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?