Gravity Sucks!

There is an unbearable pun about gravity. The fact that it is both a pun, and unbearable explains both why I know it, and why I would choose to inflict it on you. Of course, I won’t just say it, but imply that it exists – relying on the fact that right now, every bad pun about gravity is rolling through your head, and if not your fingers are twitching to Google whatever it is I am talking about – thus my purpose is achieved with minimal effort!

Regardless, the orbital movement is the final context of motion in two dimensions, our first area of study. It is clearly an outgrowth of circular motion, but it has some interesting twists of its own. For a start, the force that maintains the circular trajectory is the force of gravity – the first of the four forces of the standard model of physics. Gravity has some difference from other forces that you have thus far encountered. As an example, both Weight force (F = mg) and Elastic force (F = -kx) only involve the object that the force is affecting. Newton’s universal law of gravitation has some similarities to the weight force, but a great many more differences.

Here it is:

G is the universal constant of gravitation – 6.67 * 10 ^ (-11). m1 and m2 are the masses of the two interacting bodies, and r is the distance between them.

For a start, the force is dependent not only on the mass of the object in question, but the also the second object. Furthermore,

the force is also dependent on the square of the distance between the objects. Using the formula by substituting values is thesimplest application of knowledge – much more important is understanding how the force varies as the values are changed proportionately. The image to the right gives some of the basic variations that you must understand, replicate and explain.

As you can see, increasing the mass of one or both objects has a simple effect of multiplying the net force  – the trickier part is the effects of distance. The simplest way to state the effect is that the force is multiplied by the inverse square of the distance factor.

The more interesting thing is the relationship between the Weight force formula and Newton’s law of universal gravitation formula – considering m1 in Newton’s law as equivalent to m in the weight formula, we can determine the “g” in the weight formula – an expression which involves the mass of the attracting object, the distance from the attracting object, and the universal gravitational constant. This is problematic, because you have always been told that “g” is constant – and it is clear from this formula that it is not!

I have always told you through our studies together that I will occasionally “lie” to you – that I will make things simpler when it is necessary – well this is one of those times when we come closer to the “truth”. The acceleration due to gravity “g” is not 10 m/s/s, or even 9.8 m/s/s – it varies all across the surface of the earth – due to variations in the earth’s crust – thickness, density and distance from the centre of mass.

It also decreases the further you get from earth – which is perhaps not so much of a surprise, given that you may have seen any number of pictures of astronauts orbiting the earth in “zero gravity”. Unfortunately, this impression is wrong – the astronauts are not in zero gravity, and it is simple to know why – but I’m not going to tell you! Your best guess or theory in the comments, please!

True zero gravity does exist – where you are outside the influence of any other attracting mass – where the Newton’s law F is zero, or outside the so called gravity well. My very favourite geek site, XKCD has a cartoon showing the depths of the various “gravity wells”. Check it out at this link – and while you are there, check out a few of the “What-ifs”. They are an exceptionally amusing application of physics!

The last thing that you need to know is about the relationship of orbital speed to orbital period to centripetal acceleration – a relatively simple formula based on the circumference of a circular orbit, and the period of that orbit – substituted into the formula for centripetal acceleration, it gives a new formula, which has some useful properties – as you can see to the right.

Last issue for this post is arranging a date for next week for our last class on this topic. Please note, as discussed in class, this session will not have an outcome – it will just be a session in which we cover the content of this blog post, and a general review session so that we can start the new term with the next area of study. Please put your availability in the comments to this post, so we can negotiate the best time for all of us.

See you in class!

Explore posts in the same categories: Kinematics, Physics, Year 12

1. Prathu Khairnar Says:

I agree with that, and its not just about the “zero-gravity”, but also, “zero-friction”. they say, once a satellite is in orbit, it wont fall back to Earth, because of the “zero-friction” which would result in an eternal orbit…

ofcourse there is friction in outer space, its just that, it is “little” friction. ofcourse air is not the big cause of friction, tiny particles bouncing of the satellite – and eventually accumulating enough force to either slow or speed up (knock it of the orbit). just like any other force being applied back on Earth. only the means of giving off the force has changed. they fix this by attaching engines and gyroscopes on the space station, and every once in a while they have to increase or decrease the speed, depending on the situation.

zero gravity is related comparison. even if travel far from Earth, and assume that only Earth exist in the universe and nothing else for a while, and applied “F= (GMm)/r^2”. we see that no matter how far i am from Earth, i still have a distance, a proper distance, that i cannot disregard. i would still be exerting force on the earth, and the earth to me. only difference is it is very very very small.

now lets bring all the stuff back in the universe, and we can see, that forces are being applied everywhere, and there has to be something pulling us somewhere or pushing (if u were to stand near a supernova).

We are heavily affected by the Earth’s gravity(because we are close to it). the Earth is affected by the Sun’s gravity, because of there large masses. and hence we are affected by the Sun, because we r influenced by the Earth. the Sun is in relative orbit with the Center of the galaxy, a super massive black hole. YES!! we are in orbit with a super massive black hole, but it is soo far away, that we wont be sucked into it just yet.

okay enough with that now. i have a very interesting question sir. have u seen this video? “http://www.youtube.com/watch?v=yVkdfJ9PkRQ ” .. its about pendulum oscillations. i was just wondering if that could ever be applied to massive objects. such as planets. like all of a sudden, all planets orbit the exact speed, or something like that? i know the moon and the earth already have the same rotational speed.. hmm.. my head is a weird place >_>

• CyberChalky Says:

Hi Prathu,

Yes, I’ve seen that video before – it is one of my open-night tricks. As to explaining my question, you’re heading in the right direction, but see my response to Darcy, above.

The friction concept is a bit off – amazingly, the majority of resistance to motion of orbiting bodies is due to induced current from the magnetic field of the planetary body. You’ll learn more about this in unit 4!

2. Frances Rowlands Says:

I am free Tuesday, Wednesday, Thursday of next week.

• Frances Rowlands Says:

Similarly to what Darcy and Prathu said, regardless of the position of the astronaut, there are two masses, the universal constant of gravitation and a distance to consider.

However, the force of gravity acting on the astronauts is certainly less than force of gravity experienced by a person on Earth’s surface. This is due to the value of ‘r’ increasing, but only by a small amount. If you think about someone standing on the surface of the Earth, the value of ‘r’ would be the radius of the Earth. If you then consider an astronaut orbiting the Earth the value of ‘r’ would be then be the radius of the Earth plus the height from the Earth’s surface.

Considering that the radius of the Earth is approximately 6,300 km and the altitude of an astronaut in orbit is around 400 km from the Earth’s surface, the value of ‘r’ for the orbiting astronaut would be 6,800 km. When compared to the person standing on the Earth’s surface with an ‘r’ value of 6,300 km, this increase in the value of ‘r’ is relatively small and would have some, but not a huge, on the value of Fg.

(All of this has been done with the assumption that the masses of the orbiting astronaut and the person standing on the Earth’s surface are the same.)

• CyberChalky Says:

Hi Frances,

An interesting exploration of the mathematics of the situation, but the answer is simpler yet. There is a simpler way to demonstrate that a body in orbit is not experiencing zero gravity (hint, hint!).

• CyberChalky Says:

OK, so Monday & Friday are out so far…

• Frances Rowlands Says:

Gravity is the force that holds the moon in orbit around the Earth and, similarly, is the force that holds the astronaut in orbit. If the astronaut was experiencing zero-gravity then the astronaut would not be in orbit, because in order to be in orbit the astronaut must be acted upon by a centripetal force. In this case the force allowing the astronaut to be in orbit is gravity.

• CyberChalky Says:

Bingo!
But then there is the next problem- if there is gravity on the astronaut, why do they float around as if there (apparently) wasn’t any?
p.s. Good work, Frances!

3. I’m free all week, so whatever day everyone else wants to have the class is fine.

My best guess for why the astronauts are not actually in “zero gravity”, is basically what Prathu Khairnar said in paragraph three; that no matter how far away you are from Earth there is still a distance, two masses and the universal constant of gravitation. Therefore, though very small, there is still a value for the Force of gravity. Kind of like an asymptote, continually approaching “zero gravity” but never actually meeting it.

• CyberChalky Says:

Hi Darcy,

Both you & Prathu are heading the right way, but not on the right track yet. You need to look at the link to the XKCD infographic on escape velocity – because the concept of escape velocity specifically counters your ideal of infinite reach of gravity…

Furthermore, until (and if) you study general relativity, the situation in which those astronauts find themselves is (almost) indistinguishable (by measurement) than actually being in (true) zero gravity…

4. I’m free Monday and after 11 on Tuesday, and then I’m working the rest of the week. I know that’s really inconvenient for everyone else so if there’s another way for me to catch up I’m all for it.

I followed the same track as Darcy, Prathu and Frances (initially) that because there is always a distance, two masses and the universal constant of gravitation there is a value for the force of gravity. I’ve since read Frances’ comment about being in orbit and therefore having a centripetal force.

• CyberChalky Says:

Hi Tessa,
Looks like we may be doing Tuesday afternoon. Unless we hear from someone else by noon on Monday, I’ll lock it in.

Good to hear that you followed the discussion so far, but we have another question now: if there is gravity, as all of you have explained, why do Astronauts in orbit float around as if there apparently wasn’t any?

5. Frances Rowlands Says:

The astronaut appears to float as if there is no gravity present because they are in a state of free fall around the Earth. If the astronaut was in an orbiter and dropped an apple, the apple would still fall, it would just appear to float, just as the astronaut would appear to float around inside the orbiter. This is because the astronaut, the orbiter and the apple are all falling together, and they are falling around Earth at the same rate (with the same acceleration).

• CyberChalky Says:

Frances, that’s a fantastic answer- but still only part of the problem. If they are falling as you suggest, why don’t the fall out of orbit and eventually crash to earth? Can you explain what you mean by falling “around” the earth?

• Frances Rowlands Says:

In order for them not to fall out of orbit they would have to be travelling at a velocity so great that every time they are pulled back towards the Earth by gravity they miss Earth and continue “falling” on a circular path around Earth. They are travelling fast enough that even if they keep falling they are never going to hit the Earth. So they are effectively falling around the Earth.

I’m not sure if the wording is correct or if it makes complete sense, but that’s my current understanding of the concept.

6. Hayden Maughan Says:

I’m away Monday Tuesday and maybe Wednesday.

There is a very small amount of gravity acting upon an astronaut. Because the value of r is greater it means we are dividing by a greater number in turn producing a rather insignificant gravity force on the astronaut.

So although the attraction to the earth is small, this is the somewhat centrepetal force that holds the astronaut in orbit.

Say a satellite that orbits relatively close to earth would have a higher gravity force alowing it to easily hold its orbit around the earth. Where as an astronaut in deep space would have a very large r value producing a very small gravity force.
Am I anywhere near the right track?

• CyberChalky Says:

*Hayden – you are on *a* track, but not the track. In other words, your thinking is correct, but needs a little refinement. Check the conversation in the comments.

7. CyberChalky Says:

Hi All,

It looks like we have an irreconcilable scheduling conflict. This means that we are going to have to handle this differently. The main purpose was going to be covering orbital motion analysis and providing feedback on assessment.

My suggestion is as follows – we hold an after school session in the first week for the final section of motion. If you wish to have feedback before this I can arrange a meeting with any who are interested at a common location – maybe the school, maybe Eastlands or similar. What I won’t do is just send you your final score – I want your results to be part of process, not the only part of the process!

8. Matthew Tucker Says:

Like what Frances said,
the reason the astronaut is not experiencing “zero gravity” is because if they were experiencing zero gravity they would not remain in orbit; they would be floating around, heading in any which direction like my worst nightmare. The reason they stay in orbit is due to, like it has been said above, the force that earth exceeds on us. The best way to loom at it, i think, is to imagine the astronaut as the moon. Yes i know the moon has a force being gravity but the moon stays in orbit of earth 365 days of the year. Therefore there must be a force that is EQUALLY exerting on the moon, like the moon is to earth.
So if we take my terrible analogy and imply that onto our astronaut, we are able to give a fairly simple explanation as the why the astronaut stays in orbit, and doesn’t fly around swimming, having a great time.

I’m probably wrong but that’s what I think.

• Matthew Tucker Says:

The force that is being applied to the astronaut is what keeps it in orbit of earth. The further you go from earth the smaller force of gravity. The force of gravity will always be there however the strength of it will increase/decrease depending on how far away you are from earth.

Mr G, if you don’t understand what I mean, imagine the astronaut as the moon, and come up with some amazing physicsy reason as to why I think this, and that’s my answer. :D:D

• CyberChalky Says:

Hi Matt,

Your analogy isn’t that bad – I’d choose the term “satellite” rather than “moon”, but you cut straight to the answer. Good job! But I’d still like you to explain that if there is gravity, why does there appear not to be gravity? The astronauts “float” in space, don’t they?

• Matthew Tucker Says:

“why does there appear not to be gravity? The astronauts “float” in space, don’t they?”

There appears to be zero gravity because the astronauts, like a space station are moving so fast horizontally that the gravitational force that earth exerts on the astronauts isn’t great enough to pull the them down to earth. Earth is definitely puling on the astronauts, which is keeping them in orbit, but the speed at which the astronaut is travelling is just to strong for the force of gravity. and thus the astronaut experiences “weightless”

• CyberChalky Says:

Hi Matt,

You’re a bit off the track there. “sideways” velocity will have no effect on the amount of downwards force. Perhaps a hint will help – if you know the force is there – and thus as much on the astronauts as the vehicle – why would there appear to be no force on them *relative* to the vehicle?

9. Rowan Champion Says:

My view on the topic is that the astronauts aren’t truly in zero gravity as they are still close enough to the earth for it to have some amount of for acting on them but, similar to what others said the force is so small due to the distance that they are away from the planet that it doesn’t effect them in a noticeable way. not too sure whether am on the right track here on not.

10. trentp95 Says:

What I understood from the topic is that gravity changes due to different variables such as position on the earth and distance from the earth. This is why the astronaut appears to be in zero gravity, as he is a considerable distance from the earth gravity is very miniscule and the affects of the force acting upon him do not appear present so we would think that he is in zero gravity.