# Can a person on the equator jump up higher into the air at sunset than at sunrise?

I am wondering if a person on the equator can jump up slightly higher into the air at sunset in comparison to jumping up into the air at sunrise.

The reason for this would be that at sunset the Earth is moving away from a person during the time period that he/she is up in the air while at sunrise the Earth is moving towards the person during the time period that he/she is up in the air.

Can a person on the equator jump up higher into the air at sunset than at sunrise?

EDIT

I found the following drawing that was used in a question titled 'Why do we say Earth rotates from west to east?' which is posted on the Earth Science Stack Exchange website. This drawing illustrates the position of a person at the time of sunset and sunrise as the Earth travels in an orbit around the Sun.

• "at sunset the Earth is moving away from a person during the time period that he/she is up in the air while at sunrise the Earth is moving towards the person during the time period that he/she is up in the air." – That's not true. What's your reasoning behind that statement? Jul 4 at 14:12
• The Earth is moving in the same direction relative to the person at sunrise and sunset. The difference is its direction relative to the Sun, not the person. Jul 4 at 14:16
• I can see where you're coming from. I'm not sure if those directions are completely correct, though. At midnight, the Earth's surface is moving about the Earth's axis in the same direction as the Earth's motion about the sun, and at noon, the Earth's surface is moving about the Earth's axis in the opposite direction as the Earth's motion about the sun. At sunrise, the Earth's surface is moving towards the sun (perpendicular to Earth's direction of motion), and at sunset, the Earth's surface is moving away from the sun (again, perpendicular). (I'm making a few simplifying assumptions.) Jul 4 at 23:15
• @TannerSwett I think the situation the OP is trying to set up is one where, when you jump, you are pushing off in the direction tangential to the orbit (either in the direction of Earth's orbital motion, or in the opposite direction). I don't think the OP is taking rotation into consideration. E.g. you are standing either on the leading or the retreating edge of the Earth, and the Sun is positioned somewhere to the side (because you're standing on the terminator line). Jul 4 at 23:23
• I guess, technically, the highest you can jump (ignoring any atmospheric variables) is on the equator at noon during a total solar eclipse...
– Will
Jul 4 at 23:45

Can a person on the equator jump up higher into the air at sunset than at sunrise?

A person can jump higher at the equator, but whether it is sunset or sunrise will have no influence. But because of centrifugal force, you actually weigh a very small amount less at the equator, meaning that if you jumped with the same amount of force at the equator as you did at say the north pole, you will travel a tiny amount higher.

Also, due to tidal forces, the height you can jump when the moon is directly overhead (and directly under your feet) will be slightly higher than when it's not, and we can make a similar argument for the sun.

The reason for this would be that at sunset the Earth is moving away from a person during the time period that he/she is up in the air while at sunrise the Earth is moving towards the person during the time period that he/she is up in the air.

No, since you and the ground make up a stationary frame of reference. Your velocity with respect to the ground (when you are standing) is zero.

• Because of tidal forces, you can also jump a microscopic amount higher when the moon is directly overhead or directly under your feet, averaging for the position of the sun. Likewise for the sun, averaging for the position of the moon. Maximum jump height is thus attainable during a new moon at noon or midnight at the equator.
– g s
Jul 4 at 3:15
• You also weigh less at the equator because you're further from the earth's center of gravity (this is indirectly the result of centrifugal forces as well). I'm not sure which of these effects is greater Jul 4 at 13:26
• @DreamConspiracy it's complicated, but the lowest surface gravity is around Sri Lanka at 9.7773 m/s² Jul 4 at 21:59
• @gs wouldn't the maximum be during a solar eclipse at noon? Jul 5 at 10:56
• Lets not forget the radiation pressure exerted by sunlight!
– Aron
Jul 6 at 3:00

Imagine you are sitting in a train. And you try to jump from one end of the train towards the other end, and mark the spot on the train floor where you land. Now, do you think this spot will change based on whether the train is stationary or moving with uniform speed in one direction or in another direction?

The answer is no. It won't change, because irrespective of what uniform speed the train is moving at, in the frame of the train, you are stationary when you stand at one end of the train.

Similarly, in case of earth frame, in this context, you are stationary with respect to it, whether at sunset or sunrise.

• Just a clarification: what you say @silverrahul only applies if the train has a constant velocity, both in direction and intensity. If the train is accelerating or decelerating during the jump, the jumper will notice an effect.
– GRB
Jul 4 at 8:52

The difference, if any, would be difficult to measure.

All else being equal, the temperature might be lower after a night than a day. The air would be slightly more dense. You would weigh slightly less, and could jump slightly higher.

• And if you've been working hard all day, you might be slightly more tired by sunset. Jul 5 at 5:43

People have interpreted your question in various ways, but I think the core of what you're asking is what happens if you jump straight ahead from a leading edge of a (uniformly) moving object, as opposed to jumping "backwards" at the other end.

So, within the context of this answer, a platform in uniform motion is assumed, and no other effects are considered.

Even before you jump, you already have Earth's velocity, by virtue of being bound to it (you are held by gravity, and you are moving together with the Earth).

So when you jump, you're not "suspended" in space, stationary with respect to the Sun. The Earth doesn't "catch up to you". When you jump off of the "leading edge", you just add a bit of velocity on top of what you already have, and on the other side, you just subtract a tiny portion of that velocity. Assuming (approximately) uniform motion of the Earth for the duration of the experiment, the motion is such that this exactly cancels out, and the result is the same as if the Earth was stationary to begin with. That's just the classical (Galilean) relativity of motion.

Earth's orbital speed is about 107000 km/h (67000 mph). If the Earth suddenly stopped with respect to the Sun, on the leading edge, it would look as if you jumped incredibly high on one side, and on the other, you'd get squashed.

This is analogous to a scenario where you're standing on a train, with the train moving in a straight line with constant velocity. You've already accelerated together with the train, so, to a stationary observer, you already have the same velocity as the train.

If you jump up, you land in the same place, but to an observer standing next to the tracks, you travel the same horizontal distance as the train. If you jump forward, for you it looks like you've traveled some small distance forward, but to a stationary observer, you traveled just slightly faster then the train itself. If you jump backwards, it's similar. To you, it's a small distance. To a stationary observer, your speed was slightly lower than that of the train.

It's only when the train starts to accelerate (or decelerate, or turn) that you notice any difference, compared to the train standing still.

P.S. The reasoning behind my interpretation of the question:

I looks to me like the situation the OP is trying to set up is one where the person that is jumping is pushing off in the direction tangential to the orbit (either in the direction of Earth's orbital motion, or in the opposite direction). It seems like the OP is ignoring velocity due to rotation. E.g. The person is standing either on the leading or the retreating edge of the Earth's sphere, with the Sun positioned somewhere to their side (they are standing on the shadow terminator line).

• @ Filip Milovanović, yes, the person would be pushing off from the ground and moving in a direction that is tangential to the orbit. Jul 5 at 0:26

No, all other things being equal, you would not be able to jump higher at either sunrise or sunset. The equatorial bulge from the centrifugal force of the Earth's rotation does make gravity slightly less at the equator than at the poles.

However, at the Tropic lines when directly under the Sun at mid day or directly opposite the sun in the middle of the night, you would have a very slight (nearly immeasurable) advantage from the solar tides. While the lunar tidal force is greater, the Sun does produce tidal effects.

• Re "The equatorial bulge from the centrifugal force of the Earth's rotation does make gravity slightly less at the equator than at the poles.". No it doesn't, the surface gravity strength is related to the local mass, not the equator, see NASA's Earth gravity visualisation. There is no correlation at all between gravity strength and latitude. The strongest gravity is where the biggest masses are, such as mountain ranges. The weakest is where the least mass is, e.g. deep ocean trenches.
– RobG
Jul 5 at 13:54
• @RobG In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator. From; en.wikipedia.org/wiki/Gravity_of_Earth#Latitude Jul 5 at 17:33
• @RobG In addition to the centrifugal force, the equatorial bulge increases the radius from the center of the Earth affecting the g force. Jul 5 at 17:37
• Centrifugal force ≠ gravity. Gravity is not uniform: on the equator, there are areas of high gravity in Indonesia due to its high mountains, and low gravity in the Indian Ocean. The increased centrifugal force at the equator opposes the gravitational force so gravity is perceived to be slightly lower that it is. Distance to the Earth centre is an approximation of gravitational force, see your wikipedia reference: en.wikipedia.org/wiki/… .
– RobG
Jul 6 at 1:10

No. The jump takes place with respect to the Earth's frame of reference. If we assume that the jump takes place in a sufficiently small time span (compared to the rotational and orbital period of Earth) and the liftoff velocity is constant, then there will be no difference which point on the equator it takes place, at least in the Newtonian context.

• "The jump takes place in the Earth's frame of reference." – Well, the jump takes place in every frame of reference, doesn't it? Maybe it would be more accurate to say that the jump is easiest to analyze in the Earth's frame of reference. Jul 4 at 14:14
• @TannerSwett No, that was not what I meant. I meant that the person performing the jump is jumping with respect to the Earth. The Earth is traveling at nearly 30 km/s around the Sun, so the same action could hardly be called a "jump" in the Sun's frame. Jul 5 at 12:37

The reason for this would be that at sunset the Earth is moving away from a person during the time period that he/she is up in the air while at sunrise the Earth is moving towards the person during the time period that he/she is up in the air.

This is not correct. In fact, at sunset the athlete when standing still is moving with respect to the sun in the direction of his feet at speed that the earth is orbitting the sun, and so is the earth. As dawn, the athlete is moving in the direction of his head, and so is the earth. I have ignored, for simplicity, the rotation of the earth on its axis. During the jump, the motion of the earth around the sun has no effect on the athlete that it doesn't have on the earth.

The athlete and the earth are both in free fall in the gravitational field of the sun. The only gravitational field that affects the height of the jump is that of the earth. If the sun suddenly lost its gravitational field, it would not affect the height of the jump, because both the athlete and the earth would fly off on a tangent, but the tangents would be the same.

Having said that, there are some other things to do with the sun that could cause a change in the height of the jump. 1. At sunset it is likely hotter, and so air resistance would be lower making for a higher jump but fluid upthrust would also be lower, making for a lower jump. Difficult to say which effect would be greater. 2. There could be some relativistic effect that I don't know about that would cause a higher or lower jump. 3. Tidal effects from the sun won't directly affect the height of the jump, because the jumps are each perpendicular to the sun's gravitational field, tending to make the athlete wider (from front to back if he is facing the sunset and dawn), not taller. 4. On other hand the tidal effect of the sun changes the shape of the solid earth measurably, which is known as land tides, and so perhaps at dawn the ground is accelerating downwards and at sunset it is accelerating upwards, but this seems unlikely because the tidal effect of the sun affects noon and midnight equally (almost equally) and therefore if the ground is sinking at dawn it should be expected to be also sinking at sunset.

In conclusion, there is no reason that I can think of to believe the height of the jump will be either more or less. But there must be some difference, either more or less, in principle, in the real world, due to a difference in temperature or some other effect. Certainly the reason given in the question is not valid. But I upvoted, and attempted to answer, the question because it is a fun question indeed.

• Nice answer, +1. A little formatting advice: if you put two spaces at the end of a line, it produces a line break without making a new paragraph. Jul 4 at 19:25
• if there were a tidal effect, the moon would dominate it.
– JEB
Jul 5 at 3:22
• It is better to avoid significant trailing space. Use a backslash (\) or an HTML break (<br/>) instead. (The former was introduced with CommonMark.) Jul 5 at 5:51
• @PeterMortensen What is trailing space? Jul 12 at 20:25
• @NoName What is the advantage of that? Jul 12 at 20:28

Due to the fact that the Earth is spinning and the shape of the Earth being roughly an oblate spheroid the effective value of $$g$$ is smaller at the equator than it is at the poles, so in theory perform a high jump on the equator (and on top of a mountain).

When a high jump is executed the athlete needs a component of vertical velocity and horizontal velocity.
Vertical speed is good for height and horizontal speed is good for getting over the bar.
The ground and the bar are rotating about the axis of the Earth at the same speed so if it assumed that there is no air resistance the rotation of the Earth has no effect on the performance of the high jumper.

Now what about the effect of the Sun?
While in the air the gravitational attraction of the Sun can influence the horizontal speed of the jumper but note that this will be a very, very small effect. While in the air, the jumper needs to be pulled towards the bar, i.e., increasing the horizontal speed relative to the bar, which may mean that the vertical speed could be correspondingly increased so the jumper stays off the ground for longer?
Think of the nonsense case where the jumper jumps vertically upwards and the Sun pulls the jumper over the bar!
So when should a jump be executed?
As long as the jump is towards the Sun whether it be sunrise or sunset the effect is the same.

One could also consider a jump at midday when the Sun will help with the vertical motion of the jumper.

A similar analysis can be done for the Moon (and Sun).

You are thinking in terms of this:

Recall Newton’s Universal Law of Gravitation states that any two masses have a mutual gravitational attraction $$Gm_1m_2/r^2$$. A point mass m=1 at P will therefore feel gravitational attraction towards both masses M, and a total gravitational field equal to the vector sum of these two forces, illustrated by the red arrow in the figure.

If $$m_1$$ and $$m_2$$ are the sun and the earth center of mass and the point P is the person jumping if compared at noon and midnight, there would be a very small (because of the distances) difference in the field when the vectors add, because an earth diameter would come between the two locations, considering everything else the same. All the extra effects given in the other answers would also hold, so this would not be measurable.

Sunset and sunrise are at the same distance from the sun, if you consider the earth a symmetric spherical body at the equator, so the gravitational fields should be the same and no difference due to fields should exist.

Most people are taller in the morning, and also better rested, these two effects will probably lead to higher jumps. I'm now going to ignore these effects.

The reason for this would be that at sunset the Earth is moving away from a person during the time period that he/she is up in the air while at sunrise the Earth is moving towards the person during the time period that he/she is up in the air.

In an inertial frame of reference absolute motion can be ignored. So no that has no effect.

If you want a time when gravity is weaker it's at high tide. If you ignore the moon the solar tidal peaks are noon and midnight, but the lunar tides are much stronger. so consult an almanac for the high tide, or approximate it and jump when the moon is overhead or underfoot.

The tidal effect is quite small, so the difference may not be measurable.

If you think by jumping you disconnect from earth and let it "slide" towards/away from you, consider that the linear speed of the earth due to its rotation is

1.5 Mach

You would feel this if it happened. Instead, you are going the same 1.5 Mach with it, and feel nothing.