# Does an "escape velocity" exist in this situation (see details)

Asked by PhiNotPi (12607) December 16th, 2011

Here is the scenario: There is an almost completely empty universe. It does not have to be completely empty, but any other mass/energy must be small enough that it has not effect on how events play out. In the universe, there is one planet and one spaceship on the planet. When the spaceship is launched, it instantly accelerates to any given velocity and then stops acceleration forever.

Here is the question: How would one calculate the minimum possible velocity that the spaceship needs to be able into space forever, without ever coming to a halt?

Here’s the catch: Since the gravity of the planet is felt everywhere in the universe, the spaceship always experiences the forever-decreasing but never zero effect of the planets gravity. The spaceship is always slowing down, but always less and less as time goes by. Does an escape velocity exist in the first place?

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Past edit time, but “to be able into space forever” should be “to be able to travel into space forever”. There are also some other small errors.

PhiNotPi (12607)

@PhiNotPi “There is an almost completely empty universe. It does not have to be completely empty, but any other mass/energy must be small enough that it has not effect on how events play out.”

A completely empty universe is not a universe… That would be a theoretical void.

As to mass, no matter how small, there is no possible way for a universe to exist with small enough mass that it doesn’t effect how events play out. Any event that occurs in a closed system will be effected to some degree by the compounds of that system. It can’t not effect it.

@PhiNotPi “In the universe, there is one planet and one spaceship on the planet.”

There’s your mass… I suppose the materials used to build the spaceship were harvested from the one planet in the universe. Thus the planet has less mass than it did before the spaceship was created. The planet will also be projecting gravitational warpage upon the surrounding space/void/thingy that you propose it exists within.

@PhiNotPi “When the spaceship is launched, it instantly accelerates to any given velocity and then stops acceleration forever.”

Every action has an equal and opposite reaction. Therefor the acceleration propellent causes/effect upon the closed system it resides within.

@PhiNotPi “How would one calculate the minimum possible velocity that the spaceship needs to be able into space forever, without ever coming to a halt?”

You’d need some pretty specific data to form an equation with… mass of both planet and spaceship, sampling of average mass per cubic area of space/void/thingy, rate of entropic decay… and then determine if your space/void/thingy has an expansion rate (like our universe) or if it is in a state of contraction (if that theory is entertained), or does your universe oscillate (which requires entropic forces to be less than .001%)?

@PhiNotPi “Here’s the catch: Since the gravity of the planet is felt everywhere in the universe, the spaceship always experiences the forever-decreasing but never zero effect of the planets gravity.”

Decide the path… vertical, eliptical… start with accurate measurements of mass, and do the calculation

@PhiNotPi “Does an escape velocity exist in the first place?”

Only to the point where the gravitational pull of the planet is greater than the gravitational pull of the spacecraft. Get far enough away, and the spacecraft mass should overcome the influence of the planet mass.

That is the question that determines whether or not the universe will reverse its expansion and collapse into itself, or simply expand into the infinite. Depends on the mass of all the objects involved.

saint (3970)

The universe can’t collapse. The mechanical efficiency of the universe is 1/100,000,000 of a percent. An engine will not oscillate below a 1% mechanical efficiency. The universe is 8 orders of magnitude short of that limit.

@RealEyesRealizeRealLies The whole “almost empty universe” thing was me trying to prevent people from bringing other planets and stars into the question. It was really me trying to have the point of eliminating “what if” scenarios.

The “instant acceleration and no more” was me trying to get the point across as to what escape velocity is, that the objects motion must not be influenced by other “what if” forces that someone could pull into the situation. Perhaps I should have eliminated the existence of the spaceship and have relied on Fluther users to know what escape velocity is (which they do).

So, this appears to be a case of when eliminating what if’s actually created more loopholes in the question. I hope you understand.

PhiNotPi (12607)

Hmmm…the force the ship feels will decrease with the square of the distance between it and the planet, and so its rate of deceleration will always be decreasing with the same proportion. This appears to me to be a complex problem. I’ll hit it again tomorrow when I’m fresh. Interesting Q as always.

Mariah (23087)

Why does ”...the forever-decreasing but never zero effect of the planet’s gravity” throughout distant space bother you? When you integrate a 1/R^2 force out to infinity, you get a finite result that represents potential energy. To escape Earth a projectile must have enough kinetic energy to overcome its gravitational potential, which is when

(½)mv^2 {kinetic energy} = GMm/r {potential energy}

hence
Ve = sqrt( 2GM/r )

where m = mass of projectile; M = mass of planet; G = grav. constant; r = distance from planet; Ve = escape velocity; sqrt(x) = square root of x.

Note that Ve does not depend on m, the mass of the ballistic projectile—only on the mass M of the planet. For Earth it’s about 11.2 km/sec = 25,050 miles/hour. Also Coulomb’s Law of electrostatic force is mathematically analogous, so you could similarly calculate the escape velocity of an electron leaving the surface of a static-charged party balloon.

Although the Earth is always pulling a little bit gravitationally on the projectile, at great distances the force is too minuscule to matter. Escape velocities exist for all 1/R^2 forces as well as for higher inverse powers of R (distance).

gasman (11261)

Seems to me that we have a test case now. While it;s a much more complex Universe we are dealing with than the one @PhiNotPi posits, still, it can be broken down to that level of simplicity to make understanding its behavior more easy toi grasp. And it’s behavior is that, even through mass attraction “should” be eventually drawing everything from the big bang back to bang entrap for a big crunch, observations show the opposite is happening. The objects flying away from the original point of the big bang are accelerating their dispersion. So I believe your space ship would do the sane,

ETpro (34378)

I think the original question, as posed, is about elementary local kinematics and need not involve cosmology or big bang expansion. By local I mean, let’s say, a radius of one light-year.

As for a finite integral of an infinite gravitational field, it is reminiscent of Zeno’s paradoxes, which (in modern formulation) involve infinite series converging to finite sums, no longer considered a paradox.

gasman (11261)

I don’t see how your example differs from the real universe we live in and yes there will be a definite escape velocity. The only difference is that inertia may not exist in an empty universe.

flutherother (25315)

@gasman The OP included some other matter here and there in the universe. In such a case, I believe my answer stands. But if we posited a universe with just a single planet having no sun but life capable of building a spaceship, we have an interesting equatino to solve.

As I understand it, space-time would not then exist beyond the ionosphere of the planet in question. But a spaceship blasting beyond that outer shell would pull space.time with it. Frankly, I have no idea how the laws of this Universe’s physics would apply to that situation; and I don’t think anyone else does either.

ETpro (34378)

@ETpro Well, I’m aware of Mach’s principle that the distant galaxies establish a local inertial frame (so we know how much an ice skater’s arms fly outward during a spin) which might not be the case with empty space. Plus Einstein’s view that the shape of spacetime is warped by embedded masses. So in some sense you have to take into account all other existing bodies.

Still, physical reasoning requires judicious simplifications & I took the nub of OP’s question to be in dealing with the extended effects of a finite gravitational field between two bodies as distance goes to infinity, comparing it with Zeno’s paradox. Maybe I didn’t get the question.

I much prefer 2-body problems. I don’t even like 3, and when you’re dealing with, let’s say, 10^80 or so – fugetaboutit !

gasman (11261)

Sorry, I forgot about this question. @gasman is right. The force on the ship at any given time is proportionate to 1/r^2. 1/r^2 is an example of a p series with p>1, which means that it converges and has a finite integral. Force integrated with respect to distance is the definition of work. So this means the planet does a finite amount of work on the ship, and the ship can therefore escape it by expending a larger amount of energy.

Correct me if I’m mistaken. I am a newbie physics student.

Mariah (23087)

For a spherically-symmetric body, escape velocity is calculated by the formula

Ve = sq root of 2GM/r

where G is the universal gravitational constant (G=6.67×10–11 m3 kg-1 s-2), M the mass of the planet, star or other body, and r the distance from the center of gravity.

In this equation atmospheric friction or air drag, is not taken into account. A rocket moving out of a gravity well does not actually need to attain escape velocity to do so, but could achieve the same result at walking speed with a suitable mode of propulsion and sufficient fuel. Escape velocity only applies to ballistic trajectories.

The term escape velocity is actually a misnomer, as the concept refers to a scalar speed which is independent of direction whereas velocity is the measurement of the rate and direction of change in position of an object.

By the way, “Ve” with respect to Earth’s gravity is 11.2km/s (approx. 40,320 km/h, or 25.000 mph).

Hope that helps!

Esedess (2981)

@Mariah Exactly right.

@Esedess Glad our formulas agree! “Escape speed” is indeed more accurate than velocity, so long as we understand it to mean in a radial outward direction—ignoring tangential components.

gasman (11261)

@gasman Haha. I’ll go with that! I was just being a wise-ass right there anyways. lol
Go~ team physics! =J

Esedess (2981)

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