_{orbit}^{2}$ ... The minimum energy required to escape Sun's gravitational pull
$\frac{1}{2}mv_{f}^2$ ... The final energy of probe with residual speed $v_{f}$

Yes. This is a good list of energies needed. Though the explanation on the middle item is suspect.

$$
mgR_{e} + \frac{1}{2}mv_{orbit}^{2} + \frac{1}{2}mv_{f}^{2} = \frac{1}{2}mv_{0}^{2}
$$

I do not believe that this is correct. The calculation jumps improperly jumps from one frame of reference to another. Let me explain by counter-example.

Suppose that the final required velocity $v_f$ is zero. Suppose that we start out in circular orbit around the sun along the same orbital path as the Earth. So we can ignore the escape energy from the Earth.

Then the equation that you have derived would tell us that:
$$\frac{1}{2}mv_{orbit}^2 = \frac{1}{2}mv_0^2$$ or, more simply:$$v_{orbit} = v_0$$.
But that is obvious nonsense. If $v_0$ is equal to orbital velocity then our net sun-relative launch velocity is double orbital velocity. Doubling orbital velocity quadruples orbital kinetic energy. We only needed to double orbital kinetic energy to escape.

The problem is an unnoticed change of reference frame. Orbital escape energy is properly calculated from the sun's rest frame. Launch relative velocity is properly calculated from the launch rest frame. You cannot validly equate the two without performing a shift of reference frame calculation.

 

[haruspex]

Orbital escape energy is properly calculated from the sun's rest frame. Launch relative velocity is properly calculated from the launch rest frame. You cannot validly equate the two without performing a shift of reference frame calculation.

Well spotted, thanks.
Do you have any insight into the trajectory question? I worry that treating the two stages as independent requires a trajectory that is not achievable without a redirectable engine.

 

[jbriggs444]

Do you have any insight into the trajectory question? I worry that treating the two stages as independent requires a trajectory that is not achievable without a redirectable engine.

I am not sure what the "trajectory question" is.

My take is that we are assuming a single impulsive burn regardless. Whether minimum velocity: eastward at midnight near perihelion to take advantage of Earth's rotation and orbital velocity. Or maximum velocity: westward at midnight near perihelion so that the rotation and orbital velocity force us to expend excess launch velocity pointlessly.

Or have I missed something?

 

[haruspex]

eastward at midnight near perihelion to take advantage of Earth's rotation and orbital velocity.

My worry is that the trajectory is still influenced by Earth's gravity until the probe has effectively escaped the Earth. At first, that would swing it towards the Sun. Does that matter?

 

[jbriggs444]

My worry is that the trajectory is still influenced by Earth's gravity until the probe has effectively escaped the Earth. At first, that would swing it towards the Sun. Does that matter?

At the speeds that we are talking about, I do not think that it would matter. However, I think that I understand what you are getting at. Let me try to put it in my words.

If the Sun and Earth were motionless then the situation would be uncomplicated. They would have a combined gravitational field that is unchanging over time. It would be a conservative field. It would have a well defined potential everywhere. We could easily tot up an escape energy requirement.

But the Sun and Earth are in relative motion. The combined gravitational field is not constant over time. It does not have a well defined potential everywhere. Hence things like gravitational slingshot maneuvers.

You are suggesting that our attempt to use an argument based on gravitational potential is misguided for this reason. That they are only valid as a reasonable approximation.

 

[theblazierbroom]

But that is obvious nonsense. If is equal to orbital velocity then our net sun-relative launch velocity is double orbital velocity. Doubling orbital velocity quadruples orbital kinetic energy. We only needed to double orbital kinetic energy to escape.

Ah, I see what happened. The kinetic energy to escape the orbit of the sun has to equal the potential energy of the circular orbit, not just the additional energy.

$$
T = U_{circular \, orbit}
$$
$$
T = m v_{orbit}^{2}
$$

The problem is an unnoticed change of reference frame. Orbital escape energy is properly calculated from the sun's rest frame. Launch relative velocity is properly calculated from the launch rest frame. You cannot validly equate the two without performing a shift of reference frame calculation.

Maybe for part b you should waste the maximum by making it anti-parallel.

Adjusting to have the frame of reference of the sun, and knowing that the maximum required energy will be launching in the anti-parallel direction, the velocity in the sun's frame of reference will be $v_{initial \, sun} = v_{0} - 18.5 \, \textrm{mi/s}$. Then, we have:

$$
mgR_{e} + m v_{orbit}^{2} + \frac{1}{2}m v_{f}^2= \frac{1}{2} m (v_{0} - 18.5 \, \textrm{mi/s})^{2}
$$
$$
18.5 \, \textrm{mi/s} + \sqrt{2gR_{e} + 2 (18.5 \, \textrm{mi/s})^{2} + (10 \, \textrm{mi/s})^{2}}= v_{0}
$$
$$
v_{0} = 47.3 \, \textrm{mi/s}
$$

I get the right answer (solution is 47.1 mi/s), but is this the correct line of thinking?

I feel I'm adjusting for the frame of reference wrong. Because this would suggest that if we go in the parrallel direction, we would subtract 18.5 mi/s from $\sqrt{2gR_{e} + 2 (18.5 \, \textrm{mi/s})^{2} + (10 \, \textrm{mi/s})^{2}}$, and then $v_{0} = 10.35 \, \textrm{mi/s}$. This is less than what I derived in part (a), $v_{0} = 12.1 \, \textrm{mi/s}$, which doesn't consider the orbit of Earth.

 

[haruspex]

Sorry, no further thoughts.
There’s nothing else in your course notes that might be relevant?

 

[theblazierbroom]

Sorry, no further thoughts.
There’s nothing else in your course notes that might be relevant?

I think this is it! Thanks for the help.