Thread: Lengthening and shortening years.

A strange thought occurred to me today in school, I thought of the sun and orbits, and that if you were to:

1. Jump on the night side of the Earth, you were lengthening your orbit around the sun and so making the distance you had to travel slightly further to orbit the sun, and then landing and having experienced a longer year than another person who didnt jump.

2. Jump up on the day side of the Earth, you would be shortening your orbit around the sun and so making the distance to travel slightly shorter to orbit the sun and then landing and experiencing a shorter year than the person who did not jump.

Albeit this change is only a tiny fraction of a second, but it is still real, please tell me if my idea is a real fact.

Originally Posted by Devon Keogh

A strange thought occurred to me today in school, I thought of the sun and orbits, and that if you were to:

1. Jump on the night side of the Earth, you were lengthening your orbit around the sun and so making the distance you had to travel slightly further to orbit the sun, and then landing and having experienced a longer year than another person who didnt jump.

You're traveling a further distance but you're also moving faster.

2. Jump up on the day side of the Earth, you would be shortening your orbit around the sun and so making the distance to travel slightly shorter to orbit the sun and then landing and experiencing a shorter year than the person who did not jump.

You're traveling a shorter distance, but you're also moving slower.

Originally Posted by Devon Keogh

A strange thought occurred to me today in school, I thought of the sun and orbits, and that if you were to:

1. Jump on the night side of the Earth, you were lengthening your orbit around the sun and so making the distance you had to travel slightly further to orbit the sun, and then landing and having experienced a longer year than another person who didnt jump.

2. Jump up on the day side of the Earth, you would be shortening your orbit around the sun and so making the distance to travel slightly shorter to orbit the sun and then landing and experiencing a shorter year than the person who did not jump.

Albeit this change is only a tiny fraction of a second, but it is still real, please tell me if my idea is a real fact.

To make this work, you would have to turn off the Earth's gravity. In reality, you would in a few seconds come back down to roughly the same spot where you started. That spot would finish the year the same way it would have anyway, and your year would be normal length.

However, clever Gedanken [thought] experiments don't have to be "in reality." So let us imagine that at the top of your jump, you turn the Earth's gravity off. If you pick the direction of your jump just right, you will continue around the sun in the independent circular orbit I expect you were imagining. In that case, your conclusion is correct.

If you turn off gravity just before you jump, things are a bit more complicated. You are now in an elliptical orbit. If you were jumping on the night side, you will be moving a bit faster than you would in a circular orbit, and your position will drift outwards of the Earth's orbit. My instincts say that your results are still correct, but the part of the orbit closest to the sun is in fact a bit closer to the sun than you were when you jumped. My instinct says it isn't enough closer to change your conclusion, but it would be nice to work out the numbers.

I might add, please don't really do this experiment. When you turned off the Earth's gravity, the Earth's atmosphere would disperse into space, and I rather enjoy breathing.

Hmm, well I was thinking that all loose matter on Earth orbits itself around the sun and is pulled by the Earth, but it would shorten/lengthen your year as you would now be closer/further and so travelling very slightly faster/slower, and so you would hit the ground giving you enough time in the sky to shorten/lengthen your year.

Am I correct in this?

Originally Posted by Devon Keogh

Hmm, well I was thinking that all loose matter on Earth orbits itself around the sun and is pulled by the Earth, but it would shorten/lengthen your year as you would now be closer/further and so travelling very slightly faster/slower, and so you would hit the ground giving you enough time in the sky to shorten/lengthen your year.

Am I correct in this?

For all practical purposes, the time between the beginning and end of your jump is the same for you and the spot you land on, so the spot's year and yours are the same length. If you take into account the tiny, tiny effect of special relativity, your "year" is shorter in both cases since your clock moves more slowly as seen by the spot than the spot's clock does, and the spot is closer to being in a single inertial frame during the time of the jump than you are. Your jump becomes a rather complicated example of the twin paradox of special relativity.

The bit of relativity above is an example of why you should "never ask a physicist a simple question; he will make it complicated in an instant."

Originally Posted by mvb

Originally Posted by Devon Keogh

Hmm, well I was thinking that all loose matter on Earth orbits itself around the sun and is pulled by the Earth, but it would shorten/lengthen your year as you would now be closer/further and so travelling very slightly faster/slower, and so you would hit the ground giving you enough time in the sky to shorten/lengthen your year.

Am I correct in this?

For all practical purposes, the time between the beginning and end of your jump is the same for you and the spot you land on, so the spot's year and yours are the same length. If you take into account the tiny, tiny effect of special relativity, your "year" is shorter in both cases since your clock moves more slowly as seen by the spot than the spot's clock does, and the spot is closer to being in a single inertial frame during the time of the jump than you are. Your jump becomes a rather complicated example of the twin paradox of special relativity.

The bit of relativity above is an example of why you should "never ask a physicist a simple question; he will make it complicated in an instant."

But as you jump, you are further away from the sun at night and closer at day than the Earth, so you have more or less time to travel to reach the same point, making your year longer or shorter?

Originally Posted by Devon Keogh

But as you jump, you are further away from the sun at night and closer at day than the Earth, so you have more or less time to travel to reach the same point, making your year longer or shorter?

Nope. All you do is change what you did during the year. As long as you are attached to the earth, only the places you are at the beginning and at the end of the year count.

Well, I was not thinking of the human year, but the actual physical and precise year, where everything is down to the smallest and most exact timing.

Originally Posted by Devon Keogh

Well, I was not thinking of the human year, but the actual physical and precise year, where everything is down to the smallest and most exact timing.

I like the way you think, even when I disagree with you. Let me see if I can make my point a bit more precise.

Working on a precise definition of a body's year: Consider an object in orbit around the sun. Ignoring perturbations due to other objects, the orbit is in a single plane. The line in the plane from the sun to the object rotates around the sun. A year is the time that it takes this line to from a set angle in the plane all the way around and back to the same angle.

Using this definition, the only thing you need to know about the object's motion is what time it passes that fixed line. Anything it does in between will not affect the length of its year unless it changes the passage times. Thus your jumping around on the earth in the middle of the year does nothing significant to the Earth's year or to yours, unless you can change the time you get to the crucial angle. Your jumping gets completely lost in the effects of other, more massive, things moving around continually.

On the other hand, if you can arrange to be on the leading side of the Earth at the beginning of the year and the trailing side at the end of the year, you will definitely change the length of your year, as I have defined it above.

I hope I have gotten this a bit clearer.

Originally Posted by mvb

Originally Posted by Devon Keogh

Well, I was not thinking of the human year, but the actual physical and precise year, where everything is down to the smallest and most exact timing.

I like the way you think, even when I disagree with you. Let me see if I can make my point a bit more precise.

Working on a precise definition of a body's year: Consider an object in orbit around the sun. Ignoring perturbations due to other objects, the orbit is in a single plane. The line in the plane from the sun to the object rotates around the sun. A year is the time that it takes this line to from a set angle in the plane all the way around and back to the same angle.

Using this definition, the only thing you need to know about the object's motion is what time it passes that fixed line. Anything it does in between will not affect the length of its year unless it changes the passage times. Thus your jumping around on the earth in the middle of the year does nothing significant to the Earth's year or to yours, unless you can change the time you get to the crucial angle. Your jumping gets completely lost in the effects of other, more massive, things moving around continually.

On the other hand, if you can arrange to be on the leading side of the Earth at the beginning of the year and the trailing side at the end of the year, you will definitely change the length of your year, as I have defined it above.

I hope I have gotten this a bit clearer.

I can make my thinking a bit clearer as well,
I will make and post some diagrams.

Originally Posted by mvb

Originally Posted by Devon Keogh

A strange thought occurred to me today in school, I thought of the sun and orbits, and that if you were to:

1. Jump on the night side of the Earth, you were lengthening your orbit around the sun and so making the distance you had to travel slightly further to orbit the sun, and then landing and having experienced a longer year than another person who didnt jump.

2. Jump up on the day side of the Earth, you would be shortening your orbit around the sun and so making the distance to travel slightly shorter to orbit the sun and then landing and experiencing a shorter year than the person who did not jump.

Albeit this change is only a tiny fraction of a second, but it is still real, please tell me if my idea is a real fact.

To make this work, you would have to turn off the Earth's gravity. In reality, you would in a few seconds come back down to roughly the same spot where you started. That spot would finish the year the same way it would have anyway, and your year would be normal length.

However, clever Gedanken [thought] experiments don't have to be "in reality." So let us imagine that at the top of your jump, you turn the Earth's gravity off. If you pick the direction of your jump just right, you will continue around the sun in the independent circular orbit I expect you were imagining. In that case, your conclusion is correct.

If you turn off gravity just before you jump, things are a bit more complicated. You are now in an elliptical orbit. If you were jumping on the night side, you will be moving a bit faster than you would in a circular orbit, and your position will drift outwards of the Earth's orbit. My instincts say that your results are still correct, but the part of the orbit closest to the sun is in fact a bit closer to the sun than you were when you jumped. My instinct says it isn't enough closer to change your conclusion, but it would be nice to work out the numbers.

It's pretty straight forward to do so. The orbital energy of an object is constant and can be found by the sum of its kinetic energy and gravitational potential energy (relative to the body it is orbiting), or



Where m is the mass of the object, v its orbital speed, M the mass of the body it is orbiting and r the radius of its orbit.

For elliptical orbits r and v are always changing, but such that sum remains constant.

This energy can also be found by



Where a is the semi-major axis, or average orbital distance, of the orbit.

The period of the orbit can be found by



Do a little vector addition of your jump speed and the Earth's orbital speed, you can get your new orbital speed, with that and your distance from the Sun you can find your orbital energy, which in turn, allows you to solve for 'a', which you can use to find your orbital period.

Start with two people at the same distance from the Sun. One jumps away from the Sun. You add his outward speed to his starting orbital speed( at right angles to each other) and then use this new orbital speed to solve for the new period. Since his speed will have increased, so will his orbital energy (quick note: for an object in a closed orbit, its orbital energy will be negative, and an increase in energy means a change in the positive direction). This in turn leads to a larger value for 'a' and a longer period.

You do the same for the other person who jumps inward. Assuming that they jump at the same speed, their new orbital speeds will be the same (though in different directions). This will result in their having the same orbital periods. Thus jumping inward or outward result in a longer orbital period.

However, Persons jumping from the day and night side of the Earth are not the same distance from the Sun, the dayside person is some 12756 km closer. ( assuming the Sun is directly overhead)

Thus for him "r" is 6378 km smaller than it is for the Earth itself, and it is 6378 km larger for the nightside person. A smaller "r" leads to a more negative orbital energy and a shorter orbital period. Conversely, a larger "r" results in a longer period. This means that even without jumping (and assuming we ignore the velocity contributions made by the Earth's rotation), If you suddenly turned off gravity, a person on the dayside would enter a faster orbit and a person on the nightside a slower one.

This brings up an interesting point. We already noted that someone jumping towards the Sun would enter a faster orbit, so if you jumped at just the right speed, then the two effects would cancel out and you would end up with an orbital period equal to that of the Earth. ( my calculations show this to be ~ 276 m/s, which is, of course, just a tad bit more than what a person could jump at). Jumping at anything greater would cause you to enter a faster orbit even when starting from the dayside.

If you account for Earth's rotation, then you will have add this to the orbital speed for the nightside and subject it for the dayside. (~0.464 km/sec)

Visit this website for the diagram and tell me if the diagram is correct...

http://reshade.com/resize-image/orbital-length-diagram-1-51787426

How do I resize images?

Originally Posted by Devon Keogh

How do I resize images?

In general?
I use Microsoft Office '03, still and it has a photo editor that makes it very easy.
I also use Irfanview- a very versatile photo and video editor and display tool.
What you use to display images will depend on how- maybe you can download Irfanview and can walk you through it.

Originally Posted by Neverfly

Originally Posted by Devon Keogh

How do I resize images?

In general?
I use Microsoft Office '03, still and it has a photo editor that makes it very easy.
I also use Irfanview- a very versatile photo and video editor and display tool.
What you use to display images will depend on how- maybe you can download Irfanview and can walk you through it.

I made the diagram with Microsoft Paint

Originally Posted by mvb

Originally Posted by Devon Keogh

Well, I was not thinking of the human year, but the actual physical and precise year, where everything is down to the smallest and most exact timing.

I like the way you think, even when I disagree with you. Let me see if I can make my point a bit more precise.

Working on a precise definition of a body's year: Consider an object in orbit around the sun. Ignoring perturbations due to other objects, the orbit is in a single plane. The line in the plane from the sun to the object rotates around the sun. A year is the time that it takes this line to from a set angle in the plane all the way around and back to the same angle.

Using this definition, the only thing you need to know about the object's motion is what time it passes that fixed line. Anything it does in between will not affect the length of its year unless it changes the passage times. Thus your jumping around on the earth in the middle of the year does nothing significant to the Earth's year or to yours, unless you can change the time you get to the crucial angle. Your jumping gets completely lost in the effects of other, more massive, things moving around continually.

On the other hand, if you can arrange to be on the leading side of the Earth at the beginning of the year and the trailing side at the end of the year, you will definitely change the length of your year, as I have defined it above.

I hope I have gotten this a bit clearer.

But the seasons are just tilts in the Earths axis, so how does this work? I dont see how tilts in the orbital body can lengthen or shorten the year when you jump...

Originally Posted by Devon Keogh

But the seasons are just tilts in the Earths axis, so how does this work? I dont see how tilts in the orbital body can lengthen or shorten the year when you jump...

Oops, sorry. I didn't mean to jump at those times. What I meant was that if you are on side of the Earth which is leading the way in the orbit at the moment that the year changes you get to the proper angle sooner than the center of the Earth does. Thus your year changes before the Earth's does. Similarly you can make your year change later by being on the opposite side of the Earth. I was just trying to emphasize that it is where you are at year's end that matters and not what you have done in the meantime.

At this point I believe that you were originally thinking of getting into a different orbit, which would be like turning off Earth's gravity when you were at the top of the jump. If you fixed your jump so that your speed along the orbit was exactly that of the an orbit parallel to the Earth's [similar-shaped ellipse], then your original conclusions were correct. Jump up at midnight, longer year. Jump up at noon, shorter year. But you would have to wait a full year before returning for the process to work.

Edit: correct mistake in quoting past post.

Originally Posted by Devon Keogh

Hmm, well I was thinking that all loose matter on Earth orbits itself around the sun and is pulled by the Earth, but it would shorten/lengthen your year as you would now be closer/further and so travelling very slightly faster/slower, and so you would hit the ground giving you enough time in the sky to shorten/lengthen your year.

Am I correct in this?

It is more correct to say that the loose matter orbits the Earth and is slightly perturbed by the Sun's gravity.

Originally Posted by mvb

Originally Posted by Devon Keogh

But the seasons are just tilts in the Earths axis, so how does this work? I dont see how tilts in the orbital body can lengthen or shorten the year when you jump...

Oops, sorry. I didn't mean to jump at those times. What I meant was that if you are on side of the Earth which is leading the way in the orbit at the moment that the year changes you get to the proper angle sooner than the center of the Earth does. Thus your year changes before the Earth's does. Similarly you can make your year change later by being on the opposite side of the Earth. I was just trying to emphasize that it is where you are at year's end that matters and not what you have done in the meantime.

At this point I believe that you were originally thinking of getting into a different orbit, which would be like turning off Earth's gravity when you were at the top of the jump. If you fixed your jump so that your speed along the orbit was exactly that of the an orbit parallel to the Earth's [similar-shaped ellipse], then your original conclusions were correct. Jump up at midnight, longer year. Jump up at noon, shorter year. But you would have to wait a full year before returning for the process to work.

Edit: correct mistake in quoting past post.

In my opinion your above conclusion is correct, except for having to wait for a year, please explain?
And anyway, could you not jump at anytime being that it was not exactly dawn nor dusk and get infinitesimally small results but results all the same?

I was just trying to emphasize that if you return to Earth your "year" was longer only for the brief period you were in the other orbit.

Originally Posted by mvb

I was just trying to emphasize that if you return to Earth your "year" was longer only for the brief period you were in the other orbit.

But would you not slow down in your second orbit, causing you to return to Earth at a different point on the earth, albeit a very small difference (not counting outside influences)?

Originally Posted by Janus

Originally Posted by Devon Keogh

Hmm, well I was thinking that all loose matter on Earth orbits itself around the sun and is pulled by the Earth, but it would shorten/lengthen your year as you would now be closer/further and so travelling very slightly faster/slower, and so you would hit the ground giving you enough time in the sky to shorten/lengthen your year.

Am I correct in this?

It is more correct to say that the loose matter orbits the Earth and is slightly perturbed by the Sun's gravity.

Well, being that the sun is the "parent" orbit of which everything else in the solar system revolves around while possibly also orbiting another object that is itself orbiting the sun.

Here is a diagram of my thoughts on orbits

https://www.dropbox.com/s/hw2lqvzkzz...iagram%201.png

Originally Posted by Devon Keogh

Originally Posted by Janus

Originally Posted by Devon Keogh

Hmm, well I was thinking that all loose matter on Earth orbits itself around the sun and is pulled by the Earth, but it would shorten/lengthen your year as you would now be closer/further and so travelling very slightly faster/slower, and so you would hit the ground giving you enough time in the sky to shorten/lengthen your year.

Am I correct in this?

It is more correct to say that the loose matter orbits the Earth and is slightly perturbed by the Sun's gravity.

Well, being that the sun is the "parent" orbit of which everything else in the solar system revolves around while possibly also orbiting another object that is itself orbiting the sun.

Here is a diagram of my thoughts on orbits

https://www.dropbox.com/s/hw2lqvzkzz... Diagram 1.png

This motion is one of many cases where the overall picture is a bit misleading, and you learn a lot by looking at the details. From a great distance, it would be hard to distinguish the motion of your object on the earth from the motion of the earth. When you look closely, there are 365 small loops in the motion of the object in each year, one for each day that the earth carries it around in a circle. If you are trying to find the object from 1,000 million miles away, you can ignore these loops at the beginning of the search.

However, when you get close, the speed of the object going around those little loops is huge compared to the speed of the center of those loops going around the sun. So when you are close, it makes more sense to ignore the movement of the earth around the sun.

Now what does that mean for basic physics? Newton's law of acceleration is the basic law of classical motion. In this situation, the law reads that "The force on an object is equal to its mass times the change of its velocity per unit time," or

Force = mass X change in direction of the speed / length of time taken to change

and you have to use short time intervals so that the velocity changes only by a small amount in order to prevent the force from depending on the time interval you use to measure it. With some work you find that the force keeping the object on the earth is much larger than the force keeping its large scale motion to a circle around the sun. This is why Janus's comment is correct, and the largest force is the one keeping the object in motion around the earth's rotational axis.

[Incidentally, in a diagram of this motion that includes the more-important but small-scale details, you need to show the earth with its poles nearly perpendicular to its orbit if you want a simple picture. In the overall picture, your point was better displayed as if the poles of the earth were in the plane of its orbits, as it appears to be in your diagram.]

All this is not an easy concept, and I hope I have made it a little easier to understand.

Ok, imagine you are floating in space on a protoplanet with 1.5/6 gravity of earth, and you jump.

Depending on the day or night side your year will change ever so slightly, am I correct?
So on Earth, with the gravity 4x this, would you not get the same effects but ÷4?

This is my train of thought on the subject.

(Being these are the same distance from the sun (Just an example))

I don't like the setup here. You are on a planet, just like you were on the Earth before. The force on you exerted by the planet, its gravitational attraction, is still by far the strongest force in the problem, and your path will be almost completely determined by the positions of you and the planet. So just as before, you are going to come back down to the planet, very close to where you were standing before you jumped, and you will then continue around the sun in contact with the planet taking the same amount of time as the planet does. Not even multiple jumps will make any difference you can measure.

Now if you are interested in taking a space ship, placing it in orbit on the opposite of the sun with you at its center and at the height you would have jumped to, we can figure out what your orbital period will be. The change will be pretty tiny, actually. But I don't like the jumping model because it is misleading. To get anything but zero you have to ignore the most important object in the whole picture. Being close enough to an object beats the size of the object pretty quickly in gravitational calculations.

I would also like to get in the habit of using "orbital period" instead of "year." Technically, "year" refers only to the orbital period of the earth. When we begin comparing your orbital periods in different circumstances, there is a chance of getting meanings messed up if you use "year" in more than one role.

So, are you OK with the new way of putting the problem, or should we discuss the situation a little. I want to make sure we address the issue you are thinking about with jumping, but I am not sure I have got it right. We seen to be talking past each other a bit here.

Originally Posted by mvb

I don't like the setup here. You are on a planet, just like you were on the Earth before. The force on you exerted by the planet, its gravitational attraction, is still by far the strongest force in the problem, and your path will be almost completely determined by the positions of you and the planet. So just as before, you are going to come back down to the planet, very close to where you were standing before you jumped, and you will then continue around the sun in contact with the planet taking the same amount of time as the planet does. Not even multiple jumps will make any difference you can measure.

Now if you are interested in taking a space ship, placing it in orbit on the opposite of the sun with you at its center and at the height you would have jumped to, we can figure out what your orbital period will be. The change will be pretty tiny, actually. But I don't like the jumping model because it is misleading. To get anything but zero you have to ignore the most important object in the whole picture. Being close enough to an object beats the size of the object pretty quickly in gravitational calculations.

I would also like to get in the habit of using "orbital period" instead of "year." Technically, "year" refers only to the orbital period of the earth. When we begin comparing your orbital periods in different circumstances, there is a chance of getting meanings messed up if you use "year" in more than one role.

So, are you OK with the new way of putting the problem, or should we discuss the situation a little. I want to make sure we address the issue you are thinking about with jumping, but I am not sure I have got it right. We seen to be talking past each other a bit here.

As you wish, we will use orbital period.

But using logic does it not seem that jumping would extend your orbit for a tiny amount of time and slow you down in relation to the sun then you will land on a slightly different point.

It is like jumping up on a line around the sun (orbiting the sun), when you land, you will be on a different point on that orbital line.

Example: https://www.dropbox.com/s/eoo4uxkzve... Diagram 1.png

Originally Posted by Devon Keogh

But using logic does it not seem that jumping would extend your orbit for a tiny amount of time and slow you down in relation to the sun then you will land on a slightly different point.

It is like jumping up on a line around the sun (orbiting the sun), when you land, you will be on a different point on that orbital line.

Technically yes, if you could jump exactly straight up, took account of the rotation of the earth, corrected for the wind, and didn't move for a [real] year after landing. But you can't measure the other aspects of the jump well enough to detect the orbital effect.


Now for the effect of changing a real orbit by the amount you could jump on the planet. The height you can reach is given by

h = constants / (acceleration of gravity) = C/g

On a planet with 1/4 the gravity of the earth , you would jump to new h = 1 / (1/4) = 4h. So, as you said, you would jump four times as high. The orbital period of an orbit through the position you get to is a bit harder to do. The orbital period is

T = constants X square root (orbital radius cubed)

= C sqrt(R³) = C R^3/2

and all the constants depend on properties of the sun, not the planet.

Now, how much can you change the radius of the orbit by jumping? The radius of the new orbit is the sum of your initial distance from the center of the planet plus the distance you jump in a direction away from the sun. The size of your jump is in fact small compared with every other distance in the problem, and under these conditions the effect of the jump is proportional to the first power of the jump. So your understanding is correct: The effect of moving to an orbit that you could touch by jumping, but at a place where the planet wasn't affecting your motion, is 4 times as much on the smaller planet as it is on earth.

I wouldn't be surprised if you had questions about this even though I agree with your result. If the questions involve the mathematics, please let me know what math you are comfortable with so that I can adjust the description to fit.

Originally Posted by mvb

Originally Posted by Devon Keogh

But using logic does it not seem that jumping would extend your orbit for a tiny amount of time and slow you down in relation to the sun then you will land on a slightly different point.

It is like jumping up on a line around the sun (orbiting the sun), when you land, you will be on a different point on that orbital line.

Technically yes, if you could jump exactly straight up, took account of the rotation of the earth, corrected for the wind, and didn't move for a [real] year after landing. But you can't measure the other aspects of the jump well enough to detect the orbital effect.


Now for the effect of changing a real orbit by the amount you could jump on the planet. The height you can reach is given by

h = constants / (acceleration of gravity) = C/g

On a planet with 1/4 the gravity of the earth , you would jump to new h = 1 / (1/4) = 4h. So, as you said, you would jump four times as high. The orbital period of an orbit through the position you get to is a bit harder to do. The orbital period is

T = constants X square root (orbital radius cubed)

= C sqrt(R³) = C R^3/2

and all the constants depend on properties of the sun, not the planet.

Now, how much can you change the radius of the orbit by jumping? The radius of the new orbit is the sum of your initial distance from the center of the planet plus the distance you jump in a direction away from the sun. The size of your jump is in fact small compared with every other distance in the problem, and under these conditions the effect of the jump is proportional to the first power of the jump. So your understanding is correct: The effect of moving to an orbit that you could touch by jumping, but at a place where the planet wasn't affecting your motion, is 4 times as much on the smaller planet as it is on earth.

I wouldn't be surprised if you had questions about this even though I agree with your result. If the questions involve the mathematics, please let me know what math you are comfortable with so that I can adjust the description to fit.

This seems right, thanks for your patience and helpfulness

There is some truth in what you say, although the difference in distance to the sun isn't so great. Earth's rotation - Wikipedia, the free encyclopedia
But yes, Atomic clocks run slower or faster depending on how high they are positioned, even if it is only about a meter.

Originally Posted by Devon Keogh

Ok, imagine you are floating in space on a protoplanet with 1.5/6 gravity of earth, and you jump.

Depending on the day or night side your year will change ever so slightly, am I correct?
So on Earth, with the gravity 4x this, would you not get the same effects but ÷4?

This is my train of thought on the subject.

(Being these are the same distance from the sun (Just an example))

Here the trouble is you're not taking into account the different speeds you will be moving at.

If you are on the night side of this planet, standing still, you are at this moment moving faster than you would be if you were on the day side of the planet standing still. Why? Because you're closer to the center of the circle defined by that planet's orbit.

Did you ever play on a "Merry-go-round" as a child? They've mostly been removed from parks these days due to safety issues, but some of us are old enough to remember. If you stand near the center of the Merry-go-round as it spins, the centripetal force on you is less. Why? Because if you're nearer the center of a spinning circle, you are moving more slowly than if you are at the outer edge.

Same principle here.

Also, there is one more thing: Different gravity.

When you jump off the dark side of the planet, you are being pulled downward both by the planet's gravitation, and the Sun's gravitation. It's a slight difference, but so is the difference in how far you jumped.

When you jump off the light side of the planet, you are being pulled downward by the planet's gravitation, but instead of being pulled downward also by the Sun's gravitation, it actually pulls you upward sightly.

Originally Posted by kojax

Originally Posted by Devon Keogh

Ok, imagine you are floating in space on a protoplanet with 1.5/6 gravity of earth, and you jump.

Depending on the day or night side your year will change ever so slightly, am I correct?
So on Earth, with the gravity 4x this, would you not get the same effects but ÷4?

This is my train of thought on the subject.

(Being these are the same distance from the sun (Just an example))

Here the trouble is you're not taking into account the different speeds you will be moving at.

If you are on the night side of this planet, standing still, you are at this moment moving faster than you would be if you were on the day side of the planet standing still. Why? Because you're closer to the center of the circle defined by that planet's orbit.

Did you ever play on a "Merry-go-round" as a child? They've mostly been removed from parks these days due to safety issues, but some of us are old enough to remember. If you stand near the center of the Merry-go-round as it spins, the centripetal force on you is less. Why? Because if you're nearer the center of a spinning circle, you are moving more slowly than if you are at the outer edge.

Same principle here.

Also, there is one more thing: Different gravity.

When you jump off the dark side of the planet, you are being pulled downward both by the planet's gravitation, and the Sun's gravitation. It's a slight difference, but so is the difference in how far you jumped.

When you jump off the light side of the planet, you are being pulled downward by the planet's gravitation, but instead of being pulled downward also by the Sun's gravitation, it actually pulls you upward sightly.

You are forgetting something, no matter how small, your year will still change?

Originally Posted by Devon Keogh

You are forgetting something, no matter how small, your year will still change?

But here's the thing, the vast majority of the difference between the jump point and landing point is due to the Earth's gravity and rotation rate and not the Sun's graivity. The trajectory you take is determined by the Earth. When you jump straight up, you are essentially entering an orbit around the Earth. It is an orbit that intersects with the Earth's surface at two points, but an orbit none the less. The shape of this orbit is determined by the speed of the ground under your feet (relative to the center of the Earth), the Earth's gravity and the vertical speed at which you leave the ground. It will take the shape of a very narrow ellipse with on focus at the center of the Earth.

What we are concerned with is the segment of that orbit that is above ground. In particular, the angle between take off and landing as measured from the center of the Earth and the time elapsed between these two points. You then compare the distance between the two points to how much the take off point moves due to the Earth's rotation in the same time. The difference between these two results gives you the net difference between jump off point and landing point with respect to the ground.

As to the actual math, first you add the vertical leap velocity to the rotation velocity. Assuming you can vertically leap at a speed of 3m/s (giving you a vertical jump of a little under 1/2 meter.)
and you are standing on the equator, giving a velocity of due to rotation of 464 m/s, you get an orbital velocity of

= 464.01 m/s

At takeoff.

With this and the vis-viva equation:



You can solve for 'a', the semi-major axis of the orbit. ('r' is distance from the focus of the orbit at take off, or in this example the radius of the Earth, and M its mass)

Once you have 'a', you can find the period of the orbit:




Then we can find the "Areal velocity" with the Earth's radius and ground velocity due to rotation.



Areal velocity is also related by:



Where e is the eccentricity of the orbit.

This and the information arrived at above allows us to solve for 'e'.

Now we use the equation



To find the "true anomaly" (The angle from perigee) at the point where this orbit intersects the surface of the Earth.

This subtracted from 180° gives you the angle from jumping point to apex (apogee) of the trajectory. Twice this gives the full angle traveled from jumping to landing. The distance traveled is

determined from this and the circumference of the Earth.

This does not take into account how much the Earth rotated while you in the air. For that, you have to determine the time you were in the air.

For this first you need to find the "eccentric anomaly" (E)

which is related to the true anomaly by:



Then we need the "mean anomaly" (M), which is found by




This can then be used to find the time taken to go from perigee to the surface intersection point:



This subtracted from 1/2 the period (P/2) gives half of your "in the air" time, and doubling this answer give your full "in the air" time. Multiplying this by the Earth's rotation speed

gives you how much the Earth has turned, which, when compared with the trajectory distance, give you the distance traveled with respect to the ground.

(much of the above was cut and pasted from this post:

Curious about effect of earth rotation on air travel times?

in another thread, and slightly reworded for this thread. )

The above analysis is still valid even if started with the assumption of a tidally locked Earth (one that keeps one side to the Sun perpetually.) In this case, we take the Earth's rotation as being one year long. This makes the speed at the surface due to rotation at a point directly opposite the Sun to be 1.27 m/s, which gives an orbital velocity of 3.26 m/s. From here, we follow the same steps to determine the difference between jump off and landing points.