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And, say the apple appeared out of nowhere stationary (relative to the Sun) at that distance, would the movement of the Earth leave the apple behind or would the apple follow?

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- Thread starter Yazz
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And, say the apple appeared out of nowhere stationary (relative to the Sun) at that distance, would the movement of the Earth leave the apple behind or would the apple follow?

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adjacent

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How long would it take for an apple to fall to the Earth if it were as far away as the Moon and how do you calculate that?

The force of gravity on an object is $$F=\frac{Gm_1m_2}{r^2}$$

The force will vary as it comes near the Earth, so you would need Calculus.

If you know the force, you know the acceleration,then you can use the SUVAT equations

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QuantumPion

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And, say the apple appeared out of nowhere stationary (relative to the Sun) at that distance, would the movement of the Earth leave the apple behind or would the apple follow?

An apple stationary with respect to the sun at the distance of the moon would have a speed relative to the earth equal to earth's orbital speed (30 km/s) which is in far excess of escape velocity. Therefore the apple would fall into the sun and not the earth.

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Matterwave

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An apple stationary with respect to the sun at the distance of the moon would have a speed relative to the earth equal to earth's orbital speed (30 km/s) which is in far excess of escape velocity. Therefore the apple would fall into the sun and not the earth.

Depends on the angle relative to the earth of where the apple is.

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Yes and no. While the apple might not fall "into" the Sun, it will fall so close to the Sun that it will still be burnt to a crisp.Depends on the angle relative to the earth of where the apple is.

Assuming that the apple's perigee is at least 100 km above the surface of the Earth (any closer and it will be burnt to a crisp by the Earth's atmosphere) means that after grazing the atmosphere and escaping the Earth's gravitational clutches, the apple will be in a heliocentric orbit with a perihelion distance of 0.8 solar radii above the surface of the Sun.

http://www.wolframalpha.com/input/?...ntial+of+the+sun))+-+1+au)/(solar+radius)+-+1

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