# A small object on the opposite side of the earth's orbit with 0 velocity,“in line” with the core - would it collide with the earth?

If a small object (say mass 1 kg), at the opposite side of the earth's orbit (six months away), was placed in the same orbital position as the earth's core, at zero velocity in all directions, would the object collide with the earth or would it have been drawn by the sun's gravity away from a collision with earth?

A naive answer would simply calculate how far "inward" the object would travel in six months, when accelerated by the sun's gravity, and if the distance covered is greater than the earth's radius, a collision would be averted.

But I started thinking about the earth's own gravitational influence as it orbited closer to the object.

How would we get an answer to this question?

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Well, what forces are acting on your object (e.g. solar gravity) and what is your frame of reference for "zero velocity" ? If referencing the sun, then it'll fall straight towards the sun with possible digression if perchance Venus or Mercury comes reasonably close :-) –  Carl Witthoft Mar 19 '14 at 16:02
the object would reach the Sun in about 65 days: en.wikipedia.org/wiki/Free-fall_time –  DavePhD Mar 19 '14 at 16:11
Your "naive" answer isn't naive at all, as @DavePhD points out. If you worked out that scenario and it looked like a collision was still possible/probable, then you'd need to go to less "naive" considerations. But often the simplest explanation is correct :) –  Kyle Mar 19 '14 at 16:15
It looks like my intuition about the acceleration due to solar gravity was off by an order of magnitude: I would have guessed the free-fall time to the sun to be a few years. –  Jtrx Mar 20 '14 at 11:15

The acceleration due to the Sun's gravity at the orbit of the Earth is given by:

$$a = \frac{GM}{r^2}$$

where $M$ is the mass of the Sun ($1.9891 \times 10^{30} \mathrm{kg}$), $r$ is the Earth-Sun distance ($\approx 1.5 \times 10^{11} \mathrm{m}$) and $G$ is Newton's constant ($6.6738 \times 10^{-11} \mathrm{Nm^2kg^{-2}}$). This gives:

$$a \approx 0.0059 \mathrm{ms^{-2}}$$

The radius of the Earth is about $6.378 \times 10^6$ m. If we assume the acceleration is constant, which is a pretty good approximation over a distance that is small compared to the Earth-Sun distance, the distance moved in a time $t$ is:

$$s = \tfrac{1}{2} a t^2$$

so:

$$t = \sqrt{\frac{2s}{a}}$$

Putting in our values for $s$ and $a$ we get $t \approx 46,500$ seconds, which is about 13 hours. So it only takes 13 hours for the object to move inwards far enough not to collide with the Earth.

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Use Newton's law of gravity to calculate the force. That force will depend on the distance of the object and the sun. Then use $F = m a$ to calculate the acceleration on the object. Use $d = at^2$ to calculate the distance that is traveled within 6 months.

This is inaccurate since the force of gravity will change with the distance. The previous formulas assume that it is constant. However, within the thickness of earth, the force can be taken as constant for this. So if the distance is greater than the thickness of earth, the actual traveled distance will only be bigger. But that does not change the conclusion to your question.

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A bit sloppily worded, but yeah: a first-order calculation of the gravitational force exerted by the sun does show that the object is well away from Earth's orbit in 6 months. –  Carl Witthoft Mar 19 '14 at 17:06