# Homework Help: Minimum Velocity of an object thrown directly up to never fall down?

1. May 5, 2014

### Nathanael

EDIT: This is in the wrong section isn't it? How do I move it to the General Physics section? (My bad.)

This isn't a homework problem (I'm not in a physics class) so hopefully this isn't the wrong section.

My question is about finding the minimum velocity needed for an object (directly upwards) to never return to Earth.
(**Ignoring air resistance and other subtleties such as other sources of gravity and such**)

I have an idea for finding it, but I've gotten stuck. My idea was to make it "choppy" (turn acceleration/velocity into being piece-wise-linear) and then take the limits as the piece size gets infinitely small.

I've turned it into an equation that I think would solve it, but the problem is that it's an infinite sum and I can't make progress towards evaluating it because I can't figure out how to generalize the nth term in terms of n. Perhaps that makes this a bit more of a math problem, (but most physics is, to an extent,) so sorry if this is in the wrong section (this is my first post).

I've posted a picture of my "solution" (I don't know where it is but I'll assume it will appear).

The equation in the red circle (at the top) is my "solution"

G is the universal constant of gravity, M is the mass of Earth, r is the radius of Earth and V is the great thing I'm trying to solve for.

My idea behind this equation:
b(subscript)1 is the change in velocity over the first height interval (Δh) because Δh divided by v is the time spent moving through that interval, and a(subscript)1 is the acceleration at that height. Then for the next height interval you replace v with v+change-in-v (in other words, replace v with v+b(subscript)1) and apply the same logic (and so on for all terms). Taking the infinite sum of this would give you the change in velocity as you got infinitely far. Taking the limit as Δh→0 would make it "more true" (because acceleration is continuous, not "choppy") and setting it equal to -v would mean you only reach zero velocity at an "infinite distance" (and that would therefore make v the minimum velocity, right?)

My questions are:
1.) Is the logic/equation correct? Would it give you the appropriate minimum velocity?
(I think it would, but then again, I ALWAYS think my logic is right, and it very often isn't.)

2.) Is the equation solvable? Is this an adequate approach to the problem? Can you solve it for v?
(I think you can, since there's only constants and v, but the infinite sum might cause problems, I'm not very mathematically advanced so I don't know)

3.) Is it possible to generalize b(subscript)n in terms of n? Is it possible to evaluate the sum without generalizing it in terms of n? (If so how?) Are there other methods of evaluating the sum?

Question 3.) is the "wall" that i've run into. I can't solve the problem because I can't figure out how to generalize b(subscript)n in terms of n. (I've generalized it in the only concise way I know how, but I don't think it's solvable like as it is. It would take some genius maths I think.)

Can anyone help with these questions? Thank you and sorry for the lengthy post.

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Last edited: May 5, 2014
2. May 5, 2014

### Nathanael

Also, I'm up for discussing alternative methods.

I have one other method in mind, but I think I need to know more calculus or differential equations. The other method goes like this:

Create a function for the impact velocity depending on the initial distance away (assuming initial velocity of zero) then take the limit of that function as the initial distance (the independent variable) approaches infinity.

I THINK this would give you the same answer "v" (even though it seems like a different process) because I'm assuming it would be "symmetrical" (meaning, if you start at zero velocity at an infinite distance, then your final ("impact") velocity would be the same as the initial velocity (at earth) that ended up being zero at an infinite distance). But I don't quite know that it's true that that would give you the same answer, it just seems to be so.

But, to generalize the impact velocity as a function of the initial distance, (and therefore to solve it using this method,) I would need to integrate (integration is just the "continuous sum," correct?) the acceleration against time, but I don't know the acceleration as a function of time, I only know the acceleration as a function of the distance (newtons law of gravity). So I would need to know the distance as a function of time, but that would depend on the acceleration, and thus starts the endless circle.

I don't know how to integrate functions yet (except by anti-derivative) let alone integrate a "circular function" like that one.
(I call it a "circular function" because you need to plug your distance into the acceleration to see how your distance changes then continuously replug it in over and over at every instant of time to solve it. (I can picture 3d system (time/accl./dist.), but I don't know how to go through the circle with numbers and especially not continuously, that takes some hardcore math skill) In theory, it would be solvable, because all the information is there (I think?) but I have no idea how to tackle that sort of problem. I think that these so-called "circular functions" are the nature of what are called "differential equations," right?)

So, this second method is not really a feasible method to solving the problem (at least not for me and my limited mathematical abilities).

Last edited: May 5, 2014
3. May 5, 2014

### ehild

Your idea, that the initial speed is the same as the impact speed if the stone returns, is correct. That comes from conservation of energy. It holds for gravity.
Also, you think correctly that the minimum initial speed corresponds to infinite distance of travel and infinity is never reached.
So you can find the minimum upward speed to determine the energy of the stone on the surface of the Earth: E= KE +PE= const, KE=0.5mv2, and the potential energy is PE=-GMm/r where M is the mass of Earth, m is the mass of the stone and r is the distance from the Earth centre. At the surface of the Earth, r=R, the radius of Earth. At infinity, r→ ∞ PE=0, and KE should be also zero, so KE+PE=0. It should be zero also on the surface of Earth. From that, you get the speed.

Your next question is if the stone never returns if it goes to infinity. That looks plausible but you want to prove it mathematically.

The motion of planets and satellites are described by the Universal Law of Gravitation. From that, the planets move along closed orbit (ellipse) if their energy is negative. And it was observed by Kepler and also can be derived mathematically that the relation between the time period T of the orbits and the semimajor axis "a" of the ellipse is T2/a3=constant for the same central mass, Earth in this case. if a is infinite, T is also infinite.
Throwing up a stone, it goes along a straight line and back. You can consider it as a very thin ellipse, with zero minor axis.

You can also derive the time dependence directly, and there were such problems solved here in PF, but it involves quite hard integration.

ehild

4. May 5, 2014

### Nathanael

So you're saying that an up-down motion can be viewed as a special case of an elliptical orbit, and because of this, (and because of the T2/a3=c relationship,) that the only way to never return (a.k.a. to have an infinite period) is to have an infinite distance? Is that a legitimate proof?
I'm not a proof kind of person (frankly because I'm not clever enough to find them) but this is quite a brilliantly simple way of proving that.
(I am completely lost when it comes to proving something that I only know from intuition.)

KE+PE=Constant... that's a law of physics (I feel dumb )... (I'm very new to physics, I didn't think about this)

Is that where the formula KE=0.5mv2 comes from?

So you're saying that you can literally just find the v that satisfies the equation -GMm/r + mv2/2 = 0 and that will give you the same answer as the limit of my infinite sum (assuming no mistakes in my equation) and it will give you the same answer as solving that differential equation????
-GMm/r + mv2/2 = 0 ............ Wow. That's amazing, what a genius use of the KE+PE=Constant law, that is amazingly simple.... I always try to find the simplest way to solve something, but that, that's just remarkably simple.

Will that seriously give you the answer?? Or am I misunderstanding what you meant?
I did that and got 11,180 meters per second, that's quite a bit higher than I expected. (Although I suppose expectations are pretty useless in physics, and it is a some-what reasonable answer)

The idea that initially began me thinking about this problem was that I thought, "(In a hypothetical universe with just Earth and an object,) No matter how unimaginably far away an object is it will always eventually fall to the earth (Or should I say the Earth and the object will fall to each other)." Then I thought, "I wonder if, if the object is moving away from the Earth at any speed, will it still always fall to the Earth?"
One part of me said Yes, because, even though the gravity will be very weak, it will be acting against the objects motion for infinitely long. If it is acting for infinitely long, it shouldn't matter how weak it is.
Then another part of me said No, it wouldn't fall to Earth, because the rate at which the acceleration is slowing down will be far faster than the rate at which the speed is slowing down.
But then another part of me said "But it will still be a finite deceleration for an infinite time, so it will still EVENTUALLY slow to zero velocity then turn around and come back."
But then part of me said "but the velocity could be decreasing for infinitely long yet still never get below a certain value, (like an asymptote) because the acceleration is decreasing so much faster that there must be some limit involved (like an asymptote)"

This train of thought eventually led to the question I asked in my original post. I concluded that if the minimum velocity to never fall back down were infinite, then an object moving away from Earth at ANY speed and ANY distance would *EVENTUALLY* fall back to Earth, and if the minimum velocity were finite, then it didn't necessarily have to fall back to Earth (I'm still talking about this hypothetical Universe with just Earth and the object).
I also concluded that the minimum velocity would in fact be finite (and you could potentially never return to earth if you're moving away quick enough (which at a reasonable distance, probably wouldn't need to be very quick))

However, all my "conclusions" are actually just intuitions....
So I want to ask you (and anyone else) how would you approach these conflicting ideas? (since you're an actual physicist and I'm just a naive kid)

Last edited: May 5, 2014
5. May 5, 2014

### ehild

Nathanael,

It is very good that you are trying to understand Physical phenomena, and your method is very similar to those of ancient Greeks.... But a lot of things have been discovered since then. Physics became 90 percent Mathematics. There are some basic laws, as Newton's Laws in Mechanics, which, combined with Calculus (founded also by Newton) describe quite accurately how a thrown stone, a spaceship, a satellite, the planets move. Applying the laws of Physics, we were able to send satellites orbiting around the Earth and seeing even your car with the GPS; keeping the International Space Station (ISS) on orbit and sending supply to the people working on board there. You can see the ISS with naked eye moving among the stars on the heaven if you look at right place in right time, as the motion of the ISS can be accurately predicted. It was also possible sending rockets to the Mars or even father, to the Saturn, landing on its moon Titan, and sending back data and pictures from its surface. All our present life depends on tools which were invented and created by applying the laws of Physics and Maths.
You need to learn the basics of Physics and Maths. Do it systematically, starting from the beginning.

ehild

6. May 5, 2014

### ehild

Well, it can be proved formally, by setting up and solving equations. That was an explanation without Maths.

You asked about an object thrown up vertically. It would return as it falls back on the same way as it started, if it is launched from one of the poles. Otherwise, it would have some side-way component of velocity, equal to the velocity of the surface of Earth.
If the stone could penetrate and fall through the Earth without loss of energy, it would oscillate away and back forever.
Allowing enough horizontal velocity, the stone never falls down, but orbits above the surface of the Earth. For such circular orbit, the force of gravity equals the centripetal force for the circular motion. The minimum horizontal speed is mv2/R=GmM/R2. But the gravitational force at the surface of Earth is equal to mg=9.81m. So the minimum orbital velocity is just v=√(gR)

A Physicist can not exist without intuition, but he needs also a solid knowledge.
KE is abbreviation of kinetic energy and PE stands for potential energy. The kinetec energy of a small particle is 0.5 mv2. The kinetic energy of an extended body is the sum of the KE-s of all its particles. The potential energy is related to force. If the force is conservative, the sum of the potential energy and the kinetic energy of the body is constant.

You have the equation mass times acceleration = force. That means a differential equation for the position vector. If you solve it, you get the same relation between distance and velocity as you get from conservation of energy. Actually, conservation of energy is derived from Newton's second equation F=ma by integration.

I suggest to get a Calculus-based Mechanics book and read... You will find how the motion in gravitational field can be solved.

ehild