# Acceleration, distance and speed

Tags:
1. Aug 16, 2016

### Dream Relics

Okay, so admittedly my background and knowledge base is pretty weak compared to most of you guys on here. I have been reading as much stuff stuff as I can find because I find physics and cosmology so very interesting. I just found today a physics tutorial series on youtube. The beginning lessons are about the equations for figuring acceleration, distance and time and that sort of thing, and a little bit so far on vector and scalar. But while watching it a hypothetical situation occurred to me. But it may well be that there are a lot of established limits on certain things that I am unaware of so forgive me if this is a silly question.

The question is this. If the rate of acceleration from gravity is a constant at 9.8 m/s/s? Then presumably, if someone dropped something from whatever the highest height is above that earth from which something would fall without just floating away or going into orbit or whatever else might happen, then from that height there is a known maximum velocity that can be achieved by the falling object, of course forgetting about wind resistance and that sort of thing. So then I wondered, is the height of an influencing gravity field around a massive object always the same no matter how massive the object is, or if the object is more and more massive and the therefore the gravity more intense does the height away from the surface of that object of the gravity at the point where it would pull an object in increase? If it does increase, then is it possible to have a gravity field with enough height away from the surface that a falling object would then have enough time as it is falling to accelerate all the way up to the speed of light? it seems like if acceleration is always constant under gravity and there was sufficient distant for the object to cover before it made contact then that could happen. or if not, why not? I am really looking forward to having it explained to me why that won't work. So please tear me apart.

Though this one other thought occurs to me, if that can happen, then maybe somewhere beyond the observable universe there are super supermassive black holes that are doing just that to the distant galaxies, and that is what is causing the acceleration of the expansion all the galaxies out there.

2. Aug 17, 2016

### benny_91

Whoa...Pretty good question. I don't think this can be answered using classical physics. You might need relativity. Or may be this could just be possible. Regarding your final paragraph I think all blackholes make other heavenly bodies revolve around them just like the sun makes the earth revolve around it using its gravitational force. The earth does not fall onto the sun. Similarly our solar system and the other stellar systems of the Milky Way revolve around a massive blackhole which is at the the center of our galaxy. So nothing goes and falls on the other.
Regarding the first question I too am eagerly waiting for a General Relativity scholar to answer it!!

3. Aug 17, 2016

### TahirGorgen

4. Aug 17, 2016

### TahirGorgen

Sorry i am new. For the replay with the hole quote.

Btw the constance is just near the surface of earth.

5. Aug 17, 2016

### TahirGorgen

It's verry difficult to understand. As in coriolis you have an upside push and an towards pull. If there where no atmosphere, the density of other matter, te constance would not be set at 9.8 you would keep accerelating till the surface. Due to the G of earth and the density of matter -without resistance-.

6. Aug 17, 2016

### TahirGorgen

Higher knowledge: so the mass/density of an planet (without resistance) for it's orbitingspeed, does not matter, because kg/weight is just an illusion of G.

7. Aug 17, 2016

### Borg

If you don't have any wind resistance, then the object continues to accelerate. There wouldn't be a maximum velocity because the object will continue to accelerate until it strikes the planet. In Earth's case, there is atmosphere so you cannot ignore wind resistance. Then there is a maximum velocity which is called the terminal velocity. That velocity is highly dependant on the shape of the object that is falling through the atmosphere.
There is no point where objects just "float away" as you've described. Gravity does not have a limit for it's ability to affect other objects - its effect just becomes weaker as you get farther away. If you had a universe with just the earth and an object light years away, there would still be a very small effect on the object due to the earth's gravity. The acceleration wouldn't be 9.8 m/s2 but it wouldn't technically be zero either.

8. Aug 17, 2016

### jbriggs444

The acceleration from gravity is only a approximately constant when your distance from the gravitating object is approximately constant. For everyday purposes, sitting where we do on the surface of the Earth, our distance from the center of the earth is approximately constant. So we see the acceleration from gravity as approximately constant -- 9.8 m/s2

Gravity follows an inverse-square law. The farther away you get, the weaker the influence. Double the distance and you divide the force by four. Multply the distance by ten and you divide the force by one hundred. That law applies easily for point-like objects. For big objects, one can treat them as a collection of small pieces. But there is an important result -- Newton's spherical shell theorem. It turns out that for a spherically symmetric object like the earth, one can pretend that all the mass is concentrated in the center and the inverse square law will continue to work just fine (for objects that are not underground).

In mathematical terms, this becomes Newton's universal law of gravitation:

$F=G\frac{mM}{r^2}$

where F is the attractive force of gravity between you (mass m), the planet (mass M}, r is your distance from the center of the planet and G is Newton's universal gravitational constant. If you prefer, you can divide by your mass (m) to obtain:

$a = \frac{F}{m} = G\frac{M}{r^2}$

If you want to obtain the impact velocity when falling from a given height, that can be done. Start by taking the integral of gravitational force as r varies from your starting position (measured as a distance from the center of the planet) to its ending position at the surface of the planet. This is an integral of force times incremental distance. Force times distance is work -- a change in energy. This integral gives you the kinetic energy gained as the object falls. Evaluating the integral yields:

Impact Energy $= \frac{1}{2}mv^2 = GMm(\frac{1}{r} - \frac{1}{h})$

Impact velocity $= \sqrt{2GM(\frac{1}{r} - \frac{1}{h})}$

It turns out that the impact velocity falling from a point infinitely high up is still a finite value. Reversing the trajectory and playing it backwards, this impact velocity is the initial speed that you would need to escape to infinity. This is more commonly known as "escape velocity".

Yes, one can have objects massive enough and compact enough in radius so that their escape velocity from the surface exceeds the speed of light. Yes, this means that no trajectory originating at the surface can escape to infinity. However, the above equations are not accurate in this regime. One needs to adopt a different model of gravitation where it is not a Newtonian force and is instead a feature of curved space-time in order to obtain accurate predictions. You would be talking about a black hole.

Last edited: Aug 17, 2016
9. Aug 17, 2016

### PeroK

In classical physics, the gravity of a large object could accelerate an object to beyond the speed of light.

But, if the mass of an object is great enough and as the velocity of an object approaches the speed of light then classical physics is no longer accurate. A new model of gravity and motion (General Relativity) is needed.

In this case, the speed of a falling object would never exceed the speed of light.

10. Aug 17, 2016

### Khashishi

There's no highest height. There's gravity up into outer space and beyond, though it does get weaker with distance.
No, the gravitational acceleration is the same for a feather or for a lead ball. The mass of the object doesn't matter, assuming it is negligible compared to the mass of the Earth.
No. A falling object won't accelerate above the escape velocity (if it was dropped). Google that if you are unsure. Unless you are falling into a black hole, escape velocity is less than c. (The case of falling into a black hole is more complicated, and would require us to digress about coordinates.)

No. Black holes and mass in general should be causing a contraction of the universe, not an expansion. Something else is going on. (We call it dark energy, but it isn't really understood.)

11. Aug 18, 2016

### benny_91

Someone mentioned a formula for impact velocity which is same as escape velocity. So can't we adjust the values of the terms in the formula (such as mass of the pulling body and its radius) to get the impact velocity equal to the speed of light? I have heard that mass increases with speed but even then that would not change its acceleration which means the nature of motion will be same even at such high velocities. I think to apply Newtonian mechanics in this case is inappropriate. If we try to answer this it should be using Relativity.

12. Aug 18, 2016

### Dream Relics

13. Aug 18, 2016

### Dream Relics

I have to figure out how to do that point by point excerpt quotations that I see other people doing in replies. How does that work?

14. Aug 18, 2016

### Dream Relics

Well yes, I was talking about black hole, or at least I was talking about anything with enough mass, like a black hole, or something maybe with slightly less mass than a black hole but still a lot of mass. I guess the way I am visualizing this, which may be wrong, is a sort of analogy between the way the atmosphere of the earth defines a line between the earth and space. That in a similar fashion, we know that gravity decreases in strength as you move higher up away from the earth. That then there must be sort of an event horizon for gravity in the sense that the pull becomes weak enough at some distance to not pull something down the way it does for us here on earth. even the highest flying jets, and satellites can crash down to the earth, but where is the point where the gravity becomes weak enough that an object won't get pulled back to earth. And if there is such a point does the distance where that point occurs relative to the surface of the massive body creating the gravity increase the more massive the body?

15. Aug 18, 2016

### Staff: Mentor

Insert a [/quote] tag followed by a [quote] tag at the position where you want to put your comment, and put your comment in between them. I suggest you also hit return/enter after the new [/quote] and after your comment, to make it easier for you to see what you're doing. (The forum software will format it in its own way when you post it.)

I suggest you also delete the parts of the quoted material that you're not responding to directly. If people want to refresh their memory of the complete post, they can simply look at it up-thread.

16. Aug 18, 2016

### jbriggs444

There is no such boundary. Gravity becomes weaker with distance but never vanishes entirely.

17. Aug 18, 2016

### benny_91

There is no such point where gravitational field of a body becomes completely 0. We can only say that the distance of that point from the center of the body under consideration tends to infinity. However far we go from that point we will always experience a force that gets smaller and smaller in magnitude with distance but never becomes zero. This is similar to the notion that if we continuously divide a 1 meter line segment into unequal parts of finite length and then continue this same process with the smaller part we get after each division, we will never get to point where the length of the smaller part is 0. Always finite but tends to 0.

18. Aug 18, 2016

### Borg

There are two ways. Selecting the Reply link on the bottom right corner of a post will populate your post with a section like the one here.

If you left-click and select a section of a post, you will get an option like this:

Selecting that Reply option will quote just that section like this:

19. Aug 18, 2016

### Dream Relics

and after your comment, to make it easier for you to see what you're doing. (The forum software will format it in its own way when you post it.) [/quote] [/quote]

[/quote]his is a test. And thank you. Everyone is so nice on here and quick to respond. It is frustrating to be interested in physics as just a layman. There are so many moving parts. In fact I think they are all moving parts. You think you have a picture of something that is clear. And then you realize, nope. [/quote]

hhmm' when you say quote tag you don't mean """".

20. Aug 18, 2016

### Dream Relics

Ah, that method seems to work. Thanks.