What Speed Does a Spacecraft Need to Escape the Solar System from Earth's Orbit?

[Bashyboy]

Homework Statement

(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth's orbit?

(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth's orbit?

Homework Equations

The Attempt at a Solution

I am not exactly sure how to solve this problem. Should I use an energy approach? Do I need to calculate the escape speed for each planet, because it will pass by each one, and then sum all of the velocities together?

 

[voko]

What does "escape" really mean?

Regarding your solution, consider, for example, a trajectory perpendicular to the plane of the ecliptic. Would it pass by each planet? Secondly, consider that your speed is that sufficient to escape Jupiter; would passing by Neptune need any more speed?

 

[phinds]

The answer is posted online. I remember looking it up. Do you know how to use Google search?

 

[Bashyboy]

The answer is posted online. I remember looking it up. Do you know how to use Google search?

I am quite capable to use Google's search engine; however, I'd like to figure out the answer without copying and pasting.

 

[Bashyboy]

Regarding your solution, consider, for example, a trajectory perpendicular to the plane of the ecliptic. Would it pass by each planet?

I posted a picture of the two different trajectories. The line in red is the one I believe you are talking about. The line in blue is the one I was thinking about. Does the red line describe the trajectory you were speaking about?

Would it pass by each planet? Secondly, consider that your speed is that sufficient to escape Jupiter; would passing by Neptune need any more speed?

Well, it would seem like it would. Wouldn't the gravity of Jupiter pull on the projectile as it passed but it, thus reducing its speed? It then being possible that the speed is reduced by Jupiter's gravitational pull, the reduced speed of the projectile might then be not enough to escape Neptune's gravitational pull, to a place where Neptune's gravitational pull is negligible.

EDIT: Forgot to post picture.

 

[voko]

Yes, the red line would be one perpendicular to the plane of the ecliptic.

You have not said what you think "escape" means. Let's get this clear before we talk about anything else.

 

[Bashyboy]

From my understanding, escape speed is the speed necessary to reach a point away from the planet where the gravitational pull is so small it is assumed to be zero, implying that gravitational potential energy is also zero. It is at an "infinite" distance. Does that seem correct?

Oh, so the red line is correct. So, that's why we only have to consider the escape speed of the sun and earth, because it doesn't have to pass by the other planets; and because the projectile is at an "infinite" distance from Earth's adjacent planets, that their gravitational pulls are negligible?

 

[voko]

From my understanding, escape speed is the speed necessary to reach a point away from the planet where the gravitational pull is so small it is assumed to be zero, implying that gravitational potential energy is also zero. It is at an "infinite" distance. Does that seem correct?

I think you are slightly confused. Zero potential energy does not follow from zero force of gravity. Recall that the "zero" level of any potential energy is completely arbitrary. It just happens that the "zero" level of potential energy is usually set at the infinity, and the energy is negative elsewhere. What's the formula for it?

Anyway, "escape" means "get infinitely away" and "reach zero potential energy". What about the projectile's speed and the kinetic energy "at the infinity"?

Oh, so the red line is correct. So, that's why we only have to consider the escape speed of the sun and earth, because it doesn't have to pass by the other planets; and because the projectile is at an "infinite" distance from Earth's adjacent planets, that their gravitational pulls are negligible?

Being infinitely away does not free us from having to deal with the planet's gravity. In the end, we still have to deal with the Sun's gravity in the same situation.

 

[Bashyboy]

Yes, but what happens to Newton's Law of Gravitation as you go "infinitely" far? It goes to zero. In fact, it goes to zero more quickly than the Gravitational Potential Energy Function.

 

[voko]

Yes, but what happens to Newton's Law of Gravitation as you go "infinitely" far? It goes to zero. In fact, it goes to zero more quickly than the Gravitational Potential Energy Function.

This is all correct; but how does this help you?

 

[Bashyboy]

It allows me to make the reasonable assumption that, because all of the other planets are far enough away, we don't have to consider all of them. Although the sun is far away, its mass is still large enough for it to affect the projectiles escape speed.

 

[voko]

I do not see any physical basis for this assumption. "Far enough away" and "mass large enough" sound very much like wishful thinking.

Forget about the entire solar system for a minute. How does one determine the escape speed for just one gravitating body?

 

[Bashyboy]

Then how else would you explain the solution. By the way, I found it on the internet. To find the escape speed in this particular case, you sum the escape speed of the sun and earth. We factor in the gravitational pull of the sun, even though we are launching from, yet we don't factor in the gravitational pull Venus or Mars. Why, because they are far enough away and have small enough mass, that there gravitational pull doesn't affect the escape speed. I am sure you could show mathematically, giving us a basis for our assumption, that the magnitude of Venus and Mars when compared to the Sun and Earth would be insignificant in comparison. That's the only way I can see how to explain the solution.

 

[voko]

Again, I recommend that you start with just one body (a planet or the Sun). Understand how the mathematical formula for the escape speed is obtained in this simplest case. Then think how you could treat the entire solar system.

 

[Bashyboy]

Now that I re-look at this solution for part (a), it doesn't seem correct. http://answers.yahoo.com/question/index?qid=20091123234313AAPfdJt What do you think you? Why is it wrong?

 

[voko]

I would like you to work out the solution for the simple problem (projectile escaping one body) first. Then we can discuss the more complex problem.

 

[Bashyboy]

Well, you'd use the formula $v_e = \sqrt{\frac{2GM}{r}}$, where M is the mass of the object the projectile is trying to escape, and r is the radius of that object.

 

[voko]

That is not correct. r is not the radius, it is something else. Do you understand how this formula is obtained?

 

[Bashyboy]

According to Wikipedia, r the distance from the center of gravity. I am not really certain of the meaning of that, though.

 

[Bashyboy]

Perhaps we could go over your simplified problem first, voko? Moreover, would it be to much to ask to go over the particulars of the formula with me?

 

[voko]

I would say we should do the simple problem first. Re-read the beginning of the discussion regarding kinetic/potential energy consideration. Can you use those to derive the simple formula?

 

[Bashyboy]

Well, a spaceship launched from Earth will only possesses potential energy due to gravity and kinetic energy due to motion. I know our system is the Earth and the space ship, so do we need to consider the energy of the earth? The escape velocity will allow use to reach a distance, an "infinite distance, where we are far enough from the Earth so that the affects of gravity have effectively diminished to zero by the time we reach that point, implying the final gravitational energy will be zero. Furthermore, the escape velocity will also allow us to reach that same point mentioned in the preceding sentence, so that the gravitational force opposing the space ship's motion will reduce it zero, implying kinetic energy is zero. Hence: $U_{ig}+K_i=0$. Is this correct?

 

[voko]

Total energy, meaning the sum of kinetic and potential energy, is constant (conservation of energy). What is total energy at the infinity? What is total energy in some vicinity of a gravitating body?

 

[Bashyboy]

At infinity, the total energy is zero, right? As far as a body being within a finite vicinity of a gravitating body, I believe by convention the energy is negative; although, I don't know why we have chose it to be negative, what makes it so much more convenient?

 

[voko]

The convention is that the POTENTIAL energy at infinity is zero. This does not mean the total energy must be zero at infinity. It could be assumed zero because, physically, we are looking for the minimal escape velocity, and being at infinity with greater than zero kinetic energy obviously requires more energy at the origin.

Anyway, what is the formula for total energy at the vicinity of a gravitating body?

 

[Bashyboy]

I'm not sure what the formula is for that particular case.

 

[voko]

Formula for kinetic energy?

Formula for potential energy?

Their sum?

 

[Bashyboy]

Oh, so: $E_{mech}=U_g + K\implies E_{mech} = -\frac{Gm_1m_2}{r}+1/2mv_{tan}$
Is that correct?

 

[voko]

Exactly. Now you know that total energy is conserved, so must be equal to total energy at infinity.

 

[Bashyboy]

Well, I know that gravitational potential energy will fade to zero, resulting in the kinetic energy term having to compensate for this. Will the term have to change somehow?

 

[voko]

Review #25.