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Johannes
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While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during athe jump you reachspend your time on average roughly one meter above earth's surface, your velocity lags 460 m/s times 1/6.4x10^6 (the denominator corresponding to earth's radius in meters) which equates to about 70 micrometer per second. So, when jumping exactly vertically, after a second you land roughly 70 micrometer west of where you started.

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during a jump you reach

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during the jump you spend your time on average roughly one meter above earth's surface, your velocity lags 460 m/s times 1/6.4x10^6 (the denominator corresponding to earth's radius in meters) which equates to about 70 micrometer per second. So, when jumping exactly vertically, after a second you land roughly 70 micrometer west of where you started.

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Johannes
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While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during a jump you reach

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during a jump you reach

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Johannes
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While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both undergoingcontinuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the durationheight of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis, the. The tidal forceeffects due to the Milky Way, etc can safely be ignored. But in

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both undergoing a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as part of the effect would be in changing the duration of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis, the tidal force due to the Milky Way, etc. But in any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

In any case, the effect is of sub-micrometer size (and not 3 mm as stated in an earlier answer).

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